NCERT Exemplar for Class 12 Maths Chapter-4 (Book Solutions)

Solved Examples 

Short Answer (S. A)

Example 1: If $\begin{vmatrix} 2x & 5 \\ 8 & x \\ \end{vmatrix}$ = $ \begin{vmatrix} 6 & 5 \\ 8 & 3 \\ \end{vmatrix}$, then find $x.$

Ans: Here, we have $\begin{vmatrix} 2x & 5 \\ 8 & x \\ \end{vmatrix}$=$ \begin{vmatrix} 6 & 5 \\ 8 & 3 \\ \end{vmatrix}$

⇒ ${{2}}{{{x}}^{{2}}}{{ - 40 = 18 - 40}}$

⇒ ${{2}}{{{x}}^{{2}}}{{ = 18}}$

⇒ ${{{x}}^{{2}}}{{ = }}\dfrac{{{{18}}}}{{{2}}}$

⇒ ${{{x}}^{{2}}}{{ = 9}}$

⇒ ${{x = }}\sqrt {{9}} $

⇒ ${{x =  \pm 3}}$

Hence, value of $x$ is ${{ \pm 3}}{{.}}$

Example 2: If △ $ = \begin{vmatrix}  1  &  x  &  {{x^2}} \\   1  &  y  &  {{y^2}} \\   1  &  z  &  {{z^2}} \\ \end{vmatrix}$, △1 = $\begin{vmatrix}  1  &  1  &  1 \\  {yz} & {zx} & {xy} \\  x & y & z \\ \end{vmatrix}$, then prove that

△+△1=0.

Ans: here, we have △1 $= \begin{vmatrix}  {{1}} & {{1}} & {{1}} \\   {yz} & {zx} & {xy} \\   x & y & z  \\ \end{vmatrix}$

Now, interchanging rows and columns, we get

⇒ △1 $ = \begin{vmatrix} {{1}} & {yz} & x \\   {{1}} & {zx} & y \\   {{1}} & {xy} & z  \\ \end{vmatrix}$

⇒ △1 $ = \dfrac{1}{xyz}$ $ \begin{vmatrix}   x & {xyz} & {{x^2}} \\  y & {xyz} & {{y^2}} \\ z & {xyz} & {{z^2}}  \\ \end{vmatrix}$

Taking common $xyz$ from ${C_2},\;$ we get

⇒ △1 $=\dfrac{1}{xyz} \times\,{{xyz}}$ $\begin{vmatrix}  {{x}} & {{1}} & {{{{x}}^{{2}}}} \\ {{y}} & {{1}} & {{y}^{2}} \\ {{z}} & {{1}} & {{z}^{2}}  \\ \end{vmatrix}$

Now, interchanging ${C_1}$ and ${C_2}$, we get

⇒ △1 ${{ = }}\left( {{{ - 1}}} \right)$ $\begin{vmatrix} {{1}} & {{x}} & {{{{x}}^{{2}}}} \\  {{1}} & {{y}} & {{{{y}}^{{2}}}} \\ {{1}} & {{z}} & {{{{z}}^{{2}}}} \\ \end{vmatrix}$

⇒ △1 = -△ 

⇒ △ + △1 = 0 

Hence proved.

Example 3: Without expanding, Show that △ $ = \begin{vmatrix} {cose{c^2}\theta } & {co{t^2}\theta } & 1 \\ {co{t^2}\theta } & {cose{c^2}\theta } & { - 1} \\ {42} & {40} & 2 \\ \end{vmatrix} = 0$

Ans: here, we have △ ${{ = }}\begin{vmatrix} {{{cose}}{{{c}}^{{2}}}{{\theta }}} & {{{co}}{{{t}}^{{2}}}{{\theta }}} & {{1}} \\ {{{co}}{{{t}}^{{2}}}{{\theta }}} & {{{cose}}{{{c}}^{{2}}}{{\theta }}} & {{{ - 1}}} \\   {{{42}}} & {{{40}}} & {{2}} \\ \end{vmatrix}$

Applying ${C_1} \to {C_1} - {C_2}$

⇒ △ ${{ = }}\begin{vmatrix} {{{cose}}{{{c}}^{{2}}}{{\theta  - co}}{{{t}}^{{2}}}{{\theta }}} & {{{co}}{{{t}}^{{2}}}{{\theta }}} & {{1}} \\ {{{co}}{{{t}}^{{2}}}{{\theta  - cose}}{{{c}}^{{2}}}{{\theta }}} & {{{cose}}{{{c}}^{{2}}}{{\theta }}} & {{{ - 1}}} \\ {{2}} & {{{40}}} & {{2}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{{cose}}{{{c}}^{{2}}}{{\theta  - co}}{{{t}}^{{2}}}{{\theta }}} & {{{co}}{{{t}}^{{2}}}{{\theta }}} & {{1}} \\ {{{ - }}\left( {{{cose}}{{{c}}^{{2}}}{{\theta  - co}}{{{t}}^{{2}}}{{\theta }}} \right)} & {{{cose}}{{{c}}^{{2}}}{{\theta }}} & {{{ - 1}}} \\  {{2}} & {{{40}}} & {{2}}  \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{1}} & {{{co}}{{{t}}^{{2}}}{{\theta }}} & {{1}} \\   {{{ - 1}}} & {{{cose}}{{{c}}^{{2}}}{{\theta }}} & {{{ - 1}}} \\  {{2}} & {{{40}}} & {{2}} \\ \end{vmatrix}$

⇒ △ = 0    (when two columns are same, then value of determinants is 0)

Hence proved.

Example 4: Show that △ $=\begin{vmatrix} x & p & q \\ p & x & q \\ q & q & x \\ \end{vmatrix}$ = $\left( {x - p} \right)\left( {{x^2} + px - 2{q^2}} \right).$

Ans: here, we have △ $=\begin{vmatrix} x & p & q \\ p & x & q \\ q & q & x \\ \end{vmatrix}$

Applying ${C_1} \to {C_1} - {C_2}$

⇒ △ $=\begin{vmatrix} {{{x - p}}} & {{p}} & {{q}} \\  {{{ - }}\left( {{{x - p}}} \right)} & {{x}} & {{q}} \\ {{0}} & {{q}} & {{x}} \\ \end{vmatrix}$

Taking common $\left( {x - p} \right)$ from ${C_1}$

⇒ △ ${{ = }}\left( {{{x - p}}} \right)\begin{vmatrix} {{1}} & {{p}} & {{q}} \\ {{{ - 1}}} & {{x}} & {{q}} \\ {{0}} & {{q}} & {{x}} \\ \end{vmatrix}$

Now, expanding along ${C_1}$

⇒ △ ${{ = }}\left( {{{x - p}}} \right){{\;}}\left[ {{{1}}\left( {{{{x}}^{{2}}}{{ - }}{{{q}}^{{2}}}} \right){{ + 1}}\left( {{{px - }}{{{q}}^{{2}}}} \right){{ + 0}}} \right]$

⇒ △ $ = \left( {x - p} \right)\;\left( {{x^2} + px - 2{q^2}} \right)$

Hence proved.

Example 5: If △ $=\begin{vmatrix} 0 & {b - a} & {c - a} \\ {a - b} & 0 & {c - b} \\ {a - c} & {b - c} & 0 \\ \end{vmatrix}$, then show that △ is equal to zero.

Ans: here, we have △ ${{ = }}\begin{vmatrix} {{0}} & {{{b - a}}} & {{{c - a}}} \\  {{{a - b}}} & {{0}} & {{{c - b}}} \\  {{{a - c}}} & {{{b - c}}} & {{0}} \\ \end{vmatrix}$

Taking “-1” as common from ${{{R}}_1},{{\;}}{{{R}}_2}{{\;and\;}}{{{R}}_3}$, we get 

⇒ △ ${{ = }}{\left( {{{ - 1}}} \right)^{{3}}}\begin{vmatrix} {{0}} & {{{a - b}}} & {{{a - c}}} \\ {{{b - a}}} & {{0}} & {{{b - c}}} \\ {{{c - a}}} & {{{c - b}}} & {{0}} \\ \end{vmatrix}$

Now, interchanging rows and columns, we get

⇒ △ $= \left({-1}\right)\begin{vmatrix} {{0}} & {b - a} & {c - a} \\ {a - b} & {{0}} & {c - b} \\ {a - c} & {b - c} & {{0}} \\ \end{vmatrix}$

⇒ △ = - △ 

⇒ 2△ = 0 

⇒ △ = 0

Hence proved.

Example 6: Prove that $\left( {A^{ - 1}} \right)' = \left({A'}\right)^{-1},$ where A is an invertible matrix.

Ans: Here, matrix A is an invertible matrix, so it is non – singular i.e., $\left| {{A}} \right| \ne {{0}}$.

And we know that, $\left| {{A}} \right|{{ = }}\left| {{{A'}}} \right|$. So, $\left| {{{A'}}} \right| \ne {{0}}.$

Also, we have ${{A}}{{{A}}^{{{ - 1}}}}{{ = }}{{{A}}^{{{ - 1}}}}{{A = I}}$

Taking transpose on both sides, we get 

⇒ ${\left( {{{A}}{{{A}}^{{{ - 1}}}}} \right)^{{'}}}{{ = }}{\left( {{{{A}}^{{{ - 1}}}}{{A}}} \right)^{{'}}}{{ = I'}}$

⇒ ${\left( {{{{A}}^{{{ - 1}}}}} \right)^{{'}}}{{A' = A'}}{\left( {{{{A}}^{{{ - 1}}}}} \right)^{{'}}}{{ = I}}$

Thus, ${\left( {{{{A}}^{{{ - 1}}}}} \right)^{{'}}}$ is inverse of ${{A'}}$, 

Therefore, ${\left( {{{{A}}^{{{ - 1}}}}} \right)^{{'}}}{{ = }}{\left( {{{A'}}} \right)^{{{ - 1}}}}$

Hence proved.

Long Answer (L.A.)

Example 7: If $x = \; - 4$ is a root of △  =$\begin{vmatrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \\ \end{vmatrix} = 0,$ then find the other two roots.

Ans: here, we have $\begin{vmatrix} {{x}} & {{2}} & {{3}} \\ {{1}} & {{x}} & {{1}} \\ {{3}} & {{2}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

Applying ${R_1} \to {R_1} + {R_2} + {R_3}$

⇒ $\begin{vmatrix} {{{x + 4}}} & {{{x + 4}}} & {{{x + 4}}} \\ {{1}} & {{x}} & {{1}} \\ {{3}} & {{2}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

Taking $\left( {{{x + 4}}} \right)$ common from ${R_1}$

⇒ $\left( {{{x + 4}}} \right)\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{1}} & {{x}} & {{1}} \\ {{3}} & {{2}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

Applying ${C_1} \to {C_1} - {C_3}$ and ${C_2} \to {C_2} - {C_3}$

⇒ $\left( {{{x + 4}}} \right)\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{0}} & {{{x - 1}}} & {{1}} \\ {{{3 - x}}} & {{{2 - x}}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

⇒ $\left( {{{x + 4}}} \right)\left[ {{{0 - }}\left( {{{3 - x}}} \right)\left( {{{x - 1}}} \right)} \right]{{ = 0}}$

⇒ $\left( {{{x + 4}}} \right)\left( {{{3 - x}}} \right)\left( {{{x - 1}}} \right){{ = 0}}$

⇒ ${{x}}\,{{ = }}\,{{ - 4,\;3,\;1}}$

Hence, the other two roots are 1 and 3.

Example 8: In a triangle ABC, if $\begin{vmatrix} 1 & 1 & 1 \\ {1 + \sin A} & {1 + \sin B} & {1 + \sin C} \\ {\sin A + si{n^2}A} & {\sin B + si{n^2}B} & {\sin C + si{n^2}C} \\ \end{vmatrix} = 0,\;$

then prove that $\Delta \,ABC$ is an isosceles triangle.

Ans: Here, we have $\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{{1 + sinA}}} & {{{1 + sinB}}} & {{{1 + sinC}}} \\ {{{sinA + si}}{{{n}}^{{2}}}{{A}}} & {{{sinB + si}}{{{n}}^{{2}}}{{B}}} & {{{sinC + si}}{{{n}}^{{2}}}{{C}}} \\ \end{vmatrix}{{ = 0}}$

⇒ Applying ${C_1} \to {C_1} - {C_3}$ and ${C_2} \to {C_2} - {C_3}$

⇒ $\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{{sinA - sinB}}} & {{{sinB - sinC}}} & {{{1 + sinC}}} \\ {{{sinA - sinB + si}}{{{n}}^{{2}}}{{A - si}}{{{n}}^{{2}}}{{B}}} & {{{sinB - sinC + si}}{{{n}}^{{2}}}{{B - si}}{{{n}}^{{2}}}{{C}}} & {{{sinC + si}}{{{n}}^{{2}}}{{C}}}\\ \end{vmatrix}{{ = 0}}$

Taking $\left( {{{sinA - sinB}}} \right)$ and $\left( {{{sinB - sinC}}} \right)$ from ${{{C}}_{{1}}}$ and ${{{C}}_{{2}}}$ respectively

⇒ $\left( {{{sinA - sinB}}} \right)\left( {{{sinB - sinC}}} \right){{\;}}\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{1}} & {{1}} & {{{1 + sinC}}} \\ {{{1 + sinA + sinB}}} & {{{1 + sinB + sinC}}} & {{{sinC + si}}{{{n}}^{{2}}}{{C}}} \\ \end{vmatrix}{{ = 0}}$

Now, expanding along ${R_1}$, we get

⇒ $\left( {{{sinA - sinB}}} \right){{\;\;}}\left( {{{sinB - sinC}}} \right){{\;}}\left( {{{1 + sinB + sinC - 1 - sinA - sinB}}} \right){{ = 0}}$

⇒ $\left( {{{sinA - sinB}}} \right)\left( {{{sinB - sinC}}} \right)\left( {{{sinC - sinA}}} \right){{ = }}\,{{0}}$

⇒ $\left( {{{sinA - sinB}}} \right){{ = }}\,{{0\;or\;}}\,\left( {{{sinB - sinC}}} \right){{ = 0\;or\;}}\left( {{{sinC - sinA}}} \right){{ = 0}}$

⇒ $\sin A = \sin B\;{{or\;\;}}\sin B = \sin C\;{{or}}\;\sin C = \sin A$

⇒ $A = B\;{{or}}\;B = C\;{{or}}\;C = A$

Hence, triangle ABC is an isosceles triangle.

Hence proved.

Example 9: Show that if the determinant △ = $\begin{vmatrix} 3 & { - 2} & {sin\;3\theta } \\ { - 7} & 8 & {cos\;2\theta } \\ { - 11} & {14} & 2 \\ \end{vmatrix} = 0$, then  $sin\theta  = 0\;\;$ or $\dfrac{1}{2}$

Ans: Here, we have $\begin{vmatrix} {{3}} & {{{ - 2}}} & {{{sin\;3\theta }}} \\ {{{ - 7}}} & {{8}} & {{{cos\;2\theta }}} \\ {{{ - 11}}} & {{{14}}} & {{2}} \\ \end{vmatrix}{{ = 0}}$

Applying ${{{R}}_{{2}}} \to {{{R}}_{{2}}}{{ + 4}}{{{R}}_{{1}}}$ and ${R_3} \to {R_3} + 7{R_1}$, we get

⇒ $\begin{vmatrix} {{3}} & {{{ - 2}}} & {{{sin\;3\theta }}} \\ {{5}} & {{0}} & {{{cos\;2\theta  + 4sin3\theta }}} \\ {{{10}}} & {{0}} & {{{2 + 7sin3\theta }}} \\ \end{vmatrix}{{ = 0}}$

Now, expanding along ${C_2}\;$, we get

⇒ 2 (5(${{2 + 7sin3\theta ) - 10\;}}\left( {{{cos\;2\theta  + 4sin3\theta }}} \right){{ = 0}})$

⇒ $\left[\left(10+35sin3\theta\right)-\left(10\;cos\;2\theta+40sin3\theta \right)\right] = 0$

⇒ ${{10 - 5sin3\theta  - 10\;cos\;2\theta  = 0}}$

⇒ ${{2 - sin3\theta  - 2\;cos\;2\theta  = 0}}$

⇒ 2${{ - }}\left( {{{3sin\theta  - 4si}}{{{n}}^{{3}}}{{x}}} \right){{ - 2\;}}\left( {{{1 - 2si}}{{{n}}^{{2}}}{{\theta }}} \right){{ = 0}}$

⇒ 2${{ - 3sin\theta  + 4si}}{{{n}}^{{3}}}{{x - 2 + 4si}}{{{n}}^{{2}}}{{\theta  = 0}}$

⇒ ${{ - 3sin\theta  + 4si}}{{{n}}^{{3}}}{{x + 4si}}{{{n}}^{{2}}}{{\theta  = 0}}$

⇒ ${{sin\theta }}\left( {{{ - 3 + 4si}}{{{n}}^{{2}}}{{x + 4sin\theta }}} \right){{ = 0}}$

⇒${{\;sin\theta }}\left( {{{4si}}{{{n}}^{{2}}}{{x + 4sin\theta  - 3}}} \right){{ = 0}}$

⇒ ${{\;sin\theta }}\left( {{{4si}}{{{n}}^{{2}}}{{x + 6sin\theta  - 2sin\theta  - 3}}} \right){{ = 0}}$

⇒${{\;sin\theta }}\left[ {{{2sin\theta }}\left( {{{2sin\theta  + 3}}} \right){{ - 1}}\left( {{{2sin\theta  + 3}}} \right)} \right]{{ = 0}}$

⇒${{\;sin\theta \;}}\left( {{{2sin\theta  + 3}}} \right)\left( {{{2sin\theta  - 1}}} \right){{ = 0}}$

⇒${{\;sin\theta  = 0\;or\;\;}}\left( {{{2sin\theta  + 3}}} \right){{ = 0\;or\;}}\left( {{{2sin\theta  - 1}}} \right){{ = 0}}$

⇒${{\;sin}}\,{{\theta }}\,{{ = }}\,{{0\;or\;\;sin}}\,{{\theta }}\,{{ = }}\,{{ - }}\,\dfrac{{{3}}}{{{2}}}{{\;\;or\;sin}}\,{{\theta }}\,{{ = }}\,\dfrac{{{1}}}{{{2}}}{{\;}}$

As we know that ${{sin}}\,{{\theta }}\, \in \,\left[ {{{ - 1,\;1}}} \right]$

Thus, ${{sin}}\,{{\theta }}\,{{ = }}\,{{0\;or\;sin}}\,{{\theta }}\,{{ = }}\,\dfrac{{{1}}}{{{2}}}$

Hence proved.

Objective Type Questions 

Choose the correct answer from the given four options in each of Examples 10 and 11.

Example 10: Let △  = $\begin{vmatrix} {Ax} & {{x^2}} & 1 \\ {By} & {{y^2}} & 1 \\ {Cz} & {{z^2}} & 1  \\ \end{vmatrix}$ and △1 = $\begin{vmatrix} A & B & C \\ x & y & z \\ {zy} & {zx} & {xy} \\ \end{vmatrix}$, then 

(A) △1 = △                                                 

(B) △ $ \ne $ △1          

(C) △ - △1 = 0 

(D) None of these 

Ans: The correct answer is option (C).

Here, we have △ ${{ = }}\begin{vmatrix} {{{Ax}}} & {{{{x}}^{{2}}}} & {{1}} \\ {{{By}}} & {{{{y}}^{{2}}}} & {{1}} \\ {{{Cz}}} & {{{{z}}^{{2}}}} & {{1}} \\ \end{vmatrix}{{\;}}$ and △1  = $\begin{vmatrix} A & B & C \\ x & y & z \\ {zy} & {zx} & {xy} \\ \end{vmatrix}$       

⇒ △1  $=\begin{vmatrix} A & B & C \\  x & y & z \\ {zy} & {zx} & {xy} \\ \end{vmatrix}$             

Now, interchanging rows and columns, we get

⇒ △1 $= \begin{vmatrix} A & x & {zy} \\ B & y & {zx} \\ C & z & {xy} \\ \end{vmatrix}$

⇒ △1 $= \dfrac{1}{xyz}\begin{vmatrix} {{{Ax}}} & {{{{x}}^{{2}}}} & {{{xyz}}} \\ {{{By}}} & {{{{y}}^{{2}}}} & {{{xyz}}} \\ {{{Cz}}} & {{{{z}}^{{2}}}} & {{{xyz}}} \\ \end{vmatrix}$

Taking $xyz$ common from ${C_3},$ we get

⇒ △1 $= \dfrac{1}{xyz}\times{{xyz}}$ $\begin{vmatrix} {{{Ax}}} & {{x}^{2}} & {{1}} \\ {By} & {{y}^{2}} & {{1}} \\ {Cz} & {{z}^{2}} & {{1}} \\ \end{vmatrix}$

⇒ △1 = △ 

⇒ △ - △1 = 0 

Hence, option (C) is the correct answer.

Example 11: If $x\;,\;y\; \in R,$ then the determinant

△ = $\begin{vmatrix} {\cos x} & { - \sin x} & 1 \\ {\sin x} & {\cos x} & 1 \\ {\cos \left( {x + y} \right)} & { - \sin \left( {x + y} \right)} & 0 \\ \end{vmatrix}$ lies in the interval 

(A) $\left[ { - \sqrt 2 ,\;\;\;\sqrt 2 } \right]$                                            

(B) $\left[ { - 1,\;\;\;1} \right]$

(C) $\left[ { - \sqrt 2 ,\;\;\;1} \right]$                                                

(D) $\left[ { - 1,\;\; - \sqrt 2 } \right]$

Ans: The correct answer is option (A).

Here, we have △ ${{ = }}\begin{vmatrix} {{{cosx}}} & {{{ - sinx}}} & {{1}} \\ {{{sinx}}} & {{{cosx}}} & {{1}} \\ {{{cos}}\left( {{{x + y}}} \right)} & {{{ - sin}}\left( {{{x + y}}} \right)} & {{0}} \\ \end{vmatrix}$

Applying ${R_3} \to {R_3} - {R_1}\cos y + \;{R_2}\sin y$

⇒ △ ${{ = }}\begin{vmatrix} {{{cosx}}} & {{{ - sinx}}} & {{1}} \\ {{{sinx}}} & {{{cosx}}} & {{1}} \\ {{0}} & {{0}} & {{{siny - cosy}}} \\ \end{vmatrix}$

Now, expanding along ${R_3}$, we get

⇒ △ $ = (\sin y - \cos y)\left( {co{s^2}x + si{n^2}x} \right)\;$

⇒ △ $ = (\sin y - \cos y)$

⇒ △ ${{ = }}\sqrt {{2}} \left( {\dfrac{{{1}}}{{\sqrt {{2}} }}{{siny - }}\dfrac{{{1}}}{{\sqrt {{2}} }}{{cosy}}} \right)$

⇒ △ ${{ = }}\sqrt {{2}} \left( {{{cos}}\dfrac{{{\pi }}}{{{4}}}{{siny - sin}}\dfrac{{{\pi }}}{{{4}}}{{cosy}}} \right)$

⇒ △ ${{ = }}\sqrt {{2}} {{sin}}\left( {{{y - }}\dfrac{{{\pi }}}{{{4}}}} \right){{\;}}$

As we know that, ${{ - 1}} \leqslant {{sin}}\left( {{{y - }}\dfrac{{{\pi }}}{{{4}}}} \right){{\;}} \leqslant {{1}}$

⇒ ${{ - }}\sqrt {{2}}  \leqslant \sqrt {{2}} {{sin}}\left( {{{y - }}\dfrac{{{\pi }}}{{{4}}}} \right){{\;}} \leqslant \sqrt {{2}} $

⇒ \[{{ - }}\sqrt {{2}}  \leqslant {{\Delta }} \leqslant \sqrt {{2}} \]

Hence, option (A) is the correct answer. 

Fill in the blank in each of the examples 12 to 14.

Example 12: If A, B, C are the angles of a triangles, then △ = $\begin{vmatrix} {si{n^2}A} & {cot\;A} & 1 \\ {si{n^2}B} & {cot\;B} & 1 \\ {si{n^2}C} & {cot\;C} & 1 \\ \end{vmatrix} = $ ………..

Ans: Here, we have △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;A}}} & {{1}} \\ {{{si}}{{{n}}^{{2}}}{{B}}} & {{{cot\;B}}} & {{1}} \\ {{{si}}{{{n}}^{{2}}}{{C}}} & {{{cot\;C}}} & {{1}} \\ \end{vmatrix}$

Applying ${R_2} \to {R_2} - {R_1},$ ${R_3} \to {R_3} - {R_1}$, we get

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;A}}} & {{1}} \\  {{{si}}{{{n}}^{{2}}}{{B - si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;B - cot\;A}}} & {{0}} \\ {{{si}}{{{n}}^{{2}}}{{C - si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;C - cot\;A}}} & {{0}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;A}}} & {{1}} \\ {{{sin}}\left( {{{B + A}}} \right){{sin}}\left( {{{B - A}}} \right)} & {\dfrac{{{{cosB}}}}{{{{sinB}}}}{{ - }}\dfrac{{{{cosA}}}}{{{{sinA}}}}} & {{0}} \\ {{{sin}}\left( {{{C + A}}} \right){{sin}}\left( {{{C - A}}} \right)} & {\dfrac{{{{cosC}}}}{{{{sinC}}}}{{ - }}\dfrac{{{{cosA}}}}{{{{sinA}}}}} & {{0}} \\ \end{vmatrix}$

Here, $A + B + C = \pi $

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;A}}} & {{1}} \\ {{{sin}}\left( {{{\pi  - C}}} \right){{sin}}\left( {{{B - A}}} \right)} & {\dfrac{{{{cosB}}{{.sinA - cosA}}{{.sinB}}}}{{{{sinB}}{{.sinA}}}}} & {{0}} \\ {{{sin}}\left( {{{\pi  - B}}} \right){{sin}}\left( {{{C - A}}} \right)} & {\dfrac{{{{cosC}}{{.sinA - cosA}}{{.sinC}}}}{{{{sinC}}{{.sinA}}}}} & {{0}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {\dfrac{{{{cosA}}}}{{{{sinA}}}}} & {{1}} \\ {{{sinC}}{{.sin}}\left( {{{B - A}}} \right)} & {\dfrac{{{{sin}}\left( {{{A - B}}} \right)}}{{{{sinB}}{{.sinA}}}}} & {{0}} \\ {{{sinB\;}}{{.sin}}\left( {{{C - A}}} \right)} & {\dfrac{{{{sin}}\left( {{{A - C}}} \right)}}{{{{sinC}}{{.sinA}}}}} & {{0}} \\ \end{vmatrix}$

Taking $\sin \left( {B - A} \right)$ and $\sin \left( {C - A} \right)$ common from ${R_2}$ and ${R_3}$ respectively.

⇒ △ = $sin\left({B - A}\right){sin\left({C-A}\right)}$ $\begin{vmatrix} {{sin}^{2}{A}} & {\dfrac{cosA}{sinA}} & 1 \\ {{{sinC}}} & {\dfrac{-1}{sinB.sinA}} & 0 \\ {sinB} & {\dfrac{-1}{sinC.sinA}} & 0 \\ \end{vmatrix}$

Taking  $\dfrac{{{1}}}{{\sin A}}$ common from ${C_2},\;$we get

⇒ △ $= sin\left({B-A}\right)sin\left({C-A}\right)$ $\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cosA}}} & {{1}} \\ {{{sinC}}} & {\dfrac{{{{ - 1}}}}{{{{sinB}}}}} & {{0}} \\ {{{sinB}}} & {\dfrac{{{{ - 1}}}}{{{{sinC}}}}} & {{0}} \\ \end{vmatrix}$

Now, expanding along ${R_1},\;$we get

⇒ △ ${{ = sin}}\left( {{{B - A}}} \right){{sin}}\left( {{{C - A}}} \right){{\;}}\left( {{{ - 1 + 1}}} \right)$

⇒ △ ${{ = sin}}\left( {{{B - A}}} \right){{sin}}\left( {{{C - A}}} \right){{\; \times 0}}$

⇒ △ ${{ = 0}}$

Thus, △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{A}}} & {{{cot\;A}}} & {{1}} \\ {{{si}}{{{n}}^{{2}}}{{B}}} & {{{cot\;B}}} & {{1}} \\ {{{si}}{{{n}}^{{2}}}{{C}}} & {{{cot\;C}}} & {{1}} \\ \end{vmatrix}{{ = 0}}{{.}}$

Example 13: The determinant △ = $\begin{vmatrix} {\sqrt {23}  + \sqrt 3 } & {\sqrt 5 } & {\sqrt 5 } \\ {\sqrt {15}  + \sqrt {46} } & 5 & {\sqrt {10} } \\ {3 + \sqrt {115} } & {\sqrt {15} } & 5 \\ \end{vmatrix}$ is equal to …....

Ans: Here, we have △ ${{ = }}\begin{vmatrix} {\sqrt {{{23}}} {{ + }}\sqrt {{3}} } & {\sqrt {{5}} } & {\sqrt {{5}} } \\  {\sqrt {{{15}}} {{ + }}\sqrt {{{46}}} } & {{5}} & {\sqrt {{{10}}} } \\ {{{3 + }}\sqrt {{{115}}} } & {\sqrt {{{15}}} } & {{5}} \\ \end{vmatrix}$

Taking $\sqrt 5 \;$common from ${C_2}$ and ${C_3},$ we get

⇒ △ $ = \sqrt {{5}} {{ \times }}\sqrt {{5}} \begin{vmatrix} {\sqrt {{{23}}} {{ + }}\sqrt {{3}} } & {{1}} & {{1}} \\ {\sqrt {{{15}}} {{ + }}\sqrt {{{46}}} } & {\sqrt {{5}} } & {\sqrt {{2}} } \\ {{{3 + }}\sqrt {{{115}}} } & {\sqrt {{3}} } & {\sqrt {{5}} } \\ \end{vmatrix}$

Applying ${{{C}}_{{1}}} \to {{{C}}_{{1}}}{{ - }}\sqrt {{3}} {{{C}}_{{2}}}{{ - }}\sqrt {{{23}}} {{{C}}_{{3}}}$ 

⇒ △ $=5 \begin{vmatrix} {{0}} & {{1}} & {{1}} \\ {{0}} & {\sqrt{5}} & {\sqrt{2}} \\ {0} & {\sqrt{3}} & {\sqrt{5}} \\ \end{vmatrix}$

⇒ △ $ = 0$    (All the elements of a row is 0)

Thus, The determinant △ ${{ = }}\begin{vmatrix} {\sqrt {{{23}}} {{ + }}\sqrt {{3}} } & {\sqrt {{5}} } & {\sqrt {{5}} } \\ {\sqrt {{{15}}} {{ + }}\sqrt {{{46}}} } & {{5}} & {\sqrt {{{10}}} } \\ {{{3 + }}\sqrt {{{115}}} } & {\sqrt {{{15}}} } & {{5}} \\ \end{vmatrix}$ is equal to 0.

Example 14: The value of the determinant

△ $= \begin{vmatrix} {si{n^2}23^\circ } & {si{n^2}67^\circ } & {\cos 180^\circ } \\ { - si{n^2}67^\circ } & { - si{n^2}23^\circ } & {co{s^2}180^\circ } \\ {\cos 180^\circ } & {si{n^2}23^\circ } & {si{n^2}67^\circ } \\ \end{vmatrix} = $ …..

Ans: Here, we have △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{cos180^\circ }}} \\ {{{ - si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{ - si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{co}}{{{s}}^{{2}}}{{180^\circ }}} \\ {{{cos180^\circ }}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}\left( {{{90 - 23}}} \right){{^\circ }}} & {{{ - 1}}} \\ {{{ - si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{ - si}}{{{n}}^{{2}}}\left( {{{90 - 67}}} \right){{^\circ }}} & {{1}} \\ {{{ - 1}}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{co}}{{{s}}^{{2}}}{{23^\circ }}} & {{{ - 1}}} \\ {{{ - si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{ - co}}{{{s}}^{{2}}}{{67^\circ }}} & {{1}} \\ {{{ - 1}}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} \\ \end{vmatrix}$

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$

⇒ △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{23^\circ  + co}}{{{s}}^{{2}}}{{23^\circ  - 1}}} & {{{co}}{{{s}}^{{2}}}{{23^\circ }}} & {{{ - 1}}} \\ {{{ - (si}}{{{n}}^{{2}}}{{67^\circ  + co}}{{{s}}^{{2}}}{{67^\circ ) + 1}}} & {{{ - co}}{{{s}}^{{2}}}{{67^\circ }}} & {{1}} \\ {{{ - 1 + si}}{{{n}}^{{2}}}{{23^\circ  + si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix}{{{1 - 1}}} & {{{co}}{{{s}}^{{2}}}{{23^\circ }}} & {{{ - 1}}} \\ {{{ - 1 + 1}}} & {{{ - co}}{{{s}}^{{2}}}{{67^\circ }}} & {{1}} \\ {{{ - 1 + 1}}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}}  \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{0}} & {{{co}}{{{s}}^{{2}}}{{23^\circ }}} & {{{ - 1}}} \\ {{0}} & {{{ - co}}{{{s}}^{{2}}}{{67^\circ }}} & {{1}} \\ {{0}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}}  \\ \end{vmatrix}$

⇒ △ $ = 0$  (All the elements of a column is 0)

Hence, △ ${{ = }}\begin{vmatrix} {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{cos180^\circ }}} \\ {{{ - si}}{{{n}}^{{2}}}{{67^\circ }}} & {{{ - si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{co}}{{{s}}^{{2}}}{{180^\circ }}} \\ {{{cos180^\circ }}} & {{{si}}{{{n}}^{{2}}}{{23^\circ }}} & {{{si}}{{{n}}^{{2}}}{{67^\circ }}} \\ \end{vmatrix}{{ = 0}}{{.}}$

State whether the statements in Examples 15 to 18 is True or False.

Example 15: The determinant △ = $\begin{vmatrix} {cos\left( {x + y} \right)} & { - sin\left( {x + y} \right)} & {cos2y} \\ {sinx} & {cosx} & {siny} \\ { - cosx} & {sinx} & {cosy} \\ \end{vmatrix}$ is independent of $x$ only.

Ans: Here, we have △ = $\begin{vmatrix} {\cos \left( {x + y} \right)} & { - \sin \left( {x + y} \right)} & {\cos 2y} \\ {\sin x} & {\cos x} & {\sin y} \\ { - \cos x} & {\sin x} & {\cos y} \\ \end{vmatrix}$

Applying ${R_1} \to {R_1} + {R_2}\sin y + {R_3}\cos y$, we get

⇒ △ ${{ = }}\begin{vmatrix} {{0}} & {{0}} & {{{1 + cos2y}}} \\ {{{sinx}}} & {{{cosx}}} & {{{siny}}} \\  {{{ - cosx}}} & {{{sinx}}} & {{{cosy}}} \\ \end{vmatrix}$

Now, expanding along${R_1}$, we get

⇒ △ ${{ = \;(1 + cos2y)\;}}\left( {{{si}}{{{n}}^{{2}}}{{x + co}}{{{s}}^{{2}}}{{x}}} \right)$

⇒ △ ${{ = \;(1 + cos2y)\;}}$

Thus, the value of the given determinant is independent of $x$ only.

Hence, the given statement is True.

Example 16: The value of △ = $\begin{vmatrix} 1 & 1 & 1 \\ {{}^{n}{C_1}} & {{}^{n+2}{C_1}} & {{}^{n + 4}{C_1}} \\ {{}^{n}{C_2}} & {{}^{n + 2}{C_2}} & {{}^{n + 4}{C_2}} \\ \end{vmatrix}$ is 8.

Ans: Here, we have △ ${{ = }}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{}_{{.}}^{{n}}{{{C}}_{{1}}}} & {{}_{{.}}^{{{n + 2}}}{{{C}}_{{1}}}} & {{}_{{.}}^{{{n + 4}}}{{{C}}_{{1}}}} \\ {{}_{{.}}^{{n}}{{{C}}_{{2}}}} & {{}_{{.}}^{{{n + 2}}}{{{C}}_{{2}}}} & {{}_{{.}}^{{{n + 4}}}{{{C}}_{{2}}}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {\dfrac{{{{n!}}}}{{\left( {{{n - 1}}} \right){{!}}}}} & {\dfrac{{\left( {{{n + 2}}} \right){{!}}}}{{\left( {{{n + 1}}} \right){{!}}}}} & {\dfrac{{\left( {{{n + 4}}} \right){{!}}}}{{\left( {{{n + 3}}} \right){{!}}}}} \\ {\dfrac{{{{n!}}}}{{\left( {{{n - 2}}} \right){{!\;2!}}}}} & {\dfrac{{\left( {{{n + 2}}} \right){{!}}}}{{{{n!\;2!}}}}} & {\dfrac{{\left( {{{n + 4}}} \right){{!}}}}{{\left( {{{n + 2}}} \right){{!\;2!}}}}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {\dfrac{{{{n\;}}\left( {{{n - 1}}} \right){{!}}}}{{\left( {{{n - 1}}} \right){{!}}}}} & {\dfrac{{\left( {{{n + 2}}} \right)\left( {{{n + 1}}} \right){{!}}}}{{\left( {{{n + 1}}} \right){{!}}}}} & {\dfrac{{\left( {{{n + 4}}} \right)\left( {{{n + 3}}} \right){{!}}}}{{\left( {{{n + 3}}} \right){{!}}}}} \\ {\dfrac{{{{n}}\left( {{{n - 1}}} \right)\left( {{{n - 2}}} \right){{!}}}}{{\left( {{{n - 2}}} \right){{!\;2!}}}}} & {\dfrac{{\left( {{{n + 2}}} \right)\left( {{{n + 1}}} \right){{n!}}}}{{{{n!\;2!}}}}} & {\dfrac{{\left( {{{n + 4}}} \right)\left( {{{n + 3}}} \right)\left( {{{n + 2}}} \right){{!}}}}{{\left( {{{n + 2}}} \right){{!\;2!}}}}} \\ \end{vmatrix}$

⇒ △ ${{ = }}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{n}} & {\left( {{{n + 2}}} \right)} & {\left( {{{n + 4}}} \right)} \\ {\dfrac{{{{n}}\left( {{{n - 1}}} \right)}}{{{2}}}} & {\dfrac{{\left( {{{n + 2}}} \right)\left( {{{n + 1}}} \right)}}{{{2}}}} & {\dfrac{{\left( {{{n + 4}}} \right)\left( {{{n + 3}}} \right)}}{{{2}}}} \\ \end{vmatrix}$

Taking $\dfrac{{{1}}}{{{2}}}$ as common from ${R_3}$, 

⇒ △ ${{ = }}\dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\  {{n}} & {\left( {{{n + 2}}} \right)} & {\left( {{{n + 4}}} \right)} \\ {{{n}}\left( {{{n - 1}}} \right)} & {\left( {{{n + 2}}} \right)\left( {{{n + 1}}} \right)} & {\left( {{{n + 4}}} \right)\left( {{{n + 3}}} \right)} \\ \end{vmatrix}$

Applying ${C_1} \to {C_1} - {C_3}$ and ${C_2} \to {C_2} - {C_3}$, we get

⇒ △ ${{ = }}\dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{{ - 4}}} & {{{ - 2}}} & {\left( {{{n + 4}}} \right)} \\ {{{ - 8n - 12}}} & {{{ - 4n - 10}}} & {\left( {{{n + 4}}} \right)\left( {{{n + 3}}} \right)} \\ \end{vmatrix}$

Now, expanding along ${R_1},$ we get

⇒ △ ${{ = }}\dfrac{{{1}}}{{{2}}}{{\;}}\left[ {{{ - 4}}\left( {{{ - 4n - 10}}} \right){{ + 2}}\left( {{{ - 8n - 12}}} \right)} \right]$

⇒ △ ${{ = }}\dfrac{{{1}}}{{{2}}}{{\;}}\left[ {{{16n + 40 - 16n - 24}}} \right]$

⇒ △ ${{ = }}\dfrac{{{1}}}{{{2}}}{{\; \times 16}}$

⇒ △ = 8

Hence, the given statement is True.

Example 17: If $A = \begin{bmatrix} x & 5 & 2 \\  2 & y & 3 \\ 1 & 1 & z \end{bmatrix}$, $xyz = 80$, $3x + 2y + 10z = 20$, then 

$A\;adj.A = \begin{bmatrix} {81} & 0 & 0 \\  0 & {81} & 0 \\ 0 & 0 & {81} \end{bmatrix}$ .

Ans: here we have $A = \begin{vmatrix} {{x}} & {{5}} & {{2}} \\ {{2}} & {{y}} & {{3}} \\ {{1}} & {{1}} & {{z}}\end{vmatrix}$, ${{xyz = 80}}$, ${{3x + 2y + 10z = 20}}$

As we know that, ${{{A}}^{{{ - 1}}}}{{ = }}\dfrac{{{1}}}{{\left| {{A}} \right|}}{{\;adj}}\left( {{A}} \right)$

⇒ ${{A}}{{{A}}^{{{ - 1}}}}{{ = }}\dfrac{{{1}}}{{\left| {{A}} \right|}}{{\;A\;adj}}\left( {{A}} \right)$

⇒ ${{I = }}\dfrac{{{1}}}{{\left| {{A}} \right|}}{{\;A\;adj}}\left( {{A}} \right)$

⇒ ${{\;A\;adj}}\left( {{A}} \right){{ = }}\left| {{A}} \right|$ ……… (i)

Here, $\left| {{A}} \right|{{ = x}}\left( {{{yz - 3}}} \right){{ - 5}}\left( {{{2z - 3}}} \right){{ + 2}}\left( {{{2 - y}}} \right)$

⇒ $\left| {{A}} \right|{{ = xyz - 3x - 10z + 15 + 4 - 2y}}$

⇒ $\left| {{A}} \right|{{ = xyz + 19 - }}\left( {{{3x + 2y + 10z}}} \right)$

⇒ $\left| {{A}} \right|{{ = 80 + 19 - 20}}$

⇒ $\left| {{A}} \right|{{ = 79}}$

Therefore, ${{A\;adj}}\left( {{A}} \right){{ = 79}}$, and it is constant value.

Hence, the given statement is False.

Example 18: If A = $\begin{bmatrix} 0 & 1 & 3 \\  1 & 2 & x \\ 2 & 3 & 1 \end{bmatrix}$  ${A^{ - 1}} = \begin{bmatrix} {\dfrac{1}{2}} & { - 4} & {\dfrac{5}{2}} \\ { - \dfrac{1}{2}} & 3 & { - \dfrac{3}{2}} \\ {\dfrac{1}{2}} & y & {\dfrac{1}{2}} \end{bmatrix}$, then $x = 1,\;y =  - 1.$

Ans: Here, we have ${{A = }}\begin{bmatrix} {{0}} & {{1}} & {{3}} \\ {{1}} & {{2}} & {{x}} \\ {{2}} & {{3}} & {{1}} \end{bmatrix}$  

${{A}^{-1}}= \begin{bmatrix} {\dfrac{1}{2}} & {-4} & {\dfrac{5}{2}} \\ {-}\dfrac{1}{2} & {{3}} & {-}\dfrac{3}{2} \\ {\dfrac{1}{2}} & {{y}} & \dfrac{1}{2} \end{bmatrix}$

As we know that, ${{A}}{{{A}}^{{{ - 1}}}}{{ = I}}$

⇒ $\begin{bmatrix} {{0}} & {{1}} & {{3}} \\ {{1}} & {{2}} & {{x}} \\ {{2}} & {{3}} & {{1}} \end{bmatrix}$ $\begin{bmatrix} \dfrac{1}{2} & {-4} & \dfrac{5}{2} \\ {-}\dfrac{1}{2} & {{3}} & {-}\dfrac{3}{2} \\ \dfrac{1}{2} & {{y}} & \dfrac{1}{2} \end{bmatrix}$ ${{ = }}\begin{bmatrix} {{1}} & {{0}} & {{0}} \\ {{0}} & {{1}} & {{0}} \\ {{0}} & {{0}} & {{1}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{0-\dfrac{1}{2}+\dfrac{2}{2}}} & {{{0 + 3 + 3y}}} & {{0-\dfrac{3}{2}+\dfrac{3}{2}}} \\ {{\dfrac{1}{2}-1+\dfrac{x}{2}}} & {{{ - 4 + 6 + xy}}} & {{\dfrac{5}{2}-3+\dfrac{x}{2}}} \\ {{{1 - 1 + 1}}} & {{{ - 8 + 9 + y}}} & {{5 -\dfrac{9}{2}+\dfrac{1}{2}}} \end{bmatrix}$ = $\begin{bmatrix} {{1}} & {{0}} & {{0}} \\ {{0}} & {{1}} & {{0}} \\ {{0}} & {{0}} & {{1}} \end{bmatrix}$

⇒$\begin{bmatrix} {{1}} & {3}\left({1+y}\right) & {{0}} \\ {-}\dfrac{1}{2}+\dfrac{x}{2} & {2+xy} & {-}\dfrac{1}{2}{{+}}\dfrac{x}{2} \\ {{1}} & {{{1 + y}}} & {{1}}\\ \end{bmatrix}$ = $\begin{bmatrix}  {{1}} & {{0}} & {{0}} \\ {{0}} & {{1}} & {{0}} \\ {{0}} & {{0}} & {{1}} \end{bmatrix}$

Now, comparing both sides, we get

⇒${{3}}\left( {{{1 + y}}} \right){{ = 0}}$ and ${{ - }}\dfrac{{{1}}}{{{2}}}{{ + }}\dfrac{{{x}}}{{{2}}}{{ = 0}}$

⇒ ${{1 + y = 0}}$ and ${{ - 1 + x = 0}}$

⇒ ${{y =  - 1}}$ and x = 1 

Hence, the given statement is True.

Exercise 4.3

Using the properties of determinants in Exercise 1 to 6, evaluate

1. $\begin{vmatrix} {{{{x}}^{{2}}}{{ - x + 1\;}}} & {{{x - 1}}} \\ {{{x + 1}}} & {{{\;x + 1}}} \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{{{x}}^{{2}}}{{ - x + 1\;}}} & {{{x - 1}}} \\ {{{x + 1}}} & {{{\;x + 1}}} \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{{C}}_1} - {{{C}}_2}]$, we get

${{=}}\begin{vmatrix} {{{{x}}^{{2}}}{{ - 2x + 2\;}}} & {{{x - 1}}} \\ {{0}} & {{{\;x + 1}}} \\ \end{vmatrix}$ 

${{ = }}\left( {{{{x}}^{{2}}}{{ - 2x + 2}}} \right)\left( {{{x + 1}}} \right){{ - 0\;}}\left( {{{x - 1}}} \right)$

${{ = }}{{{x}}^{{3}}}{{ - 2}}{{{x}}^{{2}}}{{ + 2x + }}{{{x}}^{{2}}}{{ - 2x + 2}}$ 

$ = {{{x}}^3} - {{{x}}^2} + {{2}}$ 

Hence, $\begin{vmatrix} {{{{x}}^{{2}}}{{ - x + 1\;}}} & {{{x - 1}}} \\ {{{x + 1}}} & {{{\;x + 1}}} \\ \end{vmatrix}{{=}}{{{x}}^{{3}}}{{ - }}{{{x}}^{{2}}}{{ + 2}}$ .

2. $\begin{vmatrix} {{\mathbf{a}} + {\mathbf{x}}} & {\mathbf{y}} & {\mathbf{z}} \\ {\mathbf{x}} & {{\mathbf{a}} + {\mathbf{y}}} & {\mathbf{z}} \\ {\mathbf{x}} & {\mathbf{y}} & {{\mathbf{a}} + {\mathbf{z}}}  \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{{a}} + {{x}}} & {{y}} & {{z}} \\ {{x}} & {{{a}} + {{y}}} & {{z}} \\ {{x}} & {{y}} & {{{a}} + {{z}}} \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

$ = {{\;}}\begin{vmatrix}{{{a}} + {{x}} + {{y}} + {{z}}} & {{y}} & {{z}} \\ {{{a}} + {{x}} + {{y}} + {{z}}} & {{{a}} + {{y}}} & {{z}} \\ {{{a}} + {{x}} + {{y}} + {{z}}} & {{y}} & {{{a}} + {{z}}} \\ \end{vmatrix}$ 

Now, take common $\left( {{{a}} + {{x}} + {{y}} + {{z}}} \right)$ from ${{{C}}_{{1}}}$

${{ = \;}}\left( {{{a + x + y + z}}} \right){{\;}}\begin{vmatrix} {{1}} & {{y}} & {{z}} \\ {{1}} & {{{a + y}}} & {{z}} \\ {{1}} & {{y}} & {{{a + z}}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

${{ = \;}}\left( {{{a + x + y + z}}} \right){{\;}}\begin{vmatrix} {{1}} & {{y}} & {{z}} \\ {{0}} & {{a}} & {{0}} \\ {{0}} & {{0}} & {{a}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

${{ = \;}}\left( {{{a + x + y + z}}} \right){{\;}}\left[ {\left( {{{a}}{{.a - 0}}} \right){{ - y}}\left( {{{0 - 0}}} \right){{ + z}}\left( {{{0 - 0}}} \right)} \right]$ 

$ = {{\;}}\left( {{{a}} + {{x}} + {{y}} + {{z}}} \right){{\;}}\left[ {{{{a}}^{{2}}}{{ - 0 - 0}}} \right]$ 

$ = {{\;}}{{{a}}^2}{{\;}}\left( {{{a}} + {{x}} + {{y}} + {{z}}} \right)$ 

Hence, $\begin{vmatrix} {{{a}} + {{x}}} & {{y}} & {{z}} \\ {{x}} & {{{a}} + {{y}}} & {{z}} \\ {{x}} & {{y}} & {{{a}} + {{z}}} \\ \end{vmatrix} = {{{a}}^2}{{\;}}\left( {{{a}} + {{x}} + {{y}} + {{z}}} \right).$  

3. $\begin{vmatrix} 0 & {{\mathbf{x}}{{\mathbf{y}}^2}} & {{\mathbf{x}}{{\mathbf{z}}^2}} \\ {{{\mathbf{x}}^2}{\mathbf{y}}} & 0 & {{\mathbf{y}}{{\mathbf{z}}^2}} \\ {{{\mathbf{x}}^2}{\mathbf{z}}} & {{\mathbf{z}}{{\mathbf{y}}^2}} & 0 \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{0}} & {{x}{y}^2} & {{x}{z}^2} \\ {{x}^2{y}} & {{0}} & {{y}{z}^2} \\ {{{{x}}^2}{{z}}} & {{{z}}{{{y}}^2}} & {{0}} \\ \end{vmatrix}$

Now, taking common ${{{x}}^2}{{\;}},{{\;}}{{{y}}^2}{{\;and\;}}{{{z}}^2}$ from ${{{C}}_1},{{\;}}{{{C}}_2}{{\;and\;}}{{{C}}_3}$ respectively.

${{ = }}{{{x}}^{{2}}}{{{y}}^{{2}}}{{{z}}^{{2}}}\begin{vmatrix} {{0}} & {{x}} & {{x}} \\ {{y}} & {{0}} & {{y}} \\ {{z}} & {{z}} & {{0}} \\ \end{vmatrix}$

Applying \[{{[}}{{{C}}_{{2}}} \to {{{C}}_{{2}}}{{ - }}{{{C}}_{{3}}}{{]}}\], we get 

$ = {{{x}}^2}{{{y}}^2}{{{z}}^2}\begin{vmatrix} {{0}} & {{0}} & {{x}} \\ {{y}} & { - {{y}}} & {{y}} \\ {{z}} & {{z}} & {{0}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

$ = {{{x}}^2}{{{y}}^2}{{{z}}^2}{{\;}}\left[ {{{0}}\left( {{{0}} - {{yz}}} \right) - {{0}}\left( {{{0}} - {{zy}}} \right) + {{x}}\left( {{{yz}} + {{yz}}} \right)} \right]$ 

$ = {{{x}}^2}{{{y}}^2}{{{z}}^2}{{\;}}\left[ {{{x}}\left( {{{yz}} + {{yz}}} \right)} \right]$ 

$ = {{{x}}^2}{{{y}}^2}{{{z}}^2}{{\;}} \times {{2xyz\;}}$ 

$ = {{2}}{{{x}}^3}{{{y}}^3}{{{z}}^3}{{\;}}$ 

Hence, $\begin{vmatrix} {{0}} & {{{x}}{{{y}}^2}} & {{{x}}{{{z}}^2}} \\ {{{{x}}^2}{{y}}} & {{0}} & {{{y}}{{{z}}^2}} \\ {{{{x}}^2}{{z}}} & {{{z}}{{{y}}^2}} & {{0}} \\ \end{vmatrix} = {{2}}{{{x}}^3}{{{y}}^3}{{{z}}^3}.$

4. $\begin{vmatrix} {3{\mathbf{x}}} & { - {\mathbf{x}} + {\mathbf{y}}} & { - {\mathbf{x}} + {\mathbf{z}}} \\ {{\mathbf{x}} - {\mathbf{y}}} & {3{\mathbf{y}}} & {{\mathbf{z}} - {\mathbf{y}}} \\ {{\mathbf{x}} - {\mathbf{z}}} & {{\mathbf{y}} - {\mathbf{z}}} & {3{\mathbf{z}}} \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{{3x}}} & { - {{x}} + {{y}}} & { - {{x}} + {{z}}} \\ {{{x}} - {{y}}} & {{{3y}}} & {{{z}} - {{y}}} \\ {{{x}} - {{z}}} & {{{y}} - {{z}}} & {{{3z}}} \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{{C}}_{{1}}}{{ + }}{{{C}}_{{2}}}{{ + }}{{{C}}_{{3}}}]$, we get

$ = {{\;}}\begin{vmatrix} {{{x}} + {{y}} + {{z}}} & { - {{x}} + {{y}}} & { - {{x}} + {{z}}} \\ {{{x}} + {{y}} + {{z}}} & {{{3y}}} & {{{z}} - {{y}}} \\ {{{x}} + {{y}} + {{z}}} & {{{y}} - {{z}}} & {{{3z}}} \\ \end{vmatrix}$ 

Now, take common $\left( {{{x}} + {{y}} + {{z}}} \right)$ from ${{{C}}_1}$

$ = {{\;}}\left( {{{x}} + {{y}} + {{z}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{ - x + y}}} & {{{ - x + z}}} \\ {{1}} & {{{3y}}} & {{{z - y}}} \\ {{1}} & {{{y - z}}} & {{{3z}}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ = {{\;}}\left( {{{x}} + {{y}} + {{z}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{ - x + y}}} & {{{ - x + z}}} \\ {{0}} & {{{2y + x}}} & {{{x - y}}} \\ {{0}} & {{{x - z}}} & {{{2z + x}}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

$ = {{\;}}\left( {{{x}} + {{y}} + {{z}}} \right){{\;}}\left[ {\left( {{{2y + x}}} \right)\left( {{{2z + x}}} \right){{ - }}\left( {{{x - y}}} \right)\left( {{{x - z}}} \right){{ - 0 + 0}}} \right]$ 

${{ = \;}}\left( {{{x + y + z}}} \right){{\;}}\left[ {{{4yz + 2xy + 2xz + }}{{{x}}^{{2}}}{{ - }}\left( {{{{x}}^{{2}}}{{ - xz - xy + yz}}} \right)} \right]$ 

${{ = \;}}\left( {{{x + y + z}}} \right){{\;}}\left[ {{{4yz + 2xy + 2xz + }}{{{x}}^{{2}}}{{ - }}{{{x}}^{{2}}}{{ + xz + xy - yz}}} \right]$ 

${{ = \;}}\left( {{{x + y + z}}} \right){{\;}}\left[ {{{3yz + 3xy + 3xz}}} \right]$ 

$ = {{\;3}}\left( {{{x}} + {{y}} + {{z}}} \right){{\;}}\left( {{{yz}} + {{xy}} + {{xz}}} \right)$ 

Hence, $\begin{vmatrix} {{{3x}}} & {{{ - x + y}}} & {{{ - x + z}}} \\ {{{x - y}}} & {{{3y}}} & {{{z - y}}} \\ {{{x - z}}} & {{{y - z}}} & {{{3z}}} \\ \end{vmatrix}{{ = 3}}\left( {{{x + y + z}}} \right)\left( {{{yz + xy + xz}}} \right){{.}}$

5. $\begin{vmatrix} {{\mathbf{x}} + 4} & {\mathbf{x}} & {\mathbf{x}} \\ {\mathbf{x}} & {{\mathbf{x}} + 4} & {\mathbf{x}} \\ {\mathbf{x}} & {\mathbf{x}} & {{\mathbf{x}} + 4}  \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{{x}} + 4} & {{x}} & {{x}} \\ {{x}} & {{{x}} + 4} & {{x}} \\ {{x}} & {{x}} & {{{x}} + 4} \\ \end{vmatrix}$

Applying $\left[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}\right]$, we get

$= \begin{vmatrix} {{{3x + 4}}} & {{x}} & {{x}} \\ {{{3x + 4}}} & {{{x + 4}}} & {{x}} \\ {{{3x + 4}}} & {{x}} & {{{x + 4}}} \\ \end{vmatrix}$ 

Now, taking common $\left( {{{3x + 4}}} \right)$ from ${{{C}}_{{1}}}$

${{ = \;}}\left( {{{3x + 4}}} \right){{\;}}\begin{vmatrix} {{1}} & {{x}} & {{x}} \\  {{1}} & {{{x + 4}}} & {{x}} \\  {{1}} & {{x}} & {{{x + 4}}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

 ${{=}}\left( {{{3x + 4}}} \right)\begin{vmatrix} {{1}} & {{x}} & {{x}} \\ {{0}} & {{4}} & {{0}} \\  {{0}} & {{0}} & {{4}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

${{ = \;}}\left( {{{3x + 4}}} \right){{\;}}\left[ {\left( {{{16 - 0}}} \right){{ - 0 + 0}}} \right]$ 

${{ = \;16}}\left( {{{3x + 4}}} \right){{\;}}$ 

Hence, $\begin{vmatrix} {{{x + 4}}} & {{x}} & {{x}} \\ {{x}} & {{{x + 4}}} & {{x}} \\ {{x}} & {{x}} & {{{x + 4}}} \\ \end{vmatrix}{{ = 16}}\left( {{{3x + 4}}} \right)$.

6. $\begin{vmatrix} {{\mathbf{a}} - {\mathbf{b}} - {\mathbf{c}}} & {2{\mathbf{a}}} & {2{\mathbf{a}}} \\ {2{\mathbf{b}}} & {{\mathbf{b}} - {\mathbf{c}} - {\mathbf{a}}} & {2{\mathbf{b}}} \\ {2{\mathbf{c}}} & {2{\mathbf{c}}} & {{\mathbf{c}} - {\mathbf{a}} - {\mathbf{b}}} \\ \end{vmatrix}$

Ans: Here, we have $\begin{vmatrix} {{{a - b - c}}} & {{{2a}}} & {{{2a}}} \\ {{{2b}}} & {{{b - c - a}}} & {{{2b}}} \\ {{{2c}}} & {{{2c}}} & {{{c - a - b}}} \\ \end{vmatrix}$

Applying $[{{{R}}_1} \to {{{R}}_1} + {{{R}}_2} + {{{R}}_3}]$, we get

$ = {{\;}}\begin{vmatrix} {{{a}} + {{b}} + {{c}}} & {{{a}} + {{b}} + {{c}}} & {{{a}} + {{b}} + {{c}}} \\ {2{{b}}} & {{{b}} - {{c}} - {{a}}} & {2{{b}}} \\ {2{{c}}} & {2{{c}}} & {{{c}} - {{a}} - {{b}}} \\ \end{vmatrix}$ 

Now, taking common $\left( {{{a}} + {{b}} + {{c}}} \right)$ from ${{{R}}_1}$

${{ = \;}}\left( {{{a + b + c}}} \right){{\;}}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{{2b}}} & {{{b - c - a}}} & {{{2b}}} \\  {{{2c}}} & {{{2c}}} & {{{c - a - b}}} \\ \end{vmatrix}$ 

Applying $[{{{C}}_2} \to {{{C}}_2} - {{{C}}_1}{{\;and\;\;}}{{{C}}_3} \to {{{C}}_3} - {{{C}}_1}]$, we get

${{ = \;}}\left( {{{a + b + c}}} \right){{\;}}\begin{vmatrix} {{1}} & {{0}} & {{0}} \\ {{{2b}}} & {{{ - }}\left( {{{a + b + c}}} \right)} & {{0}} \\ {{{2c}}} & {{0}} & {{{ - }}\left( {{{a + b + c}}} \right)} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

$ = {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right){{\;}}\left[ {{{\left( {{{a}} + {{b}} + {{c}}} \right)}^2} - {{0}}} \right]$ 

$ = {{\;\;}}{\left( {{{a}} + {{b}} + {{c}}} \right)^3}$ 

Hence, $\begin{vmatrix} {{{a - b - c}}} & {{{2a}}} & {{{2a}}} \\ {{{2b}}} & {{{b - c - a}}} & {{{2b}}} \\ {{{2c}}} & {{{2c}}} & {{{c - a - b}}} \\ \end{vmatrix}{{=}}{\left( {{{a + b + c}}} \right)^{{3}}}{{.}}$

Using the properties of determinants in Exercises 7 to 9.

7. $\begin{vmatrix} {{{\mathbf{y}}^2}{{\mathbf{z}}^2}} & {{\mathbf{yz}}} & {{\mathbf{y}} + {\mathbf{z}}} \\ {{{\mathbf{z}}^2}{{\mathbf{x}}^2}} & {{\mathbf{zx}}} & {{\mathbf{z}} + {\mathbf{x}}} \\ {{{\mathbf{x}}^2}{{\mathbf{y}}^2}} & {{\mathbf{xy}}} & {{\mathbf{x}} + {\mathbf{y}}} \\ \end{vmatrix} = 0$

Ans: Here, we have L.H.S =$\begin{vmatrix} {{{{y}}^2}{{{z}}^2}} & {{{yz}}} & {{{y}} + {{z}}} \\ {{{{z}}^2}{{{x}}^2}} & {{{zx}}} & {{{z}} + {{x}}} \\ {{{{x}}^2}{{{y}}^2}} & {{{xy}}} & {{{x}} + {{y}}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_1} \to {{x}}{{{R}}_1},{{\;\;}}{{{R}}_2} \to {{y}}{{{R}}_2}{{\;and\;}}{{{R}}_3} \to {{z}}{{{R}}_3}]$,

And it can be written as, 

⇒ L.H.S  ${{ = }}\dfrac{{{1}}}{{{{xyz}}}}\begin{vmatrix} {{{x}}{{{y}}^{{2}}}{{{z}}^{{2}}}} & {{{xyz}}} & {{{xy + xz}}} \\ {{{y}}{{{z}}^{{2}}}{{{x}}^{{2}}}} & {{{yzx}}} & {{{yz + xy}}} \\  {{{z}}{{{x}}^{{2}}}{{{y}}^{{2}}}} & {{{zxy}}} & {{{zx + zy}}} \\ \end{vmatrix}$

Taking common (xyz) from ${{{C}}_1}{{\;and\;}}{{{C}}_2}{{\;}}$, we get

⇒ L.H.S  ${{ = }}\dfrac{{{1}}}{{{{xyz}}}}{{ \times }}{\left( {{{xyz}}} \right)^{{2}}}\begin{vmatrix}  {{{yz}}} & {{1}} & {{{xy + xz}}} \\  {{{zx}}} & {{1}} & {{{yz + xy}}} \\ {{{xy}}} & {{1}} & {{{zx + zy}}} \\ \end{vmatrix}$

⇒ L.H.S ${{ = xyz}}\begin{vmatrix} {{{yz}}} & {{1}} & {{{xy + xz}}} \\ {{{zx}}} & {{1}} & {{{yz + xy}}} \\ {{{xy}}} & {{1}} & {{{zx + zy}}} \\ \end{vmatrix}$

Now, applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_3}]$, we get

⇒ L.H.S  ${{ = xyz}}\begin{vmatrix} {{{xy + yz + xz}}} & {{1}} & {{{xy + xz}}} \\ {{{xy + yz + xz}}} & {{1}} & {{{yz + xy}}} \\ {{{xy + yz + xz}}} & {{1}} & {{{zx + zy}}} \\ \end{vmatrix}$

Taking common (${{xy}} + {{yz}} + {{xz}})$ from ${{{C}}_1}$

⇒ L.H.S  ${{ = xyz}}\left( {{{xy + yz + xz}}} \right)\begin{vmatrix} {{1}} & {{1}} & {{{xy + xz}}} \\ {{1}} & {{1}} & {{{yz + xy}}} \\ {{1}} & {{1}} & {{{zx + zy}}} \\ \end{vmatrix}$

⇒ L.H.S  $ = {{xyz}}\left( {{{xy}} + {{yz}} + {{xz}}} \right) \times {{0}}$        

(when two columns are same, then value of determinants is 0)

⇒ L.H.S = 0        

⇒ L.H.S  $ = {{R}}.{{H}}.{{S}}$        

Hence Proved.

8. $\begin{vmatrix} {{\mathbf{y}} + {\mathbf{z}}} & {\mathbf{z}} & {\mathbf{y}} \\ {\mathbf{z}} & {{\mathbf{z}} + {\mathbf{x}}} & {\mathbf{x}} \\ {\mathbf{y}} & {\mathbf{x}} & {{\mathbf{x}} + {\mathbf{y}}} \\ \end{vmatrix} = 4{\mathbf{xyz}}$

Ans: Here, we have L.H.S $=\begin{vmatrix} {{{y}} + {{z}}} & {{z}} & {{y}} \\ {{z}} & {{{z}} + {{x}}} & {{x}} \\ {{y}} & {{x}} & {{{x}} + {{y}}} \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

${{ = \;}}\begin{vmatrix} {{{2}}\left( {{{y + z}}} \right)} & {{z}} & {{y}} \\  {{{2}}\left( {{{x + z}}} \right)} & {{{z + x}}} & {{x}} \\ {{{2}}\left( {{{x + y}}} \right)} & {{x}} & {{{x + y}}} \\ \end{vmatrix}$ 

Now, taking common $\left( {{2}} \right)$ from ${{{C}}_{{1}}}$

$ = {{\;2}}\begin{vmatrix} {\left( {{{y}} + {{z}}} \right)} & {{z}} & {{y}} \\ {\left( {{{x}} + {{z}}} \right)} & {{{z}} + {{x}}} & {{x}} \\ {\left( {{{x}} + {{y}}} \right)} & {{x}} & {{{x}} + {{y}}} \\ \end{vmatrix}$ 

Applying $[{{{C}}_1} \to {{{C}}_1} - {{{C}}_2}{{\;}}]$, we get

${{ = 2}}\begin{vmatrix} {{y}} & {{z}} & {{y}} \\  {{0}} & {{{z + x}}} & {{x}} \\ {{y}} & {{x}} & {{{x + y}}} \\ \end{vmatrix}{{\;}}$ 

Applying $[{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}{{\;}}]$, we get

${{ = 2}}\begin{vmatrix} {{y}} & {{z}} & {{y}} \\ {{0}} & {{{z + x}}} & {{x}} \\ {{0}} & {{{x - z}}} & {{x}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

${{ = \;2\;}}\left[ {{{y}}\left\{ {{{xz + }}{{{x}}^{{2}}}{{ - }}\left( {{{{x}}^{{2}}}{{ - xz}}} \right)} \right\}{{ - 0 - 0}}} \right]$ 

$ = {{\;2\;}}\left[ {{{y}}\left\{ {{{xz}} + {{{x}}^2} - {{{x}}^2} + {{xz}}} \right\}} \right]$ 

${{ = \;2\; \times 2xyz}}$ 

$ = {{\;4xyz}}$ 

$ = {{\;R}}.{{H}}.{{S}}$ 

Thus, L.H.S = R.H.S

Hence proved.

9. $\begin{vmatrix} {{{\mathbf{a}}^2} + 2{\mathbf{a}}} & {2{\mathbf{a}} + 1} & 1 \\ {2{\mathbf{a}} + 1} & {{\mathbf{a}} + 2} & 1 \\  3 & 3 & 1 \\ \end{vmatrix} = {\left( {{\mathbf{a}} - 1} \right)^3}$

Ans: Here, we have L.H.S ${{ = }}\begin{vmatrix} {{{{a}}^{{2}}}{{ + 2a}}} & {{{2a + 1}}} & {{1}} \\ {{{2a + 1}}} & {{{a + 2}}} & {{1}} \\ {{3}} & {{3}} & {{1}} \\ \end{vmatrix}$

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

${{ = \;}}\begin{vmatrix} {{{{a}}^{{2}}}{{ + 2a}}} & {{{2a + 1}}} & {{1}} \\ {{{ - }}\left( {{{{a}}^{{2}}}{{ - 1}}} \right)} & {{{ - }}\left( {{{a + 1}}} \right)} & {{0}} \\ {{{ - (}}{{{a}}^{{2}}}{{ + 2a - 3)}}} & {{{ - }}\left( {{{2a - 2}}} \right)} & {{0}} \\ \end{vmatrix}$ 

${{ = \;}}\begin{vmatrix} {{{{a}}^{{2}}}{{ + 2a}}} & {{{2a + 1}}} & {{1}} \\ {{{ - }}\left( {{{a - 1}}} \right)\left( {{{a + 1}}} \right)} & {{{ - }}\left( {{{a - 1}}} \right)} & {{0}} \\ {{{ - }}\left( {{{a + 3}}} \right)\left( {{{a - 1}}} \right)} & {{{ - 2}}\left( {{{a - 1}}} \right)} & {{0}} \\ \end{vmatrix}$ 

Now, taking common $\left( {{{a - 1}}} \right)$ from ${{{R}}_2}$ and ${{{R}}_3}$

${{ = \;}}{\left( {{{a - 1}}} \right)^{{2}}}\begin{vmatrix} {{{{a}}^{{2}}}{{ + 2a}}} & {{{2a + 1}}} & {{1}} \\ {{{ - }}\left( {{{a + 1}}} \right)} & {{{ - 1}}} & {{0}} \\ {{{ - }}\left( {{{a + 3}}} \right)} & {{{ - 2}}} & {{0}} \\ \end{vmatrix}$ 

Now, expanding along ${{{R}}_1}$

${{ = \;}}{\left( {{{a - 1}}} \right)^{{2}}}\left[ {{{0 - 0 + 1}}\left\{ {{{2a + 2 - }}\left( {{{a + 3}}} \right)} \right\}} \right]$ 

${{ = \;}}{\left( {{{a - 1}}} \right)^{{2}}}\left( {{{2a + 2 - a - 3}}} \right)$ 

${{ = \;}}{\left( {{{a - 1}}} \right)^{{2}}}\left( {{{a - 1}}} \right)$ 

$ = {{\;}}{\left( {{{a}} - {{1}}} \right)^3}$ 

$ = {{\;R}}.{{H}}.{{S}}$ 

Thus, L.H.S = R.H.S

Hence proved.

10. If ${\mathbf{A}} + {\mathbf{B}} + {\mathbf{C}} = 0$, then prove that $\begin{vmatrix} {{1}} & {{{cosC}}} & {{{cosB}}} \\ {{{cosC}}} & {{1}} & {{{cosA}}} \\ {{{cosB}}} & {{{cosA}}} & {{1}} \\ \end{vmatrix} = 0$

Ans: Here, we have ${{A}} + {{B}} + {{C}} = {{0}}$

Now, L.H.S = $\begin{vmatrix} {{1}} & {{{cosC}}} & {{{cosB}}} \\ {{{cosC}}} & {{1}} & {{{cosA}}} \\ {{{cosB}}} & {{{cosA}}} & {{1}} \\ \end{vmatrix}$

Expanding along ${{{R}}_1}$, we get

$= {{1}}\left( {{{1 - Co}}{{{s}}^{{2}}}{{A}}} \right) - \cos {{C}}\left( {\cos {{C}} - \cos {{A}}.\cos {{B}}} \right) + \cos {{B}}\left( {\cos {{A}}.\cos {{C}} - \cos {{B}}} \right)$

$= {{si}}{{{n}}^2}{{A}} - {{co}}{{{s}}^2}{{C}} + \cos {{A}}.\cos {{B}}.\cos {{C}} + \cos {{A}}.\cos {{B}}.\cos {{C}} - {{co}}{{{s}}^2}{{B}}$

$= {{si}}{{{n}}^2}{{A}} - {{co}}{{{s}}^2}{{B}} + {{2}}\cos {{A}}.\cos {{B}}.\cos {{C}} - {{co}}{{{s}}^2}{{C}}$

$= - \cos \left( {{{A}} + {{B}}} \right).\cos \left( {{{A}} - {{B}}} \right) + {{2}}\cos {{A}}.\cos {{B}}.\cos {{C}} - {{co}}{{{s}}^2}{{C}}$

[ ${{co}}{{{s}}^2}{{B}} - {{si}}{{{n}}^2}{{A}} = \cos \left( {{{A}} + {{B}}} \right).\cos \left( {{{A}} - {{B}}} \right)]$

$= - \cos \left( { - {{C}}} \right).\cos \left( {{{A}} - {{B}}} \right) + \cos {{C}}\left( {2\cos {{A}}.\cos {{B}} - \cos {{C}}} \right)$   

$= - \cos {{C}}\left( {\cos {{A}}.\cos {{B}} + \sin {{A}}.\sin {{B}}} \right) + \cos {{C}}\left( {2\cos {{A}}.\cos {{B}} - \cos {{C}}} \right)$

$= - \cos {{C}}\left( {\cos {{A}}.\cos {{B}} + \sin {{A}}.\sin {{B}} - 2\cos {{A}}.\cos {{B}} + \cos {{C}}} \right)$

$= - \cos {{C}}\left( {\sin {{A}}.\sin {{B}} - \cos {{A}}.\cos {{B}} + \cos {{C}}} \right)$

$= \cos {{C}}\left( {\cos {{A}}.\cos {{B}} - \sin {{A}}.\sin {{B}} - \cos {{C}}} \right)$

$= \cos {{C}}\left[ {\cos \left( {{{A}} + {{B}}} \right) - \cos {{C}}} \right]$

$= \cos {{C}}\left[ {\cos \left( { - {{C}}} \right) - \cos {{C}}} \right]$

$= \cos {{C}}\left[ {\cos {{C}} - \cos {{C}}} \right]$

= 0 

= R.H.S

Thus, L.H.S = R.H.S

Hence proved.

11. If the co-ordinates of the vertices of an equilateral triangle with sides of length $'{\mathbf{a}}'$ are $\left( {{{\mathbf{x}}_1},\;{{\mathbf{y}}_1}} \right)\;,\;\left( {{{\mathbf{x}}_2},\;{{\mathbf{y}}_2}} \right),\;\left( {{{\mathbf{x}}_3},\;{{\mathbf{y}}_3}} \right)$, then ${\begin{vmatrix} {{{\mathbf{x}}_1}} & {{{\mathbf{y}}_1}} & 1 \\ {{{\mathbf{x}}_2}} & {{{\mathbf{y}}_2}} & 1 \\ {{{\mathbf{x}}_3}} & {{{\mathbf{y}}_3}} & 1 \\ \end{vmatrix}^2} = \dfrac{{3{{\mathbf{a}}^4}}}{4}.$

Ans: Since, we know that area of a triangle with vertices $\left( {{{{x}}_1},{{\;}}{{{y}}_1}} \right){{\;}},{{\;}}\left( {{{{x}}_2},{{\;}}{{{y}}_2}} \right){{\;and\;}}\left( {{{{x}}_3},{{\;}}{{{y}}_3}} \right)$ is given by ${{\Delta \; = }}\dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}} \\ \end{vmatrix}$

$ \Rightarrow {{{\Delta }}^{{2}}}{{\; = }}\dfrac{{{1}}}{{{4}}}{\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}} \\ \end{vmatrix}^{{2}}}$ ……………… eq (i)

And we know that area of an equilateral triangle with side a, 

\[ \Rightarrow \Delta {{\;}} = \dfrac{{\sqrt {{3}} }}{{{4}}} \times {{{a}}^2}\] 

$ \Rightarrow {\Delta ^2}{{\;}} = {\left( {\dfrac{{\sqrt {{3}} }}{{{4}}} \times {{{a}}^2}} \right)^2}$ 

$ \Rightarrow {\Delta ^2}{{\;}} = \dfrac{{{3}}}{{{{16}}}} \times {{{a}}^4}$ ………………. Eq (ii)

From eq (i) and (ii), we get

$ \Rightarrow \dfrac{{{1}}}{{{4}}}{\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}} \\ \end{vmatrix}^{{2}}}{{ = }}\dfrac{{{3}}}{{{{16}}}}{{ \times }}{{{a}}^{{4}}}$ 

$ \Rightarrow {{\;}}{\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}}  \\ \end{vmatrix}^{{2}}}{{ = }}\dfrac{{{{3}}{{{a}}^{{4}}}}}{{{4}}}$ 

Hence proved.

12. Find the value of ${\mathbf{\theta }}$ satisfying $\begin{vmatrix} 1 & 1 & {{\mathbf{sin}}\;3{\mathbf{\theta }}} \\ { - 4} & 3 & {{\mathbf{cos}}\;2{\mathbf{\theta }}} \\ 7 & { - 7} & { - 2} \\ \end{vmatrix} = 0$

Ans: Here, we have $\begin{vmatrix} {{1}} & {{1}} & {{{sin\;3\theta }}} \\ {{{ - 4}}} & {{3}} & {{{cos\;2\theta }}} \\ {{7}} & {{{ - 7}}} & {{{ - 2}}} \\ \end{vmatrix}{{ = 0}}$

Applying $[{{{C}}_1} \to {{{C}}_1} - {{{C}}_2}{{\;}}]$, we get

⇒ $\begin{vmatrix} {{0}} & {{1}} & {{{sin\;3\theta }}} \\ {{{ - 7}}} & {{3}} & {{{cos\;2\theta }}} \\ {{{14}}} & {{{ - 7}}} & {{{ - 2}}} \\ \end{vmatrix}{{ = 0}}$

Taking common 7 from ${{{C}}_1}$

⇒ ${{7}}\begin{vmatrix} {{0}} & {{1}} & {{{sin\;3\theta }}} \\ {{{ - 1}}} & {{3}} & {{{cos\;2\theta }}} \\ {{2}} & {{{ - 7}}} & {{{ - 2}}}  \\ \end{vmatrix}{{ = 0}}$

⇒ $\begin{vmatrix} {{0}} & {{1}} & {{{sin\;3\theta }}} \\ {{{ - 1}}} & {{3}} & {{{cos\;2\theta }}} \\ {{2}} & {{{ - 7}}} & {{{ - 2}}} \\ \end{vmatrix}{{ = 0}}$

Now, expanding along ${{{R}}_1}$

⇒ ${{0}}\left( {{{ - 6 + 7\;cos2\theta }}} \right){{ - 1}}\left( {{{2 - 2cos2\theta }}} \right){{ + sin\;3\theta }}\left( {{{7 - 6}}} \right){{ = 0}}$

⇒ ${{ - 2 + 2cos2\theta  + sin\;3\theta  = 0}}$

⇒ \[{{ - 2 + 2}}\left( {{{1 - 2si}}{{{n}}^{{2}}}{{\theta }}} \right){{ + 3sin\theta  - 4\;si}}{{{n}}^{{3}}}{{ = 0}}\]

⇒ ${{ - 2 + 2 - 4si}}{{{n}}^{{2}}}{{\theta  + 3sin\theta  - 4\;si}}{{{n}}^{{3}}}{{ = 0}}$

⇒ ${{4\;si}}{{{n}}^{{3}}}{{ + 4si}}{{{n}}^{{2}}}{{\theta  - 3sin\theta  = 0}}$

⇒ ${{sin\theta }}\left( {{{4\;si}}{{{n}}^{{2}}}{{ + 4sin\theta  - 3}}} \right){{ = 0}}$

⇒ ${{sin\theta }}\left( {{{4\;si}}{{{n}}^{{2}}}{{ + 6sin\theta  - 2sin\theta  - 3}}} \right){{ = 0}}$

⇒ ${{sin\theta }}\left\{ {{{\;2sin\theta }}\left( {{{2sin\theta  + 3}}} \right){{ - 1}}\left( {{{2sin\theta  + 3}}} \right){{\;}}} \right\}{{ = 0}}$

⇒ ${{sin\theta }}\left( {{{2sin\theta  + 3}}} \right)\left( {{{2sin\theta  - 1}}} \right){{ = 0}}$

⇒ ${{sin\theta  = 0,\;\;sin\theta  =  - }}\dfrac{{{3}}}{{{2}}}{{\;,\;sin\theta  = }}\dfrac{{{1}}}{{{2}}}$

Now, Case 1: When $\sin {{\theta }} = {{0}}$

⇒ $\sin {{\theta }} = {{Sin\;0}}^\circ $

⇒ ${{\theta }} = {{n\pi }}$, where ${{n}} \in {{Z}}$

Case 2: When $\sin {{\theta }} =  - \dfrac{{{3}}}{{{2}}}$

Here, it is not possible because $\sin {{\theta \;}} \in \left[ {{{ - 1,\;1}}} \right]$

Case 3: when $\sin {{\theta }} = \dfrac{{{1}}}{{{2}}}$

⇒ $\sin {{\theta }} = {{Sin\;}}\dfrac{{{\pi }}}{{{6}}}$

⇒ $\sin {{\theta }} = {{Sin\;}}\dfrac{{{\pi }}}{{{6}}}$

⇒ ${{\theta }} = {{n\pi }} + {\left( { - {{1}}} \right)^{{n}}}\dfrac{{{\pi }}}{{{6}}}$ , where ${{n}} \in {{Z}}$

Hence, values of ${{\theta }}$ are ${{n\pi }}$ or ${{n\pi }} + {\left( { - {{1}}} \right)^{{n}}}\dfrac{{{\pi }}}{{{6}}}$, where ${{n}} \in {{Z}}.$

13. If $\begin{vmatrix} {4 - {\mathbf{x}}} & {4 + {\mathbf{x}}} & {4 + {\mathbf{x}}} \\ {4 + {\mathbf{x}}} & {4 - {\mathbf{x}}} & {4 + {\mathbf{x}}} \\  {4 + {\mathbf{x}}} & {4 + {\mathbf{x}}} & {4 - {\mathbf{x}}} \\ \end{vmatrix} = 0$, then find the value of ${\mathbf{x}}$.

Ans: Here, we have $\begin{vmatrix} {{{4 - x}}} & {{{4 + x}}} & {{{4 + x}}} \\ {{{4 + x}}} & {{{4 - x}}} & {{{4 + x}}} \\  {{{4 + x}}} & {{{4 + x}}} & {{{4 - x}}} \\ \end{vmatrix}{{ = 0}}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

⇒ $\begin{vmatrix} {{{12 + x}}} & {{{4 + x}}} & {{{4 + x}}} \\ {{{12 + x}}} & {{{4 - x}}} & {{{4 + x}}} \\ {{{12 + x}}} & {{{4 + x}}} & {{{4 - x}}} \\ \end{vmatrix}{{ = 0}}$

Now, taking common $\left( {{{12}} + {{x}}} \right)$ from ${{{C}}_1}$

⇒ $\left( {{{12 + x}}} \right)\begin{vmatrix} {{1}} & {{{4 + x}}} & {{{4 + x}}} \\ {{1}} & {{{4 - x}}} & {{{4 + x}}} \\ {{1}} & {{{4 + x}}} & {{{4 - x}}} \\ \end{vmatrix}{{ = 0}}$

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ \Rightarrow {{\;}}\left( {{{12 + x}}} \right)\begin{vmatrix} {{1}} & {{{4 + x}}} & {{{4 + x}}} \\ {{0}} & {{{ - 2x}}} & {{0}} \\ {{0}} & {{0}} & {{{ - 2x}}} \\ \end{vmatrix}{{ = 0}}$ 

Now, expanding along ${{{C}}_1}$

$ \Rightarrow {{\;}}\left( {{{12 + x}}} \right)\left( {{{4}}{{{x}}^{{2}}}} \right){{ = 0}}$ 

$ \Rightarrow {{\;}}\left( {{{12 + x}}} \right){{ = 0\;or\;}}\left( {{{4}}{{{x}}^{{2}}}} \right){{ = 0}}$ 

$ \Rightarrow {{\;x =  - 12\;or\;x = 0}}$ 

Hence, the values of ${{x}}$ are -12 and 0.

14. If ${{\mathbf{a}}_1},\;{{\mathbf{a}}_2},\;{{\mathbf{a}}_3},$ …… ${{\mathbf{a}}_{\mathbf{r}}}$ are in G.P, then prove that the determinant$\begin{vmatrix} {{{\mathbf{a}}_{{\mathbf{r}} + 1}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 5}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 9}}} \\ {{{\mathbf{a}}_{{\mathbf{r}} + 7}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 11}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 15}}} \\ {{{\mathbf{a}}_{{\mathbf{r}} + 11}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 17}}} & {{{\mathbf{a}}_{{\mathbf{r}} + 21}}} \\ \end{vmatrix}$ is independent of ${\mathbf{r}}.$

Ans: Let the first term of given G.P be A and common ratio be R.

Then, nth term, ${{{a}}_{{n}}} = {{A}}{{{R}}^{{{n}} - 1}}$

Therefore, 

$= \begin{vmatrix} {{{{a}}_{{{r}} + 1}}} & {{{{a}}_{{{r}} + 5}}} & {{{{a}}_{{{r}} + 9}}} \\ {{{{a}}_{{{r}} + 7}}} & {{{{a}}_{{{r}} + 11}}} & {{{{a}}_{{{r}} + 15}}} \\ {{{{a}}_{{{r}} + 11}}} & {{{{a}}_{{{r}} + 17}}} & {{{{a}}_{{{r}} + 21}}} \\ \end{vmatrix}$

$= \begin{vmatrix} {{{A}}{{{R}}^{{r}}}} & {{{A}}{{{R}}^{{{r}} + 4}}} & {{{A}}{{{R}}^{{{r}} + 8}}} \\ {{{\;A}}{{{R}}^{{{r}} + 6}}} & {{{A}}{{{R}}^{{{r}} + 10}}} & {{{A}}{{{R}}^{{{r}} + 14}}} \\ {{{A}}{{{R}}^{{{r}} + 10}}} & {{{A}}{{{R}}^{{{r}} + 16}}} & {{{A}}{{{R}}^{{{r}} + 20}}} \\ \end{vmatrix}$

Taking common ${{A}}{{{R}}^{{r}}},{{\;A}}{{{R}}^{{{r}} + 6}}$ and ${{A}}{{{R}}^{{{r}} + 10}}$from ${{{R}}_1},{{\;}}{{{R}}_2}$ and ${{{R}}_3}$ respectively.

$ = {{A}}{{{R}}^{{r}}}.{{\;A}}{{{R}}^{{{r}} + 6}}$. ${{A}}{{{R}}^{{{r}} + 10}}\begin{vmatrix} 1 & {{{{R}}^4}} & {{{{R}}^8}} \\ 1 & {{{{R}}^4}} & {{{{R}}^8}} \\ 1 & {{{{R}}^6}} & {{{{R}}^{10}}} \\ \end{vmatrix}$

= 0                      (two rows are identical, hence value of determinant is 0)

Hence, the determinant $\begin{vmatrix} {{{{a}}_{{{r}} + 1}}} & {{{{a}}_{{{r}} + 5}}} & {{{{a}}_{{{r}} + 9}}} \\ {{{{a}}_{{{r}} + 7}}} & {{{{a}}_{{{r}} + 11}}} & {{{{a}}_{{{r}} + 15}}} \\ {{{{a}}_{{{r}} + 11}}} & {{{{a}}_{{{r}} + 17}}} & {{{{a}}_{{{r}} + 21}}} \\ \end{vmatrix}$ is independent of ${{r}}.$

15. Show that the points $\left( {{\mathbf{a}} + 5,\;{\mathbf{a}} - 4} \right),\;\left( {{\mathbf{a}} - 2,\;{\mathbf{a}} + 3} \right)\;$and $\left( {{\mathbf{a}},\;{\mathbf{a}}} \right)$ do not lie on a straight line for any value of ${\mathbf{a}}$.

Ans: Here, given points are $\left( {{{a + 5,\;a - 4}}} \right){{,\;}}\left( {{{a - 2,\;a + 3}}} \right){{\;}}$ and $\left( {{{a}},{{\;a}}} \right)$.

Now, $\Delta {{\;}} = \dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}} \\ \end{vmatrix}{{\;\;}}$

⇒  ${{\Delta \; = }}\dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{{a + 5}}} & {{{a - 4}}} & {{1}} \\ {{{a - 2}}} & {{{a + 3}}} & {{1}} \\ {{a}} & {{a}} & {{1}} \\ \end{vmatrix}$

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

⇒  ${{\Delta \; = }}\dfrac{{{1}}}{{{2}}}\begin{vmatrix} {{{a + 5}}} & {{{a - 4}}} & {{1}} \\ {{{ - 7}}} & {{7}} & {{0}} \\ {{{ - 5}}} & {{4}} & {{0}} \\ \end{vmatrix}$

Now, expanding along ${{{C}}_3}$

⇒  ${{\Delta \; = }}\dfrac{{{1}}}{{{2}}}\left[ {{{ - 28 + 35}}} \right]$

⇒  ${{\Delta \; = }}\dfrac{{{7}}}{{{2}}}{{\;}}$sq. units

⇒  $\Delta {{\;}} \ne {{0}}$

Here, the area of this triangle is not equal to 0.

Hence, given points form a triangle i.e., points do not lie in a straight line.

16. Show that the $\Delta {\mathbf{ABC}}$ is an isosceles triangle if the determinant,

$\Delta \; = \;\begin{vmatrix} 1 & 1 & 1 \\ {1 + \cos {\mathbf{A}}} & {1 + \cos {\mathbf{B}}} & {1 + \cos {\mathbf{C}}} \\ {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + \cos {\mathbf{A}}} & {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}} + \cos {\mathbf{B}}} & {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{C}} + \cos {\mathbf{C}}} \\ \end{vmatrix} = 0$ 

Ans: Here, we have $\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{{1 + cosA}}} & {{{1 + cosB}}} & {{{1 + cosC}}} \\ {{{co}}{{{s}}^{{2}}}{{A + cosA}}} & {{{co}}{{{s}}^{{2}}}{{B + cosB}}} & {{{co}}{{{s}}^{{2}}}{{C + cosC}}}  \\ \end{vmatrix}{{ = 0}}$

Now, applying $\left[{{C}_{2}} \to {{C}_{2}} - {{C}_{1}}\right] \text{ and } \left[{{C}_{3}} \to {{C}_{3}} - {{C}_{1}}\right]$, we get

⇒ $\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{{1 + cosA}}} & {{{1 + cosB}}} & {{{1 + cosC}}} \\ {{{co}}{{{s}}^{{2}}}{{A + cosA}}} & {{{co}}{{{s}}^{{2}}}{{B + cosB}}} & {{{co}}{{{s}}^{{2}}}{{C + cosC}}}  \\ \end{vmatrix}{{ = 0}}$

⇒ $\begin{vmatrix} {{1}} & {{0}} & {{0}} \\ {{{1 + cosA}}} & {{{cosB - cosA}}} & {{{cosC - cosA}}} \\ {{{co}}{{{s}}^{{2}}}{{A + cosA}}} & {{{co}}{{{s}}^{{2}}}{{B - co}}{{{s}}^{{2}}}{{A + cosB - cosA}}} & {{{co}}{{{s}}^{{2}}}{{C - co}}{{{s}}^{{2}}}{{A + cosC - cosA}}} \\ \end{vmatrix}{{ = 0}}$

⇒ $\begin{vmatrix} {{1}} & {{0}} & {{0}} \\ {{{1 + cosA}}} & {{{cosB - cosA}}} & {{{cosC - cosA}}} \\ {{{co}}{{{s}}^{{2}}}{{A + cosA}}} & {{{(cosB - cosA)}}\left( {{{cosB + cosA + 1}}} \right)} & {{{(cosC - cosA)(cosC + cosA + 1)}}} \\ \end{vmatrix}{{ = 0}}$

Now, take common $(\cos {{B}} - \cos {{A}}){{\;}}$ and $(\cos {{C}} - \cos {{A}})$ from ${{{C}}_1}$and ${{{C}}_2}$ respectively, we get

⇒ ${{(cosB - cosA)(cosC - cosA)\;}}\begin{vmatrix} {{1}} & {{0}} & {{0}} \\ {{{1 + cosA}}} & {{1}} & {{1}} \\ {{{co}}{{{s}}^{{2}}}{{A + cosA}}} & {\left( {{{cosB + cosA + 1}}} \right)} & {{{(cosC + cosA + 1)}}} \\ \end{vmatrix}{{ = 0}}$

Expanding along ${{{R}}_1}$,

⇒ ${{(cosB - cosA)(cosC - cosA)\;}}\left( {{{cosC + cosA + 1 - cosB - cosA - 1}}} \right){{ = 0}}$

⇒ $(\cos {{B}} - \cos {{A}})(\cos {{C}} - \cos {{A}}){{\;}}\left( {\cos {{C}} - \cos {{B}}} \right) = {{0}}$

⇒ ${{(cosB - cosA) = 0\;or\;(cosC - cosA) = 0\;or\;}}\left( {{{cosC - cosB}}} \right){{ = 0}}$

⇒ $\cos {{B}} = \cos {{Aor\;}}\cos {{C}} = \cos {{A\;or\;}}\cos {{C}} = \cos {{B}}$

⇒ ${{B}} = {{A\;or\;C}} = {{A\;or\;C}} = {{B}}$

Hence, $\Delta {{ABC}}$ is an isosceles triangle.

17. Find ${{\mathbf{A}}^{ - 1}}$ if ${\mathbf{A}} = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ and show that ${{\mathbf{A}}^{ - 1}} = \dfrac{{{{\mathbf{A}}^2} - 3{\mathbf{I}}}}{2}$.

Ans: Here, we have $A = \begin{bmatrix} {{0}} & {{1}} & {{1}} \\ {{1}} & {{0}} & {{1}} \\  {{1}} & {{1}} & {{0}} \end{bmatrix}$

Here, minors of elements are 

⇒ ${{{M}}_{{{11}}}}{{ = }}\begin{vmatrix} {{0}} & {{1}} \\ {{1}} & {{0}} \\ \end{vmatrix}{{ =  - 1}}$,  ${{{M}}_{{{12}}}}{{ = }}\begin{vmatrix} {{1}} & {{1}} \\ {{1}} & {{0}} \\ \end{vmatrix}{{ =  - 1}}$, ${{M}_{13}} = \begin{vmatrix} {{1}} & {{0}} \\ {{1}} & {{1}} \\ \end{vmatrix} = 1$

⇒ ${{{M}}_{{{21}}}}{{ = }}\begin{vmatrix} {{1}} & {{1}} \\ {{1}} & {{0}}  \\ \end{vmatrix}{{ =  - 1}}$ ,  ${{M}_{22}}= \begin{vmatrix} {{0}} & {{1}} \\ {{1}} & {{0}} \\ \end{vmatrix} = - 1$, ${{{M}}_{{{23}}}}{{ = }}\begin{vmatrix}  {{0}} & {{1}} \\   {{1}} & {{1}}  \\ \end{vmatrix} = -1$

⇒ ${{{M}}_{{{31}}}}{{ = }}\begin{vmatrix} {{1}} & {{1}} \\ {{0}} & {{1}} \\ \end{vmatrix}{{ = 1}}$ ,  ${{{M}}_{{{32}}}}{{ = }}\begin{vmatrix}{{0}} & {{1}} \\ {{1}} & {{1}} \\ \end{vmatrix}{{ =  - 1}}$, ${{{M}}_{{{33}}}}{{ = }}\begin{vmatrix}{{0}} & {{1}} \\ {{1}} & {{0}} \\ \end{vmatrix}{{ =  - 1}}$

Now, cofactor of an element ${{{a}}_{{{ij}}}}{{\;is\;}}{{{A}}_{{{ij}}}}{{ = }}{\left( {{{ - 1}}} \right)^{{{i + j}}}}{{{M}}_{{{ij}}}}$.

⇒ ${{{A}}_{{{11}}}}{{ = }}{\left( {{{ - 1}}} \right)^{{{1 + 1}}}}{{{M}}_{{{11}}}}{{ =  - 1}}$ ,  ${{{A}}_{12}} = {{1}}$, ${{{A}}_{13}} = {{1}}$

⇒ ${{{A}}_{21}} = {{1}}$ , ${{{A}}_{22}} =  - {{1}}$, ${{{A}}_{23}} = {{1}}$

⇒ ${{{A}}_{31}} = {{1}}$ ,  ${{{A}}_{32}} = {{1}}$, ${{{A}}_{33}} =  - {{1}}$

Thus, matrix formed by cofactors, say ${{P}} = \begin{bmatrix} {{{ - 1}}} & {{1}} & {{1}} \\  {{1}} & {{{ - 1}}} & {{1}} \\ {{1}} & {{1}} & {{{ - 1}}} \end{bmatrix}$

Now, Adjoint matrix, ${{adj\;A}} = $ transpose of matrix P

⇒ ${{adj\;A}} = {\left[ {{P}} \right]^{{T}}}$

⇒ $adj A = {\begin{bmatrix} {{-1}} & {{1}} & {{1}} \\ {{1}} & {{-1}} & {{1}} \\ {{1}} & {{1}} & {{-1}} \end{bmatrix}}^{{T}}$

⇒ ${{adj\;A = }}\begin{bmatrix} {{{ - 1}}} & {{1}} & {{1}} \\ {{1}} & {{{ - 1}}} & {{1}} \\   {{1}} & {{1}} & {{{ - 1}}} \end{bmatrix}$

And $\left| {{A}} \right|{{ = }}\begin{vmatrix} {{0}} & {{1}} & {{1}} \\  {{1}} & {{0}} & {{1}} \\ {{1}} & {{1}} & {{0}} \\ \end{vmatrix}$

⇒ $\left| {{A}} \right|{{ = 0}}\left( {{{0 - 1}}} \right){{ - 1}}\left( {{{0 - 1}}} \right){{ + 1}}\left( {{{1 - 0}}} \right)$

⇒ $\left| {{A}} \right|{{ = 1 + 1 = 2}}$

Now, L.H.S $ = {{\;}}{{{A}}^{ - 1}}$

$ = {{\;}}\dfrac{{{{adj\;A}}}}{{\left| {{A}} \right|}}$

${{ = }}\dfrac{{{1}}}{{{2}}}\begin{bmatrix} {{{ - 1}}} & {{1}} & {{1}} \\ {{1}} & {{{ - 1}}} & {{1}} \\ {{1}} & {{1}} & {{{ - 1}}} \end{bmatrix}$

And R.H.S $= \dfrac{{{{{A}}^{{2}}}{{ - 3I}}}}{{{2}}}$

$= \dfrac{{{1}}}{{{2}}}$$\begin{bmatrix} {{0}} & {{1}} & {{1}} \\ {{1}} & {{0}} & {{1}} \\ {{1}} & {{1}} & {{0}} \end{bmatrix}$$\begin{bmatrix} {{0}} & {{1}} & {{1}} \\ {{1}} & {{0}} & {{1}} \\ {{1}} & {{1}} & {{0}}\end{bmatrix}$ ${{-}}\dfrac{{{1}}}{{{2}}}\begin{bmatrix} {{3}} & {{0}} & {{0}} \\ {{0}} & {{3}} & {{0}} \\ {{0}} & {{0}} & {{3}} \end{bmatrix}$

$= \dfrac{{{1}}}{{{2}}}{{\;}}\begin{bmatrix} {{2}} & {{1}} & {{1}} \\ {{1}} & {{2}} & {{1}} \\ {{1}} & {{1}} & {{2}} \end{bmatrix}$${{ - }}\dfrac{{{1}}}{{{2}}}\begin{bmatrix}{{3}} & {{0}} & {{0}} \\ {{0}} & {{3}} & {{0}} \\ {{0}} & {{0}} & {{3}} \end{bmatrix}$

$= \dfrac{{{1}}}{{{2}}}{{\;}}\begin{bmatrix} {{{ - 1}}} & {{1}} & {{1}} \\ {{1}} & {{{ - 1}}} & {{1}} \\  {{1}} & {{1}} & {{{ - 1}}} \end{bmatrix}$

Thus, L.H.S = R.H.S

Hence proved.

Long Answer (L.A)

18. If $A = \begin{bmatrix} 1 & 2 & 0 \\ { - 2} & { - 1} & { - 2} \\ 0 & { - 1} & 1 \end{bmatrix}$, find ${{\mathbf{A}}^{ - 1}}.$ Using ${{\mathbf{A}}^{ - 1}}$, solve the system of linear equations $x - 2y = 10,\;2x - y - z = 8,\; - 2y + z = 7.$

Ans: Here, we have ${{A}} = {{\;}}\begin{bmatrix} {{1}} & {{2}} & {{0}} \\ {{{ - 2}}} & {{{ - 1}}} & {{{ - 2}}} \\ {{0}} & {{{ - 1}}} & {{1}} \end{bmatrix}$ ……. Eq(i)

Now, $\left| {{A}} \right| = \begin{vmatrix} {{1}} & {{2}} & {{0}} \\ {{{ - 2}}} & {{{ - 1}}} & {{{ - 2}}} \\ {{0}} & {{{ - 1}}} & {{1}} \\ \end{vmatrix}$

⇒ $\left| {{A}} \right|{{ = 1}}\left( {{{ - 1 - 2}}} \right){{ - 2}}\left( {{{ - 2 - 0}}} \right){{ + 0}}\left( {{{2 - 0}}} \right)$

⇒ $\left| {{A}} \right|{{ =  - 3 + 4 = 1}}$

Here, minors of elements are 

⇒ ${{{M}}_{{{11}}}}{{ = }}\begin{vmatrix} {{{ - 1}}} & {{{ - 2}}} \\ {{{ - 1}}} & {{1}} \\ \end{vmatrix}{{ =  - 3}}$ ,${{{M}}_{{{12}}}}{{ = }}\begin{vmatrix} {{{ - 2}}} & {{{ - 2}}} \\ {{0}} & {{1}} \\ \end{vmatrix}{{ =  - 2}}$, ${{{M}}_{{{13}}}}{{ = }}\begin{vmatrix} {{{ - 2}}} & {{{ - 1}}} \\ {{0}} & {{{ - 1}}} 1\\ \end{vmatrix}{{ = 2}}$

⇒ ${{{M}}_{{{21}}}}{{ = }}\begin{vmatrix} {{2}} & {{0}} \\  {{{ - 1}}} & {{1}} \\ \end{vmatrix}{{ = 2}}$ ,  ${{M}_{22}}=\begin{vmatrix} {{1}} & {{0}} \\ {{0}} & {{1}} \\ \end{vmatrix}= 1$, ${{{M}}_{{{23}}}}{{ = }}\begin{vmatrix} {{1}} & {{2}} \\ {{0}} & {{{ - 1}}} \\ \end{vmatrix}{{ =  - 1}}$

⇒ ${{{M}}_{{{31}}}}{{ = }}\begin{vmatrix} {{2}} & {{0}} \\ {{{ - 1}}} & {{{ - 2}}} \\ \end{vmatrix}{{ =  - 4}}$ , ${{{M}}_{{{32}}}}{{ = }}\begin{vmatrix} {{1}} & {{0}} \\ {{{ - 2}}} & {{{ - 2}}} \\ \end{vmatrix}{{ =  - 2}}$, ${{M}_{33}}{{ = }}\begin{vmatrix} {{1}} & {{2}} \\{{{-2}}} & {{{-1}}} \\ \end{vmatrix} = 3$

Now, cofactor of an element ${{{a}}_{{{ij}}}}{{\;is\;}}{{{A}}_{{{ij}}}} = {\left( {{{ - 1}}} \right)^{{{i}} + {{j}}}}{{{M}}_{{{ij}}}}$.

⇒ ${{{A}}_{11}} = {\left( { - {{1}}} \right)^{1 + 1}}{{{M}}_{11}} =  - {{3}}$ ,  ${{{A}}_{12}} = {{2}}$, ${{{A}}_{13}} = {{2}}$

⇒ ${{{A}}_{21}} =  - {{2}}$ ,  ${{{A}}_{22}} = {{1}}$, ${{{A}}_{23}} = {{1}}$

⇒ ${{{A}}_{31}} =  - {{4}}$ ,  ${{{A}}_{32}} = {{2}}$, ${{{A}}_{33}} = {{3}}$

Thus, matrix formed by cofactors, say ${{P}} = {{\;}}\begin{bmatrix} {{{ - 3}}} & {{2}} & {{2}} \\ {{{ - 2}}} & {{1}} & {{1}} \\ {{{ - 4}}} & {{2}} & {{3}} \end{bmatrix}$

Now, Adjoint matrix, ${{adj\;A}} = $ transpose of matrix P

⇒ ${{adj\;A}} = {\left[ {{P}} \right]^{{T}}}$

⇒ ${{adj\;A}} = {\begin{bmatrix} {{{ - 3}}} & {{2}} & {{2}} \\ {{{ - 2}}} & {{1}} & {{1}} \\ {{{ - 4}}} & {{2}} & {{3}} \end{bmatrix}^{{T}}}$

⇒ ${{adj\;A}} = \begin{bmatrix}  {{{ - 3}}} & {{{ - 2}}} & {{{ - 4}}} \\ {{2}} & {{1}} & {{2}} \\ {{2}} & {{1}} & {{3}} \end{bmatrix}$

Now, ${{\;}}{{{A}}^{ - 1}}$ $ = {{\;}}\dfrac{{{{adj\;A}}}}{{\left| {{A}} \right|}}$

⇒ ${{{A}}^{ - 1}}$ $ = {{\;}}\begin{bmatrix}{{{ - 3}}} & {{{ - 2}}} & {{{ - 4}}} \\ {{2}} & {{1}} & {{2}} \\  {{2}} & {{1}} & {{3}} \end{bmatrix}$ ……….. eq(ii)

Hence, the required ${{A}^{ - 1}}$ is $\begin{bmatrix} {{-3}} & {{{-2}}} & {{{ - 4}}} \\ {{2}} & {{1}} & {{2}} \\ {{2}} & {{1}} & {{3}} \\ \end{bmatrix}$

Also, we have the system of linear equations as

${{x - 2y = 10,\;\;2x - y - z = 8,\;\; - 2y + z = 7}}$ 

These equations can be written as, 

$\begin{bmatrix} {{1}} & {{{ - 2}}} & {{0}} \\ {{2}} & {{{ - 1}}} & {{{ - 1}}} \\ {{0}} & {{{ - 2}}} & {{1}} \end{bmatrix}$$\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\begin{bmatrix} {{{10\;}}} \\ {{8}} \\  {{7}} \end{bmatrix}$ 

Let ${{C = }}\begin{bmatrix} {{1}} & {{{ - 2}}} & {{0}} \\ {{2}} & {{{ - 1}}} & {{{ - 1}}} \\ {{0}} & {{{ - 2}}} & {{1}} \end{bmatrix}$ ${{,\;X = \;}}\begin{bmatrix}  {{x}} \\ {{y}} \\ {{z}}  \end{bmatrix}$ & ${{D=}}\begin{bmatrix} {{{10}}} \\ {{8}} \\ {{7}} \end{bmatrix}$

⇒ ${{CX}} = {{D}}$

We know that, ${\left( {{{{A}}^{{T}}}} \right)^{ - 1}} = {\left( {{{{A}}^{ - 1}}} \right)^{{T}}}$

⇒ ${{{C}}^{{T}}}{{ = }}\begin{bmatrix} {{1}} & {{2}} & {{0}} \\  {{{ - 2}}} & {{{ - 1}}} & {{{ - 2}}} \\  {{0}} & {{{ - 1}}} & {{1}} \end{bmatrix}{{ = A}}$             (from eq (i))

⇒ ${{X}} = {{{C}}^{ - 1}}{{D}} = {\left( {{{{A}}^{ - 1}}} \right)^{{T}}}{{D}}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ = $\begin{bmatrix} {{{ - 3}}} & {{2}} & {{2}} \\ {{{ - 2}}} & {{1}} & {{1}} \\  {{{ - 4}}} & {{2}} & {{3}} \end{bmatrix}$ $\begin{bmatrix} {{{10\;}}} \\ {{8}} \\ {{7}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{=}}\begin{bmatrix} {{{ - 30 + 16 + 14\;}}} \\ {{{ - 20 + 8 + 7}}} \\ {{{ - 40 + 16 + 21}}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\begin{bmatrix} {{{0\;}}} \\ {{{ - 5}}} \\ {{{ - 3}}} \end{bmatrix}$

On comparing, we get

⇒ ${{x = 0,\;y =  - 5,\;z =  - 3}}$

Hence, values of ${{x}},{{\;y}}$ and ${{z}}$ are 0, $ - {{5}}$ and $ - {{3}}$ respectively.

19. Using the matrix method, solve the system of equations 3x+2y-2z=3, x+2y+3z=6, 2x-y+z=2.

Ans: Here, we have ${{3x + 2y - 2z = 3,\;x + 2y + 3z = 6,\;2x - y + z = 2}}$

These, equations can be written as, 

⇒ $\begin{bmatrix} {{3}} & {{2}} & {{{ - 2}}} \\  {{1}} & {{2}} & {{3}} \\  {{2}} & {{{ - 1}}} & {{1}} \end{bmatrix}$ $\begin{bmatrix} {{x}} \\ {{{y\;}}} \\ {{z}} \end{bmatrix} {{= }}\begin{bmatrix} {{3}} \\ {{{6}}} \\ {{2}} \end{bmatrix}$

Let ${{A = }}\begin{bmatrix} {{3}} & {{2}} & {{{ - 2}}} \\ {{1}} & {{2}} & {{3}} \\ {{2}} & {{{ - 1}}} & {{1}} \end{bmatrix}$ ${{X = }}\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ & ${{B = }}\begin{bmatrix} {{3}} \\ {{6}} \\ {{2}} \end{bmatrix}$

⇒ ${{AX}} = {{B}}$

⇒ ${{X}} = {{{A}}^{ - 1}}{{B}}$ …………….. eq (i),

Now, $\left| {{A}} \right|{{ = }}\begin{vmatrix} {{3}} & {{2}} & {{{ - 2}}} \\ {{1}} & {{2}} & {{3}} \\ {{2}} & {{{ - 1}}} & {{1}} \\ \end{vmatrix}$

⇒ $\left| {{A}} \right|{{ = 3}}\left( {{{2 + 3}}} \right){{ - 2}}\left( {{{1 - 6}}} \right){{ - 2}}\left( {{{ - 1 - 4}}} \right)$

⇒ $\left| {{A}} \right|{{ = 15 + 10 + 10}}$

⇒ $\left| {{A}} \right|{{ = 35}}$

Here, minors of elements are 

⇒ ${{M}_{11}}{{=}}\begin{vmatrix}  {{2}} & {{3}} \\  {{{ - 1}}} & {{1}} \\ \end{vmatrix}=5$, ${{{M}}_{{{12}}}}{{ = }}\begin{vmatrix}  {{1}} & {{3}} \\ {{2}} & {{1}} \\ \end{vmatrix}{{ =  - 5}}$, ${{{M}}_{{{13}}}}{{ = }}\begin{vmatrix} {{1}} & {{2}} \\ {{2}} & {{{ - 1}}} \\ \end{vmatrix}{{ =  - 5}}$

⇒ ${{{M}}_{{{21}}}}{{ = }}\begin{vmatrix} {{2}} & {{{ - 2}}} \\ {{{ - 1}}} & {{1}} \\ \end{vmatrix}{{ = 0}}$ ,  ${{{M}}_{{{22}}}}{{ = }}\begin{vmatrix} {{3}} & {{{ - 2}}} \\ {{2}} & {{1}} \\ \end{vmatrix}{{ = 7}}$, ${{{M}}_{{{23}}}}{{ = }}\begin{vmatrix} {{3}} & {{2}} \\ {{2}} & {{{ - 1}}} \\ \end{vmatrix}{{ =  - 7}}$

⇒ ${{{M}}_{{{31}}}}{{ = }}\begin{vmatrix} {{2}} & {{{ - 2}}} \\ {{2}} & {{3}} \\ \end{vmatrix}{{ = 10}}$ , ${{{M}}_{{{32}}}}{{ = }}\begin{vmatrix} {{3}} & {{{ - 2}}} \\ {{1}} & {{3}} \\ \end{vmatrix}{{ = 11}}$, ${{{M}}_{{{33}}}}{{ = }}\begin{vmatrix} {{3}} & {{2}} \\ {{1}} & {{2}} \\ \end{vmatrix}{{ = 4}}$

Now, cofactor of an element ${{{a}}_{{{ij}}}}{{\;is\;}}{{{A}}_{{{ij}}}} = {\left( { - {{1}}} \right)^{{{i}} + {{j}}}}{{{M}}_{{{ij}}}}$.

⇒ ${{{A}}_{{{11}}}}{{ = }}{\left( {{{ - 1}}} \right)^{{{1 + 1}}}}{{{M}}_{{{11}}}}{{ = 5}}$ ,  ${{{A}}_{{{12}}}}{{ = 5}}$, ${{{A}}_{{{13}}}}{{ =  - 5}}$

⇒ ${{{A}}_{{{21}}}}{{ = 0}}$ ,  ${{{A}}_{22}} = {{7}}$, ${{{A}}_{23}} = {{7}}$

⇒ ${{{A}}_{31}} = {{10}}$ ,  ${{{A}}_{32}} =  - {{11}}$, ${{{A}}_{33}} = {{4}}$

Thus, matrix formed by cofactors, say ${{P}} = {{\;}}\begin{bmatrix} {{5}} & {{5}} & {{{ - 5}}} \\ {{0}} & {{7}} & {{7}} \\ {{{10}}} & {{{ - 11}}} & {{4}} \end{bmatrix}$

Now, Adjoint matrix, ${{adj\;A}} = $ transpose of matrix P

⇒ ${{adj\;A}} = {\left[ {{P}} \right]^{{T}}}$

⇒ ${{adj\;A = }}{\begin{bmatrix} {{5}} & {{5}} & {{{ - 5}}} \\ {{0}} & {{7}} & {{7}} \\ {{{10}}} & {{{ - 11}}} & {{4}} \end{bmatrix}^{{T}}}$

⇒ ${{adj\;A = }}\begin{bmatrix} {{5}} & {{0}} & {{{10}}} \\ {{5}} & {{7}} & {{{ - 11}}} \\ {{{ - 5}}} & {{7}} & {{4}} \end{bmatrix}$

Now, ${{\;}}{{{A}}^{ - 1}}$ $ = {{\;}}\dfrac{{{{adj\;A}}}}{{\left| {{A}} \right|}}$

⇒ ${{{A}}^{-1}}$ ${{ = }}\dfrac{1}{35}\begin{bmatrix} {{5}} & {{0}} & {{{10}}} \\ {{5}} & {{7}} & {{{ - 11}}} \\ {{{ - 5}}} & {{7}} & {{4}} \end{bmatrix}$

Now, ${{X}} = {{{A}}^{ - 1}}{{B}}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\dfrac{{{1}}}{{{{35}}}}\begin{bmatrix} {{5}} & {{0}} & {{{10}}} \\ {{5}} & {{7}} & {{{ - 11}}} \\ {{{ - 5}}} & {{7}} & {{4}} \end{bmatrix}\begin{bmatrix} {{3}} \\ {{{6\;}}} \\ {{2}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}{{ = }}\dfrac{{{1}}}{{{{35}}}}\begin{bmatrix} {{{15 + 0 + 20}}} \\ {{{15 + 42 - 22\;}}} \\ {{{ - 15 + 42 + 8}}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\dfrac{{{1}}}{{{{35}}}}\begin{bmatrix} {{{35}}} \\ {{{35\;}}} \\ {{{35}}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\begin{bmatrix} {\dfrac{35}{35}} \\ {\dfrac{35}{35}} \\ {\dfrac{35}{35}} \end{bmatrix}$

⇒ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}{{ = }}\begin{bmatrix} {{1}} \\ {{{1\;}}} \\ {{1}} \end{bmatrix}$

On comparing, we get

⇒ ${{x = 1,\;y = 1,\;z = 1}}$

Hence, values of ${{x}},{{\;y}}$ and ${{z}}$ are 1, ${{1}}$ and ${{1}}$ respectively.

20. Given $A = \begin{bmatrix} 2 & 2 & { - 4} \\  { - 4} & 2 & { - 4} \\ 2 & { - 1} & 5 \end{bmatrix}$, $B = \begin{bmatrix} 1 & {-1} & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$, find BA and use this to solve the system of equations $y + 2z = 7,\;x - y = 3,\;2x + 3y + 4z = 17.$ 

Ans: Here, we have ${{A = }}\begin{bmatrix} {{2}} & {{2}} & {{{ - 4}}} \\ {{{ - 4}}} & {{2}} & {{{ - 4}}} \\  {{2}} & {{{ - 1}}} & {{5}} \end{bmatrix}{{,\;B = }}\begin{bmatrix} {{1}} & {{{ - 1}}} & {{0}} \\ {{2}} & {{3}} & {{4}} \\ {{0}} & {{1}} & {{2}} \end{bmatrix}$

⇒ $BA = \begin{bmatrix} {{1}} & {{{ - 1}}} & {{0}} \\ {{2}} & {{3}} & {{4}} \\ {{0}} & {{1}} & {{2}} \end{bmatrix}$ $\begin{bmatrix} {{2}} & {{2}} & {{{ - 4}}} \\ {{{ - 4}}} & {{2}} & {{{ - 4}}} \\ {{2}} & {{{ - 1}}} & {{5}} \end{bmatrix}$

⇒ $BA =\begin{bmatrix} {{{2 + 4 + 0}}} & {{{2 - 2 + 0}}} & {{{ - 4 + 4 + 0}}} \\ {{{4 - 12 + 8}}} & {{{4 + 6 - 4}}} & {{{ - 8 - 12 + 20}}} \\ {{{0 - 4 + 4}}} & {{{0 + 2 - 2}}} & {{{0 - 4 + 10}}} \end{bmatrix}$

⇒ ${{BA = }}\begin{bmatrix}  {{6}} & {{0}} & {{0}} \\ {{0}} & {{6}} & {{0}} \\  {{0}} & {{0}} & {{6}} \end{bmatrix}$

⇒ ${{BA = 6}}\begin{bmatrix} {{1}} & {{0}} & {{0}} \\ {{0}} & {{1}} & {{0}} \\ {{0}} & {{0}} & {{1}} \end{bmatrix}$

⇒ ${{BA = 6I}}$

⇒ ${{{B}}^{ - 1}} = \dfrac{{{A}}}{{{6}}}$

⇒ ${{B}^{-1}} = \dfrac{1}{6} \begin{bmatrix} {{2}} & {{2}} & {{{ - 4}}} \\ {{{ - 4}}} & {{2}} & {{{ - 4}}} \\ {{2}} & {{{ - 1}}} & {{5}} \end{bmatrix} $ ……………. Eq (i) 

Also, we have ${{\;x - y = 3,\;2x + 3y + 4z = 17,\;y + 2z = 7}}$

These, equations can be written as, 

⇒$\begin{bmatrix} {{1}} & {{{ - 1}}} & {{0}} \\ {{2}} & {{3}} & {{4}} \\ {{0}} & {{1}} & {{2}} \end{bmatrix}$ $\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}{{ = }}\begin{bmatrix} {{3}} \\  {{{17}}} \\ {{7}} \end{bmatrix}$

⇒$\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}{\begin{bmatrix}{{1}} & {{{ - 1}}} & {{0}} \\  {{2}} & {{3}} & {{4}} \\ {{0}} & {{1}} & {{2}} \end{bmatrix}}^{-1}$ $\begin{bmatrix}{{3}} \\ {{{17}}} \\ {{7}} \end{bmatrix}$

⇒$\begin{bmatrix} {{x}} \\  {{y}} \\ {{z}} \end{bmatrix}$ ${{ = }}\dfrac{{{1}}}{{{6}}}\begin{bmatrix} {{2}} & {{2}} & {{{ - 4}}} \\ {{{ - 4}}} & {{2}} & {{{ - 4}}} \\ {{2}} & {{{ - 1}}} & {{5}} \end{bmatrix}$ $\begin{bmatrix} {{3}} \\ {{{17}}} \\ {{7}} \end{bmatrix}$                 (from eq (i))

⇒$\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}{{ = }}\dfrac{{{1}}}{{{6}}}\begin{bmatrix} {{{6 + 34 - 28}}} \\ {{{ - 12 + 34 - 28}}} \\ {{{6 - 17 + 35}}} \end{bmatrix}$ 

⇒$\begin{bmatrix} {{x}} \\ {{y}} \\ {{z}} \end{bmatrix}{{ = }}\dfrac{{{1}}}{{{6}}}\begin{bmatrix} {{{12}}} \\ {{{ - 6}}} \\ {{{24}}} \end{bmatrix}$  

⇒$\begin{bmatrix} {{x}} \\  {{y}} \\  {{z}} \end{bmatrix}{{ = }}\begin{bmatrix} {\dfrac{{{{12}}}}{{{6}}}} \\ {{{ - }}\dfrac{{{6}}}{{{6}}}} \\ {\dfrac{{{{24}}}}{{{6}}}} \end{bmatrix}$   

⇒$\begin{bmatrix} {{x}} \\  {{y}} \\ {{z}} \end{bmatrix}$${{ = }}\begin{bmatrix}{{{\;\;\;2}}} \\ {{{ - 1}}} \\ {{4}} \end{bmatrix}$       

On comparing, we get

⇒ ${{x = 2,\;y =  - 1,\;z = 4}}$

Hence, values of ${{x}},{{\;y}}$ and ${{z}}$ are 2, $ - 1$ and $4$ respectively.

21. If ${\mathbf{a}} + {\mathbf{b}} + {\mathbf{c}} \ne 0$ and $\begin{vmatrix} {\mathbf{a}} & {\mathbf{b}} & {\mathbf{c}} \\ {\mathbf{b}} & {\mathbf{c}} & {\mathbf{a}} \\ {\mathbf{c}} & {\mathbf{a}} & {\mathbf{b}} \\ \end{vmatrix} = 0$, then prove that ${\mathbf{a}} = {\mathbf{b}} = {\mathbf{c}}.$

Ans: Here, we have ${{a}} + {{b}} + {{c}} \ne {{0}}$ and $\begin{vmatrix} {{a}} & {{b}} & {{c}} \\ {{b}} & {{c}} & {{a}} \\  {{c}} & {{a}} & {{b}} \\ \end{vmatrix} = {{0}}$

⇒ $\begin{vmatrix} {{a}} & {{b}} & {{c}} \\  {{b}} & {{c}} & {{a}} \\ {{c}} & {{a}} & {{b}}  \\ \end{vmatrix} = {{0}}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

⇒ $\begin{vmatrix} {{{a}} + {{b}} + {{c}}} & {{b}} & {{c}} \\ {{{a}} + {{b}} + {{c}}} & {{c}} & {{a}} \\ {{{a}} + {{b}} + {{c}}} & {{a}} & {{b}} \\ \end{vmatrix} = {{0}}$

Now, taking common $\left( {{{a}} + {{b}} + {{c}}} \right)$ from ${{{C}}_{{1}}}$

⇒ $\left( {{{a}} + {{b}} + {{c}}} \right)\begin{vmatrix} {{1}} & {{b}} & {{c}} \\ {{1}} & {{c}} & {{a}} \\  {{1}} & {{a}} & {{b}}  \\ \end{vmatrix}{{ = 0}}$

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ \Rightarrow {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right)\begin{vmatrix} {{1}} & {{b}} & {{c}} \\ {{0}} & {{{c - b}}} & {{{a - c}}} \\ {{0}} & {{{a - b}}} & {{{b - c}}} \\ \end{vmatrix} = {{0}}$ 

Now, expanding along ${{{C}}_1}$

$ \Rightarrow {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right)\left[ {\left( {{{c}} - {{b}}} \right)\left( {{{b}} - {{c}}} \right) - \left( {{{a}} - {{b}}} \right)\left( {{{a}} - {{c}}} \right)} \right] = {{0}}$ 

$ \Rightarrow {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right)\left[ {{{bc}} - {{{c}}^2} - {{{b}}^2} + {{bc}} - \left( {{{{a}}^2} - {{ac}} - {{ab}} + {{bc}}} \right)} \right] = {{0}}$ 

$ \Rightarrow {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right)\left( {2{{bc}} - {{{c}}^2} - {{{b}}^2} - {{{a}}^2} + {{ac}} + {{ab}} - {{bc}}} \right) = {{0}}$ 

$ \Rightarrow {{\;}}\left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{bc}} - {{{c}}^2} - {{{b}}^2} - {{{a}}^2} + {{ac}} + {{ab}}} \right) = {{0}}$ 

$ \Rightarrow {{\;}} - \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^2} + {{{b}}^2} + {{{c}}^2} - {{ab}} - {{bc}} - {{ac}}} \right) = {{0}}$ 

$ \Rightarrow {{ - }}\dfrac{{{1}}}{{{2}}}\left( {{{a + b + c}}} \right)\left( {{{2}}{{{a}}^{{2}}}{{ + 2}}{{{b}}^{{2}}}{{ + 2}}{{{c}}^{{2}}}{{ - 2ab - 2bc - 2ac}}} \right){{ = 0}}$ 

\[ \Rightarrow {{\;}}\left( {{{a + b + c}}} \right)\left[ {{{{a}}^{{2}}}{{ + }}{{{a}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ + }}{{{c}}^{{2}}}{{ + }}{{{c}}^{{2}}}{{ - 2ab - 2bc - 2ac}}} \right]{{ = 0}}\] 

$ \Rightarrow \left( {{{a + b + c}}} \right)\left[ {{{(}}{{{a}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ - 2ab}}} \right){{ + }}\left( {{{{b}}^{{2}}}{{ + }}{{{c}}^{{2}}}{{ - 2bc}}} \right){{ + }}{{{c}}^{{2}}}{{ + }}{{{a}}^{{2}}}{{ - 2ac] = 0}}$ 

$ \Rightarrow \left( {{{a}} + {{b}} + {{c}}} \right)\left[ {{{\;}}{{\left( {{{a}} - {{b}}} \right)}^2} + {{\left( {{{b}} - {{c}}} \right)}^2} + {{\left( {{{c}} - {{a}}} \right)}^2}} \right] = {{0}}$ 

Here, $\left( {{{a}} + {{b}} + {{c}}} \right) \ne {{0}}$ 

Therefore, 

$ \Rightarrow {{\;}}{\left( {{{a}} - {{b}}} \right)^2} + {\left( {{{b}} - {{c}}} \right)^2} + {\left( {{{c}} - {{a}}} \right)^2} = {{0}}$ 

It is possible when ${\left( {{{a - b}}} \right)^{{2}}}{{ = 0,\;}}{\left( {{{b - c}}} \right)^{{2}}}{{ = 0\;,\;}}{\left( {{{c - a}}} \right)^{{2}}}{{ = 0}}$

$ \Rightarrow {{a - b = 0,\;b - c = 0\;,\;c - a = 0}}$ 

$ \Rightarrow {{a}} = {{b}},{{\;b}} = {{c\;}},{{\;c}} = {{a}}$ 

Thus, ${{a}} = {{b}} = {{c}}.$

Hence proved.

22. Prove that $\begin{vmatrix} {{\mathbf{bc}} - {{\mathbf{a}}^2}} & {{\mathbf{ca}} - {{\mathbf{b}}^2}} & {{\mathbf{ab}} - {{\mathbf{c}}^2}} \\ {{\mathbf{ca}} - {{\mathbf{b}}^2}} & {{\mathbf{ab}} - {{\mathbf{c}}^2}} & {{\mathbf{bc}} - {{\mathbf{a}}^2}} \\ {{\mathbf{ab}} - {{\mathbf{c}}^2}} & {{\mathbf{bc}} - {{\mathbf{a}}^2}} & {{\mathbf{ca}} - {{\mathbf{b}}^2}} \\ \end{vmatrix}$ is divisible by $\left( {{\mathbf{a}} + {\mathbf{b}} + {\mathbf{c}}} \right)$ and find the quotient.

Ans: Here, we have $\begin{vmatrix} {{{bc}} - {{{a}}^2}} & {{{ca}} - {{{b}}^2}} & {{{ab}} - {{{c}}^2}} \\ {{{ca}} - {{{b}}^2}} & {{{ab}} - {{{c}}^2}} & {{{bc}} - {{{a}}^2}} \\ {{{ab}} - {{{c}}^2}} & {{{bc}} - {{{a}}^2}} & {{{ca}} - {{{b}}^2}} \\ \end{vmatrix}$

Applying $\left[{{C}_{1}} \to {{C}_{1}} - {{C}_{2}}\right] \text{ and }  \left[{{C}_{2}} \to {{C}_{2}} - {{C}_{3}} \right]$, we get

$= \begin{vmatrix} {{{bc}} - {{{a}}^2} - {{ca}} + {{{b}}^2}} & {{{ca}} - {{{b}}^2} - {{ab}} + {{{c}}^2}} & {{{ab}} - {{{c}}^2}} \\ {{{ca}} - {{{b}}^2} - {{ab}} + {{{c}}^2}} & {{{ab}} - {{{c}}^2} - {{bc}} + {{{a}}^2}} & {{{bc}} - {{{a}}^2}} \\ {{{ab}} - {{{c}}^2} - {{bc}} + {{{a}}^2}} & {{{bc}} - {{{a}}^2} - {{ca}} + {{{b}}^2}} & {{{ca}} - {{{b}}^2}} \\ \end{vmatrix}$ 

$= \begin{vmatrix} \left(b-a\right)\left(a+b+c\right) & \left(c-b\right)\left(a+b+c\right) & {{ab}-{{c}^{2}}} \\ \left(c-b\right)\left(a+b+c\right) & \left(a-c \right)\left(a+b+c\right) & {{bc}-{a}^{2}} \\ \left(a-c\right)\left(a+b+c\right) & \left(b-a\right)\left(a+b+c\right) & {{ca}-{b}^{2}} \\ \end{vmatrix}$ 

Taking $\left( {{{a}} + {{b}} + {{c}}} \right)$ common from ${{{C}}_1}$ and ${{{C}}_2}$

$ = {\left( {{{a}} + {{b}} + {{c}}} \right)^2}\begin{vmatrix} {\left( {{{b}} - {{a}}} \right)} & {\left( {{{c}} - {{b}}} \right)} & {{{ab}} - {{{c}}^2}} \\ {\left( {{{c}} - {{b}}} \right)} & {\left( {{{a}} - {{c}}} \right)} & {{{bc}} - {{{a}}^2}} \\ {\left( {{{a}} - {{c}}} \right)} & {\left( {{{b}} - {{a}}} \right)} & {{{ca}} - {{{b}}^2}} \\ \end{vmatrix}$ 

Applying ${{{R}}_1} \to {{{R}}_1} + {{{R}}_2} + {{{R}}_3}$

${{ = }}{\left( {{{a + b + c}}} \right)^{{2}}}\begin{vmatrix} {{0}} & {{0}} & {{{ab + bc + ca - }}{{{a}}^{{2}}}{{ - }}{{{b}}^{{2}}}{{ - }}{{{c}}^{{2}}}} \\ {\left( {{{c - b}}} \right)} & {\left( {{{a - c}}} \right)} & {{{bc - }}{{{a}}^{{2}}}} \\ {\left( {{{a - c}}} \right)} & {\left( {{{b - a}}} \right)} & {{{ca - }}{{{b}}^{{2}}}} \\ \end{vmatrix}$ 

Taking common $\left( {{{ab}} + {{bc}} + {{ca}} - {{{a}}^2} - {{{b}}^2} - {{{c}}^2}} \right)$ from ${{{R}}_1}$

${{ = }}{\left( {{{a + b + c}}} \right)^{{2}}}\left( {{{ab + bc + ca - }}{{{a}}^{{2}}}{{ - }}{{{b}}^{{2}}}{{ - }}{{{c}}^{{2}}}} \right)\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {\left( {{{c - b}}} \right)} & {\left( {{{a - c}}} \right)} & {{{bc - }}{{{a}}^{{2}}}} \\ {\left( {{{a - c}}} \right)} & {\left( {{{b - a}}} \right)} & {{{ca - }}{{{b}}^{{2}}}} \\ \end{vmatrix}$ 

${{=  -}}{\left( {{{a + b + c}}} \right)^{{2}}}\left( {{{{a}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ + }}{{{c}}^{{2}}}{{ - ab - bc - ac}}} \right)\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {\left( {{{c - b}}} \right)} & {\left( {{{a - c}}} \right)} & {{{bc - }}{{{a}}^{{2}}}} \\ {\left( {{{a - c}}} \right)} & {\left( {{{b - a}}} \right)} & {{{ca - }}{{{b}}^{{2}}}} \\ \end{vmatrix}$ 

${{ =  - }}\left( {{{a + b + c}}} \right)\left( {{{{a}}^{{3}}}{{ + }}{{{b}}^{{3}}}{{ + }}{{{c}}^{{3}}}{{ - 3abc}}} \right)\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {\left( {{{c - b}}} \right)} & {\left( {{{a - c}}} \right)} & {{{bc - }}{{{a}}^{{2}}}} \\ {\left( {{{a - c}}} \right)} & {\left( {{{b - a}}} \right)} & {{{ca - }}{{{b}}^{{2}}}} \\ \end{vmatrix}$ 

Expanding along ${{{R}}_1}$

$ =  - \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left[ {\left( {{{c}} - {{b}}} \right)\left( {{{b}} - {{a}}} \right) - \left( {{{a}} - {{c}}} \right)\left( {{{a}} - {{c}}} \right)} \right]$ 

$ =  - \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left[ {{{bc}} - {{ac}} - {{{b}}^2} + {{ab}} - \left( {{{{a}}^2} + {{{c}}^2} - {{2ac}}} \right)} \right]$ 

$ =  - \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left( {{{bc}} - {{ac}} - {{{b}}^2} + {{ab}} - {{{a}}^2} - {{{c}}^2} + {{2ac}}} \right)$ 

$ =  - \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left( { - {{{a}}^2} - {{{b}}^2} - {{{c}}^2} + {{ab}} + {{bc}} + {{ac}}} \right)$ 

$ = \dfrac{{{1}}}{{{2}}}\left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left[ {{{\;}}{{\left( {{{a}} - {{b}}} \right)}^2} + {{\left( {{{b}} - {{c}}} \right)}^2} + {{\left( {{{c}} - {{a}}} \right)}^2}} \right]$ 

Hence, the given determinant is divisible by $\left( {{{a}} + {{b}} + {{c}}} \right)$.

And quotient is $\dfrac{{{1}}}{{{2}}}\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)\left[ {{{\;}}{{\left( {{{a}} - {{b}}} \right)}^2} + {{\left( {{{b}} - {{c}}} \right)}^2} + {{\left( {{{c}} - {{a}}} \right)}^2}} \right]$.

23. If ${\mathbf{x}} + {\mathbf{y}} + {\mathbf{z}} = 0,$ prove that $\begin{vmatrix} {{\mathbf{xa}}} & {{\mathbf{yb}}} & {{\mathbf{zc}}} \\ {{\mathbf{yc}}} & {{\mathbf{za}}} & {{\mathbf{xb}}} \\ {{\mathbf{zb}}} & {{\mathbf{xc}}} & {{\mathbf{ya}}} \\ \end{vmatrix} = {\mathbf{xyz}}\begin{vmatrix} {\mathbf{a}} & {\mathbf{b}} & {\mathbf{c}} \\ {\mathbf{c}} & {\mathbf{a}} & {\mathbf{b}} \\ {\mathbf{b}} & {\mathbf{c}} & {\mathbf{a}} \\ \end{vmatrix}$.

Ans: Here, we have ${{x}} + {{y}} + {{z}} = {{0}}$

Now, L.H.S $ = \begin{vmatrix} {{{xa}}} & {{{yb}}} & {{{zc}}} \\  {{{yc}}} & {{{za}}} & {{{xb}}} \\ {{{zb}}} & {{{xc}}} & {{{ya}}} \\ \end{vmatrix}$

Expanding along ${{{R}}_1}$

$ = {{xa}}\left( {{{{a}}^2}{{yz}} - {{{x}}^2}{{bc}}} \right) - {{yb}}\left( {{{{y}}^2}{{ac}} - {{{b}}^2}{{xz}}} \right) + {{zc}}\left( {{{{c}}^2}{{xy}} - {{{z}}^2}{{ab}}} \right)$ 

$ = {{{a}}^3}{{xyz}} - {{{x}}^3}{{abc}} - {{{y}}^3}{{abc}} + {{{b}}^3}{{xyz}} + {{{c}}^3}{{xyz}} - {{{z}}^3}{{abc}}$ 

$ = {{xyz}}\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3}} \right) - {{abc}}\left( {{{{x}}^3} + {{{y}}^3} + {{{z}}^3}} \right)$ 

$ = {{xyz}}\left( {{{{a}}^3} + {{{b}}^3} + {{{c}}^3}} \right) - {{abc}} \times {{3xyz}}$    

[if ${{\;x}} + {{y}} + {{z}} = {{0}}$, ${{{x}}^3} + {{{y}}^3} + {{{z}}^3} = {{3xyz}}]$

$ = {{xyz}}\left( {{{\;}}{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)$    

Now, R.H.S $ = {{xyz}}\begin{vmatrix} {{a}} & {{b}} & {{c}} \\ {{c}} & {{a}} & {{b}} \\ {{b}} & {{c}} & {{a}} \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

$ = {{xyz}}\begin{vmatrix} {{{a}} + {{b}} + {{c}}} & {{b}} & {{c}} \\ {{{a}} + {{b}} + {{c}}} & {{a}} & {{b}} \\ {{{a}} + {{b}} + {{c}}} & {{c}} & {{a}} \\ \end{vmatrix}$ 

Taking common $\left( {{{a}} + {{b}} + {{c}}} \right)$ from ${{{C}}_1}$

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right)\begin{vmatrix} {{1}} & {{b}} & {{c}} \\ {{1}} & {{a}} & {{b}} \\ {{1}} & {{c}} & {{a}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right){{\;}}\begin{vmatrix} {{1}} & {{b}} & {{c}} \\ {{0}} & {{{a - b}}} & {{{b - c}}} \\ {{0}} & {{{c - b}}} & {{{a - c}}} \\ \end{vmatrix}$ 

Now, expanding along ${{{C}}_1}$, we get

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right)\left[ {\left( {{{a}} - {{b}}} \right)\left( {{{a}} - {{c}}} \right) - \left( {{{c}} - {{b}}} \right)\left( {{{b}} - {{c}}} \right)} \right]$ 

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right)\left[ {{{{a}}^2} - {{ac}} - {{ab}} + {{bc}} - \left( {{{\;bc}} - {{{c}}^2} - {{{b}}^2} + {{bc}}} \right)} \right]$ 

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^2} - {{ac}} - {{ab}} + {{bc}} - {{\;bc}} + {{{c}}^2} + {{{b}}^2} - {{bc}}} \right)$ 

$ = {{xyz}}\left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{{a}}^2} + {{{b}}^2} + {{{c}}^2} - {{ab}} - {{bc}} - {{\;ac}}} \right)$ 

$ = {{xyz}}\left( {{{\;}}{{{a}}^3} + {{{b}}^3} + {{{c}}^3} - {{3abc}}} \right)$ 

Thus, L.H.S = R.H.S

Hence proved.

Objective Type Questions (M.C.Q)

Choose the correct answer from given four options in each of the exercise from 24 to 37.

24. If $\begin{vmatrix} {2{\mathbf{x}}} & 5 \\ 8 & {\mathbf{x}} \\ \end{vmatrix}=$ $\begin{vmatrix} 6 & { - 2} \\ 7 & {\;\;\;3} \\ \end{vmatrix}$, then value of ${\mathbf{x}}$ is 

(A) 3                                           

(B) $ \pm 3$

(C) $ \pm 6$                                          

(D)  6

Ans: The correct answer is option (C).

Here, we have $\begin{vmatrix} {{{2x}}} & {{5}} \\ {{8}} & {{x}} \\ \end{vmatrix}{{ = }}\begin{vmatrix} {{6}} & {{{ - 2}}} \\ {{7}} & {{{\;\;\;3}}} \\ \end{vmatrix}$

⇒ ${{2}}{{{x}}^{{2}}}{{ - 40 = 18 + 14}}$

⇒ ${{2}}{{{x}}^{{2}}}{{ - 40 = 32}}$

⇒ ${{2}}{{{x}}^{{2}}}{{ = 72}}$

⇒ ${{\;}}{{{x}}^{{2}}}{{ = 36}}$

⇒ ${{x = }}\sqrt {{{36}}} $

⇒ ${{x =  \pm \;6}}$

Hence, option (C) is the correct answer. 

25. The value of determinant $\begin{vmatrix} {{\mathbf{a}} - {\mathbf{b}}} & {{\mathbf{b}} + {\mathbf{c}}} & {\mathbf{a}} \\ {{\mathbf{b}} - {\mathbf{a}}} & {{\mathbf{c}} + {\mathbf{a}}} & {\mathbf{b}} \\ {{\mathbf{c}} - {\mathbf{a}}} & {{\mathbf{a}} + {\mathbf{b}}} & {\mathbf{c}} \\ \end{vmatrix}$

(A) ${{\mathbf{a}}^3} + {{\mathbf{b}}^3} + {{\mathbf{c}}^3}$                                      

(B) $3{\mathbf{bc}}$

(C) ${{\mathbf{a}}^3} + {{\mathbf{b}}^3} + {{\mathbf{c}}^3} - 3{\mathbf{abc}}$                          

(D) none of these

Ans: Option (D) is the correct answer.

Here, we have $\begin{vmatrix} {{{a}} - {{b}}} & {{{b}} + {{c}}} & {{a}} \\ {{{b}} - {{a}}} & {{{c}} + {{a}}} & {{b}} \\ {{{c}} - {{a}}} & {{{a}} + {{b}}} & {{c}} \\ \end{vmatrix}$

Applying ${{{C}}_3} \to {{{C}}_2} + {{{C}}_3}$

$= \begin{vmatrix} {{{a}} - {{b}}} & {{{b}} + {{c}}} & {{a}} \\ {{{b}} - {{a}}} & {{{c}} + {{a}}} & {{b}} \\ {{{c}} - {{a}}} & {{{a}} + {{b}}} & {{c}} \\ \end{vmatrix}$ 

$= \begin{vmatrix} {{{a}} - {{b}}} & {{{b}} + {{c}}} & {{{a}} + {{b}} + {{c}}} \\ {{{b}} - {{a}}} & {{{c}} + {{a}}} & {{{a}} + {{b}} + {{c}}} \\  {{{c}} - {{a}}} & {{{a}} + {{b}}} & {{{a}} + {{b}} + {{c}}} \\ \end{vmatrix}$ 

Taking common $\left( {{{a}} + {{b}} + {{c}}} \right)$ from ${{{C}}_3}$, we get

$ = \left( {{{a}} + {{b}} + {{c}}} \right)\begin{vmatrix} {{{a - b}}} & {{{b + c}}} & {{1}} \\ {{{b - a}}} & {{{c + a}}} & {{1}} \\ {{{c - a}}} & {{{a + b}}} & {{1}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ = \left( {{{a}} + {{b}} + {{c}}} \right)\begin{vmatrix} {{{a - b}}} & {{{b + c}}} & {{1}} \\ {{{ - 2}}\left( {{{a - b}}} \right)} & {{{a - b}}} & {{0}} \\ {{{c - 2a + b}}} & {{{a - c}}} & {{0}} \\ \end{vmatrix}$ 

Taking common $\left( {{{a}} - {{b}}} \right)$ from ${{{R}}_2}$

${{ = }}\left( {{{a + b + c}}} \right)\left( {{{a - b}}} \right)\begin{vmatrix} {{{a - b}}} & {{{b + c}}} & {{1}} \\ {{{ - 2}}} & {{1}} & {{0}} \\ {{{c - 2a + b}}} & {{{a - c}}} & {{0}} \\ \end{vmatrix}$ 

Expanding along ${{{C}}_3}$

${{ = }}\left( {{{a + b + c}}} \right)\left( {{{a - b}}} \right)\left( {{{ - 2a + 2c - c + 2a - b}}} \right)$ 

$ = \left( {{{a}} + {{b}} + {{c}}} \right)\left( {{{a}} - {{b}}} \right)\left( {{{c}} - {{b}}} \right)$ 

Hence, option (D) is correct.

26. The area of a triangle with the vertices $\left( { - 3,\;0} \right),\;\left( {3,\;0} \right)$ and $\left( {0,k} \right)$ is 9 sq. units. The value of ${\mathbf{k}}$ will be 

(A) 9                                                  

(B) 3 

(C) -9                                                 

(D) 6

Ans: The correct answer is option (B).

we know that, area of a triangle with vertices $\left( {{{\;}}{{{x}}_1},{{\;}}{{{y}}_1}} \right),{{\;}}\left( {{{\;}}{{{x}}_2},{{\;}}{{{y}}_2}} \right)$ and $\left( {{{\;}}{{{x}}_3},{{\;}}{{{y}}_3}} \right)$ is given by 

⇒ ${{\Delta \; = }}\dfrac{{{1}}}{{{2}}}{{\;}}\begin{vmatrix} {{{{x}}_{{1}}}} & {{{{y}}_{{1}}}} & {{1}} \\ {{{{x}}_{{2}}}} & {{{{y}}_{{2}}}} & {{1}} \\ {{{{x}}_{{3}}}} & {{{{y}}_{{3}}}} & {{1}} \\ \end{vmatrix}$

⇒ ${{9\; = }}\dfrac{{{1}}}{{{2}}}{{\;}}\begin{vmatrix} {{{ - 3}}} & {{0}} & {{1}} \\ {{3}} & {{0}} & {{1}} \\ {{0}} & {{k}} & {{1}} \\ \end{vmatrix}$

⇒ ${{18\; = \;}}\begin{vmatrix} {{{ - 3}}} & {{0}} & {{1}} \\ {{3}} & {{0}} & {{1}} \\ {{0}} & {{k}} & {{1}}  \\ \end{vmatrix}$

Expanding along ${{{C}}_2}$

⇒ ${{18\; = \; - k}}\left( {{{ - 3 - 3}}} \right)$

⇒  ${{18\; = \;6k}}$

⇒ $\dfrac{{{{18}}}}{{{6}}}{{\; = \;k}}$

⇒ ${{3\; = \;k}}$

Thus, the value of ${{k}}$ is 3.

Hence, option (B) is correct.

27. The determinant $\begin{vmatrix} {{{\mathbf{b}}^2} - {\mathbf{ab}}} & {{\mathbf{b}} - {\mathbf{c}}} & {{\mathbf{bc}} - {\mathbf{ac}}} \\ {{\mathbf{ab}} - {{\mathbf{a}}^2}} & {{\mathbf{a}} - {\mathbf{b}}} & {{{\mathbf{b}}^2} - {\mathbf{ab}}} \\ {{\mathbf{bc}} - {\mathbf{ac}}} & {{\mathbf{c}} - {\mathbf{a}}} & {{\mathbf{ab}} - {{\mathbf{a}}^2}} \\ \end{vmatrix}$ equals

(A) ${\mathbf{abc}}\left( {{\mathbf{b}} - {\mathbf{c}}} \right)\left( {{\mathbf{c}} - {\mathbf{a}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right)$                 

(B) $\left( {{\mathbf{b}} - {\mathbf{c}}} \right)\left( {{\mathbf{c}} - {\mathbf{a}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right)$    

(C)  $\left( {{\mathbf{a}} + {\mathbf{b}} + {\mathbf{c}}} \right)\left( {{\mathbf{b}} - {\mathbf{c}}} \right)\left( {{\mathbf{c}} - {\mathbf{a}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right)$     

(D) None of these   

Ans: The correct answer is option (D).

Here, we have  $\begin{vmatrix} {{{{b}}^2} - {{ab}}} & {{{b}} - {{c}}} & {{{bc}} - {{ac}}} \\ {{{ab}} - {{{a}}^2}} & {{{a}} - {{b}}} & {{{{b}}^2} - {{ab}}} \\ {{{bc}} - {{ac}}} & {{{c}} - {{a}}} & {{{ab}} - {{{a}}^2}}  \\ \end{vmatrix}$        

$= \begin{vmatrix} {{{b}}\left( {{{b}} - {{a}}} \right)} & {{{b}} - {{c}}} & {{{c}}\left( {{{b}} - {{a}}} \right)} \\ {{{a}}\left( {{{b}} - {{a}}} \right)} & {{{a}} - {{b}}} & {{{b}}\left( {{{b}} - {{a}}} \right)} \\ {{{c}}\left( {{{b}} - {{a}}} \right)} & {{{c}} - {{a}}} & {{{a}}\left( {{{b}} - {{a}}} \right)} \\ \end{vmatrix}$     

Taking common $\left( {{{b}} - {{a}}} \right)$ from ${{{C}}_1}$ and ${{{C}}_3}$

$ = {\left( {{{b}} - {{a}}} \right)^2}\begin{vmatrix} {{b}} & {{{b}} - {{c}}} & {{c}} \\ {{a}} & {{{a}} - {{b}}} & {{b}} \\ {{c}} & {{{c}} - {{a}}} & {{a}} \\ \end{vmatrix}$ 

Applying ${{{C}}_2} \to {{{C}}_2} + {{{C}}_3}$

$ = {\left( {{{b}} - {{a}}} \right)^2}\begin{vmatrix} {{b}} & {{b}} & {{c}} \\  {{a}} & {{a}} & {{b}} \\  {{c}} & {{c}} & {{a}} \\ \end{vmatrix} = {{0}}$ 

Here, two columns of determinant are identical. Thus, the value of this determinant is 0.

Hence, option (D) is the correct answer.

28. The number of distinct real roots of $\begin{vmatrix} {\sin {\mathbf{x}}} & {\cos {\mathbf{x}}} & {\cos {\mathbf{x}}} \\ {\cos {\mathbf{x}}} & {\sin {\mathbf{x}}} & {\cos {\mathbf{x}}} \\ {\cos {\mathbf{x}}} & {\cos {\mathbf{x}}} & {\sin {\mathbf{x}}} \\ \end{vmatrix} = 0$ in the interval $ - \dfrac{{\mathbf{\pi }}}{4} \leqslant {\mathbf{x}} \leqslant \dfrac{{\mathbf{\pi }}}{4}$ is 

(A) 0                                                  

(B) 2

(C) 1                                                   

(D) 3

Ans: The correct answer is option (C). 

Here, we have $\begin{vmatrix} {\sin {{x}}} & {\cos {{x}}} & {\cos {{x}}} \\  {\cos {{x}}} & {\sin {{x}}} & {\cos {{x}}} \\ {\cos {{x}}} & {\cos {{x}}} & {\sin {{x}}} \\ \end{vmatrix} = {{0}}$

Applying $[{{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}]$, we get

⇒ $\begin{vmatrix} {{{2cosx + sinx}}} & {{{cosx}}} & {{{cosx}}} \\ {{{2cosx + sinx}}} & {{{sinx}}} & {{{cosx}}} \\ {{{2cosx + sinx}}} & {{{cosx}}} & {{{sinx}}} \\ \end{vmatrix}{{ = 0}}$

Taking common ($2\cos {{x}} + \sin {{x}})$ from ${{{C}}_1}$

⇒ $\left( {{{2cosx + sinx}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{cosx}}} & {{{cosx}}} \\ {{1}} & {{{sinx}}} & {{{cosx}}} \\ {{1}} & {{{cosx}}} & {{{sinx}}} \\ \end{vmatrix}{{ = 0}}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

⇒ $\left( {{{2cosx + sinx}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{cosx}}} & {{{cosx}}} \\ {{0}} & {{{sinx - cosx}}} & {{0}} \\ {{0}} & {{0}} & {{{sinx - cosx}}} \\ \end{vmatrix}{{ = 0}}$ 

Expanding along ${{{C}}_1}$

⇒ $\left( {{{2}}\cos {{x}} + \sin {{x}}} \right){{\;}}{\left( {\sin {{x}} - \cos {{x}}} \right)^2} = {{0}}$ 

⇒ $\left( {{{2cosx + sinx}}} \right){{ = 0\;or\;}}{\left( {{{sinx - cosx}}} \right)^{{2}}}{{ = 0}}$ 

⇒ ${{sinx =  - 2cosx\;or\;sinx - cosx = 0}}$ 

⇒ ${{tanx =  - 2\;or\;sinx = cosx}}$

⇒ ${{tanx =  - 2\;or\;tanx = 1}}$

Also, we have $ - \dfrac{{{\pi }}}{{{4}}} \leqslant {{x}} \leqslant \dfrac{{{\pi }}}{{{4}}}$

⇒ $ - \tan \dfrac{{{\pi }}}{{{4}}} \leqslant \tan {{x}} \leqslant \tan \dfrac{{{\pi }}}{{{4}}}$

⇒ $ - {{1}} \leqslant \tan {{x}} \leqslant {{1}}$

Thus,  $\tan {{x}}$ =$ - {{2}}$  is not possible.

Therefore, $\tan {{x}}$ =${{1}}$  

⇒ ${{x}} = \dfrac{{{\pi }}}{{{4}}}$

So, only one distinct real root exists.

Hence, option (C) is the correct answer.

29. If ${\mathbf{A}},\;{\mathbf{B}}$ and ${\mathbf{C}}$ are angles of a triangle, then the determinant $\begin{vmatrix} { - 1} & {cosC} & {cosB} \\ {cosC} & { - 1} & {cosA} \\ {cosB} & {cosA} & { - 1} \\ \end{vmatrix}$ is equal to

(A) 0                                                         

(B) -1

(C) 1 

(D) None of these 

Ans: The correct answer is option (A). 

Here, we have $\begin{vmatrix} {{{ - 1}}} & {{{cosC}}} & {{{cosB}}} \\ {{{cosC}}} & {{{ - 1}}} & {{{cosA}}} \\ {{{cosB}}} & {{{cosA}}} & {{{ - 1}}}  \\ \end{vmatrix}$

Applying $[{{{C}}_1} \to {{a}}{{{C}}_1} + {{b}}{{{C}}_2} + {{c}}{{{C}}_3}]$, we get

${{ = }}\begin{vmatrix} {{{ - a + b}}{{.cosC + c}}{{.cosB}}} & {{{cosC}}} & {{{cosB}}} \\  {{{acosC - b + c}}{{.cosA}}} & {{{ - 1}}} & {{{cosA}}} \\  {{{a}}{{.cosB + b}}{{.cosA - c}}} & {{{cosA}}} & {{{ - 1}}} \\ \end{vmatrix}$ 

Also, by projection rule in a triangle, we know that

⇒ ${{a}} = {{b}}.\cos {{C}} + {{c}}.\cos {{B}}$

⇒ ${{b}} = {{a}}\cos {{C}} + {{c}}.\cos {{A}}$

⇒ ${{c}} = {{a}}.\cos {{B}} + {{b}}.\cos {{A}}$

Using above equations in column first, we get

${{ = }}\begin{vmatrix} {{{ - a + a}}} & {{{cosC}}} & {{{cosB}}} \\ {{{b - b}}} & {{{ - 1}}} & {{{cosA}}} \\ {{{c - c}}} & {{{cosA}}} & {{{ - 1}}} \\ \end{vmatrix}$ 

${{ = }}\begin{vmatrix} {{0}} & {{{cosC}}} & {{{cosB}}} \\ {{0}} & {{{ - 1}}} & {{{cosA}}} \\ {{0}} & {{{cosA}}} & {{{ - 1}}} \\ \end{vmatrix}  = 0$ 

Since, a determinant having all elements of any column or row gives the value of determinant is zero.

Hence, option (A) is the correct answer.

30. Let ${\mathbf{f}}\left( {\mathbf{t}} \right) =\begin{vmatrix} {\cos {\mathbf{t}}} & {\mathbf{t}} & 1 \\ {2\sin {\mathbf{t}}} & {\mathbf{t}} & {2{\mathbf{t}}} \\ {\sin {\mathbf{t}}} & {\mathbf{t}} & {\mathbf{t}} \\ \end{vmatrix}$, then $\mathop{\lim }\limits_{{\mathbf{t}} \to 0} \dfrac{{{\mathbf{f}}\left( {\mathbf{t}} \right)}}{{{{\mathbf{t}}^2}}}$ is equal to 

(A) 0

(B) -1

(C) 2

(D) 3

Ans: The correct answer is option (A).

Here, we have ${{f}}\left( {{t}} \right){{ = }}\begin{vmatrix} {{{cost}}} & {{t}} & {{1}} \\  {{{2sint}}} & {{t}} & {{{2t}}} \\ {{{sint}}} & {{t}} & {{t}} \\ \end{vmatrix}$

Expanding along ${{{R}}_1}$

$ \Rightarrow {{f}}\left( {{t}} \right){{ = cost\;}}\left( {{{{t}}^{{2}}}{{ - 2}}{{{t}}^{{2}}}} \right){{ - t}}\left( {{{\;2tsint - 2tsint}}} \right){{ + 1}}\left( {{{2tsint - tsint\;}}} \right)$ 

$ \Rightarrow {{f}}\left( {{t}} \right) =  - {{{t}}^2}\cos {{t}} - {{0}} + {{t}}\sin {{t}}$ 

$ \Rightarrow {{f}}\left( {{t}} \right) =  - {{{t}}^2}\cos {{t}} + {{t}}\sin {{t}}$ 

Now, $\mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} = \mathop {{{lim}}}\limits_{{{t}} \to 0} \left( {\dfrac{{ - {{{t}}^2}\cos {{t}} + {{t}}\sin {{t}}}}{{{{{t}}^2}}}} \right)$

$ \Rightarrow \mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} = \mathop {{{lim}}}\limits_{{{t}} \to 0} \left( { - \cos {{t}} + \dfrac{{\sin {{t}}}}{{{t}}}} \right)$ 

$ \Rightarrow \mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} =  - \mathop {{{lim}}}\limits_{{{t}} \to 0} \cos {{t}} + \mathop {{{lim}}}\limits_{{{t}} \to 0} \left( {\dfrac{{\sin {{t}}}}{{{t}}}} \right)$ 

$ \Rightarrow \mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} =  - \mathop {{{lim}}}\limits_{{{t}} \to 0} \,\cos {{0}} + {{1}}$                         $\left[ {\mathop {{{lim}}}\limits_{{{t}} \to 0} \left( {\dfrac{{\sin {{t}}}}{{{t}}}} \right) = {{1}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} = {{ - 1 + 1}}$ 

$ \Rightarrow \mathop {\lim }\limits_{{{t}} \to 0} \dfrac{{{{f}}\left( {{t}} \right)}}{{{{{t}}^2}}} = {{0}}$ 

Hence, option (A) is the correct answer. 

31. The maximum value of $\Delta \; = \;\begin{vmatrix} 1 & 1 & 1 \\ 1 & {1 + sin\theta } & 1 \\ {1 + cos\theta } & 1 & 1 \\ \end{vmatrix}$ is (${\mathbf{\theta }} \in {\mathbf{R}}).$

(A) $\dfrac{1}{2}\;$                                                     

(B) $\dfrac{{\sqrt 3 }}{2}$

(C) $\sqrt 2 $                                                    

(D) $\dfrac{{2\sqrt 3 }}{4}$

Ans: Here, we have ${{\Delta \; = \;}}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{1}} & {{{1 + sin\theta }}} & {{1}} \\  {{{1 + cos\theta }}} & {{1}} & {{1}} \\ \end{vmatrix}$

Applying $\left[{{C}_{1}} \to {C}_{1}-{C}_{3} \right] \text{ and } \left[{{C}_{2}} \to {C}_{2}-{C}_{3}\right]$, we get

⇒ ${{\Delta \; = \;}}\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{0}} & {{{sin\theta }}} & {{1}} \\ {{{cos\theta }}} & {{0}} & {{1}} \\ \end{vmatrix}$

Now, expanding along ${{{R}}_1}$

⇒ ${{\Delta \; = \;1}}\left( {{{0 - sin\theta }}{{.cos\theta }}} \right)$

⇒ $\Delta {{\;}} = {{\;}} - \sin {{\theta }}.\cos {{\theta }}$

⇒ ${{\Delta \; = \; - }}\dfrac{{{2}}}{{{2}}}{{sin\theta }}{{.cos\theta }}$

⇒ ${{\Delta \; = \; - }}\dfrac{{{1}}}{{{2}}}{{sin2\theta }}$

Also, we know that, $ - {{1}} \leqslant \sin {{2\theta }} \leqslant {{1}}$

⇒ $\dfrac{{{1}}}{{{2}}} \geqslant  - \dfrac{{{1}}}{{{2}}}\sin {{2\theta }} \geqslant  - \dfrac{{{1}}}{{{2}}}$

⇒ $ - \dfrac{{{1}}}{{{2}}} \leqslant \Delta {{\;}} \leqslant \dfrac{{{1}}}{{{2}}}$

Thus, maximum value of given determinant is $\dfrac{{{1}}}{{{2}}}$

Hence, option (A) is the correct answer.

32. If ${\mathbf{f}}\left( {\mathbf{x}} \right) = \begin{vmatrix} 0 & {{\mathbf{x}} - {\mathbf{a}}} & {{\mathbf{x}} - {\mathbf{b}}} \\ {{\mathbf{x}} + {\mathbf{a}}} & 0 & {{\mathbf{x}} - {\mathbf{c}}} \\ {{\mathbf{x}} + {\mathbf{b}}} & {{\mathbf{x}} + {\mathbf{c}}} & 0 \\ \end{vmatrix}$, then

(A) $f\left( a \right) = 0$                                     

(B) $f\left( b \right) = 0$   

(C) $f\left( 0 \right) = 0$  

(D) $f\left( 1 \right) = 0$   

Ans: The correct answer is option (C).

Here, we have ${{f}}\left( {{x}} \right) = \begin{vmatrix} 0 & {{{x}} - {{a}}} & {{{x}} - {{b}}} \\ {{{x}} + {{a}}} & 0 & {{{x}} - {{c}}} \\ {{{x}} + {{b}}} & {{{x}} + {{c}}} & 0 \\ \end{vmatrix}$

⇒ ${{f}}\left( {{a}} \right) $ = $\begin{vmatrix} 0 & 0 & {{{a}} - {{b}}} \\ {2{{a}}} & 0 & {{{a}} - {{c}}} \\ {{{a}} + {{b}}} & {{{a}} + {{c}}} & 0 \\ \end{vmatrix} = 2{{a}}\left( {{{a}} - {{b}}} \right)\left( {{{a}} + {{c}}} \right) \ne 0$

⇒ ${{f}}\left( {{b}} \right) =\begin{vmatrix} 0 & {{{b}} - {{a}}} & 0 \\ {{{b}} + {{a}}} & 0 & {{{b}} - {{c}}} \\ {2{{b}}} & {{{b}} + {{c}}} & 0 \\ \end{vmatrix} =  - 2{{b}}\left( {{{b}} - {{a}}} \right)\left( {{{b}} - {{c}}} \right) \ne 0$

⇒ ${{f}}\left( 0 \right) = \begin{vmatrix} 0 & { - {{a}}} & { - {{b}}} \\ {{a}} & 0 & { - {{c}}} \\ {{b}} & {{c}} & 0 \\ \end{vmatrix} = {{abc}} - {{abc}} = 0$

⇒ ${{f}}\left( 1 \right) =\begin{vmatrix} 0 & {1 - {{a}}} & {1 - {{b}}} \\  {1 + {{a}}} & 0 & {1 - {{c}}} \\  {1 + {{b}}} & {1 + {{c}}} & 0 \\ \end{vmatrix} \ne 0$

Hence, option (C) is the correct answer.

33. If $A = \begin{bmatrix} 2 & \lambda  & { - 3} \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$, then ${{\mathbf{A}}^{ - 1}}$ exists if 

(A) $\lambda  = 2$                                        

(B) $\lambda  \ne 2$          

(C) $\lambda  \ne  - 2$  

(D) None of these

Ans: The correct answer is option (D). 

Here, we have ${{A = }}\begin{bmatrix}  {{2}} & {{\lambda }} & {{{ - 3}}} \\ {{0}} & {{2}} & {{5}} \\ {{1}} & {{1}} & {{3}} \end{bmatrix}$            

Expanding along ${{{R}}_1}$  

⇒ $\left| {{A}} \right|{{ = 2}}\left( {{{6 - 5}}} \right){{ - \lambda }}\left( {{{0 - 5}}} \right){{ - 3}}\left( {{{0 - 2}}} \right)$   

⇒ $\left| {{A}} \right|{{ = 2 + 5\lambda  + 6}}$   

We know that, ${{{A}}^{ - 1}}$ exists, if A is non- singular matrix i.e., $\left| {{A}} \right| \ne {{0}}$

Therefore, 

⇒ $\left| {{A}} \right| = {{2 + 5\lambda  + 6}} \ne {{0}}$   

⇒ ${{5\lambda  + 8}} \ne {{0}}$  

⇒ ${{5\lambda }} \ne  - {{8}}$ 

⇒ ${{\lambda }} \ne  - \dfrac{{{8}}}{{{5}}}$ 

Hence, option (D) is the correct answer. 

34. If A and B are invertible matrices, then which of the following is not correct?

(A) ${\mathbf{adj}}\;{\mathbf{A}} = \left| {\mathbf{A}} \right|.{{\mathbf{A}}^{ - 1}}$                             

(B) ${\mathbf{det}}{\left( {\mathbf{A}} \right)^{ - 1}} = {\left[ {{\mathbf{det}}\left( {\mathbf{A}} \right)} \right]^{ - 1}}$

(C) ${\left( {{\mathbf{AB}}} \right)^{ - 1}} = {{\mathbf{B}}^{ - 1}}{{\mathbf{A}}^{ - 1}}$                           

(D) ${\left( {{\mathbf{A}} + {\mathbf{B}}} \right)^{ - 1}} = {{\mathbf{B}}^{ - 1}} + {{\mathbf{A}}^{ - 1}}$                           

Ans: The correct answer is option (D).

Since, A and B are invertible matrices. So, we can say that 

${\left( {{{AB}}} \right)^{ - 1}} = {{{B}}^{ - 1}}{{{A}}^{ - 1}}$              (it is correct)

Also, ${{{A}}^{ - 1}} = \dfrac{{{1}}}{{\left| {{A}} \right|}}\left( {{{adj\;A}}} \right)$    

⇒ ${{adj\;A}} = \left| {{A}} \right|.{{{A}}^{ - 1}}$            (it is correct)

And ${{A}}{{{A}}^{ - 1}} = {{I}}$ 

⇒ $\left| {{{A}}{{{A}}^{ - 1}}} \right| = \left| {{I}} \right|$

⇒ $\left| {{A}} \right|.\left| {{{{A}}^{ - 1}}} \right| = {{1}}$

⇒ $\left| {{{{A}}^{ - 1}}} \right| = \dfrac{{{1}}}{{\left| {{A}} \right|}}$

$ \Rightarrow {{det}}{\left( {{A}} \right)^{ - 1}} = {\left[ {{{det}}\left( {{A}} \right)} \right]^{ - 1}}$    (it is correct)

Now, ${\left( {{{A}} + {{B}}} \right)^{ - 1}} = {{{B}}^{ - 1}} + {{{A}}^{ - 1}}$

L.H.S = ${\left( {{{A}} + {{B}}} \right)^{ - 1}}$

= $\dfrac{{{1}}}{{\left| {{{A}} + {{B}}} \right|}}\left( {{{adj\;A}} + {{B}}} \right)$

R.H.S = ${{{B}}^{ - 1}} + {{{A}}^{ - 1}}$

= $\dfrac{{{1}}}{{\left| {{B}} \right|}}\left( {{{adj\;B}}} \right) + \dfrac{{{1}}}{{\left| {{A}} \right|}}\left( {{{adj\;A}}} \right)$

Thus, L.H.S $ \ne $ R.H.S (it is incorrect)

Hence, option (D) is the correct answer.

35. If ${\mathbf{x}},\;\;{\mathbf{y}},\;\;{\mathbf{z}}$ are all different from zero and $\begin{vmatrix} {1 + x} & 1 & 1 \\ 1 & {1 + y} & 1 \\ 1 & 1 & {1 + z} \\ \end{vmatrix} = 0$, then value ${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}\;$ is 

(A) ${\mathbf{xyz}}$

(B) ${{\mathbf{x}}^{ - 1}}.{{\mathbf{y}}^{ - 1}}.{{\mathbf{z}}^{ - 1}}$

(C) $ - {\mathbf{x}} - {\mathbf{y}} - {\mathbf{z}}$               

(D)  $ - 1$

Ans: The correct answer is option (D).

Here, we have $\begin{vmatrix} {{{1 + x}}} & {{1}} & {{1}} \\ {{1}} & {{{1 + y}}} & {{1}} \\ {{1}} & {{1}} & {{{1 + z}}} \\ \end{vmatrix}{{ = 0}}$

Applying ${{{C}}_1} \to {{{C}}_1} - {{{C}}_2}$ and ${{{C}}_2} \to {{{C}}_2} - {{{C}}_3}$

⇒ $\begin{vmatrix} {{x}} & {{0}} & {{1}} \\ {{{ - y}}} & {{y}} & {{1}} \\  {{0}} & {{{ - z}}} & {{{1 + z}}}  \\ \end{vmatrix}{{ = 0}}$

Expanding along ${{{R}}_1}$

⇒ ${{x}}\left( {{{y}} + {{yz}} + {{z}}} \right) + {{yz}} = {{0}}$

⇒ ${{xy}} + {{xyz}} + {{xz}} + {{yz}} = {{0}}$

⇒ ${{xy}} + {{xz}} + {{yz}} =  - {{xyz}}$

⇒ $\dfrac{{{{xy}} + {{xz}} + {{yz}}}}{{{{xyz}}}} =  - {{1}}$

⇒ $\dfrac{{{1}}}{{{z}}}{{ + }}\dfrac{{{1}}}{{{y}}}{{ + }}\dfrac{{{1}}}{{{x}}}{{ =  - 1}}$

⇒ $\dfrac{{{1}}}{{{x}}}{{ + }}\dfrac{{{1}}}{{{y}}}{{ + }}\dfrac{{{1}}}{{{z}}}{{ =  - 1}}$

⇒ ${{{x}}^{ - 1}} + {{{y}}^{ - 1}} + {{{z}}^{ - 1}} =  - {{1}}$

Hence, option (D) is the correct answer.

36. The value of the determinant $\begin{vmatrix} {\mathbf{x}} & {{\mathbf{x}} + {\mathbf{y}}} & {{\mathbf{x}} + 2{\mathbf{y}}} \\ {{\mathbf{x}} + 2{\mathbf{y}}} & {\mathbf{x}} & {{\mathbf{x}} + {\mathbf{y}}} \\ {{\mathbf{x}} + {\mathbf{y}}} & {{\mathbf{x}} + 2{\mathbf{y}}} & {\mathbf{x}}  \\ \end{vmatrix}$ is 

(A) $9{x^2}\left( {x + y} \right)$ 

(B) $9{y^2}\left( {x + y} \right)$           

(C) $3{y^2}\left( {x + y} \right)$

 (D) $7{x^2}\left( {x + y} \right)$      

Ans: The correct answer is option (B). 

Here, we have  $\begin{vmatrix} {{x}} & {{{x + y}}} & {{{x + 2y}}} \\ {{{x + 2y}}} & {{x}} & {{{x + y}}} \\ {{{x + y}}} & {{{x + 2y}}} & {{x}} \\ \end{vmatrix}$     

Applying   ${{{C}}_1} \to {{{C}}_1} + {{{C}}_2} + {{{C}}_3}$ , we get

=$\begin{vmatrix} {{3}}\left({{{x+y}}}\right) & {{{x + y}}} & {{{x + 2y}}} \\ {{3}}\left({{{x + y}}}\right) & {{x}} & {{{x + y}}} \\ {{3}}\left({{{x+y}}}\right) & {{{x + 2y}}} & {{x}} \\ \end{vmatrix}$        

Taking common  ${{3}}\left( {{{x}} + {{y}}} \right)$ from ${{{C}}_1}$     

$= 3\left( {{{x + y}}} \right) \begin{vmatrix}  {{1}} & {{{x + y}}} & {{{x + 2y}}} \\ {{1}} & {{x}} & {{{x + y}}} \\ {{1}} & {{{x + 2y}}} & {{x}} \\ \end{vmatrix}$ 

Applying ${{{R}}_2} \to {{{R}}_2} - {{{R}}_1}$ and  ${{{R}}_3} \to {{{R}}_3} - {{{R}}_1}$         

$=3\left({{{x+y}}}\right)\begin{vmatrix} {{1}} & {{{x + y}}} & {{{x + 2y}}} \\ {{0}} & {{{ - y}}} & {{{ - y}}} \\ {{0}} & {{y}} & {{{ -2y}}} \\ \end{vmatrix}$ 

Expanding along ${{{C}}_1}$

${{ = 3}}\left( {{{x + y}}} \right){{\;}}\left( {{{2}}{{{y}}^{{2}}}{{ + }}{{{y}}^{{2}}}} \right)$ 

${{ = 3}}\left( {{{x + y}}} \right){{ \times 3}}{{{y}}^{{2}}}$ 

$ = {{9}}{{{y}}^2}\left( {{{x}} + {{y}}} \right)$ 

Hence, option (B) is the correct answer.

37. There are two values of ${\mathbf{a}}$ which makes determinant, $\Delta \; = \;\begin{vmatrix} 1 & { - 2} & 5 \\  2 & a & { - 1} \\ 0 & 4 & {2a} \\ \end{vmatrix} = 86,$ then sum of these number is 

(A) 4                                                                

(B) 5

(C) – 4                                                                

(D) 9

Ans: The correct answer is option (C).

Here, we have ${{\Delta \; = \;}}\begin{vmatrix} {{1}} & {{{ - 2}}} & {{5}} \\ {{2}} & {{a}} & {{{ - 1}}} \\ {{0}} & {{4}} & {{{2a}}}  \\ \end{vmatrix}{{ = 86}}$

⇒ $\begin{vmatrix} {{1}} & {{{ - 2}}} & {{5}} \\ {{2}} & {{a}} & {{{ - 1}}} \\ {{0}} & {{4}} & {{{2a}}} \\ \end{vmatrix}{{ = 86}}$

Expanding along ${{{R}}_1}$

⇒ ${{1}}\left( {{{2}}{{{a}}^{{2}}}{{ + 4}}} \right){{ + 2}}\left( {{{4a - 0}}} \right){{ + 5}}\left( {{{8 - 0}}} \right){{ = 86}}$

⇒ ${{2}}{{{a}}^{{2}}}{{ + 4 + 8a + 40 = 86}}$

⇒ ${{2}}{{{a}}^{{2}}}{{ + 8a + 44 = 86}}$

⇒ ${{2}}{{{a}}^{{2}}}{{ + 8a + 44 - 86 = 0}}$

⇒ ${{2}}{{{a}}^{{2}}}{{ + 8a - 42 = 0}}$

⇒ ${{{a}}^{{2}}}{{ + 4a - 21 = 0}}$

⇒ ${{{a}}^{{2}}}{{ + 7a - 3a - 21 = 0}}$

⇒ ${{a}}\left( {{{a + 7}}} \right){{ - 3}}\left( {{{a + 7}}} \right){{ = 0}}$

⇒ $\left( {{{a + 7}}} \right)\left( {{{a - 3}}} \right){{ = 0}}$

⇒ $\left( {{{a + 7}}} \right){{ = 0\;or\;}}\left( {{{a - 3}}} \right){{ = 0}}$

⇒ ${{a}} =  - {{7\;or\;a}} = {{3}}$

Here, values of ${{a}}$ are ${{ - 7}}$ and ${{3}}{{.}}$

Therefore, sum of these values = ${{ - 7 + 3 =  - 4}}$

Hence, option (C) is the correct answer.                                                   

Fill in the blanks 

38. If A is a matrix of order, $3 \times 3,$ then $\left| {3{\mathbf{A}}} \right| = $ …………………

Ans: Here, A is a matrix of order, ${{3 \times 3}}$.

And we know that, $\left| {{{\lambda \;A}}} \right| = {{{\lambda }}^{{n}}}\left| {{A}} \right|$ , here n is the order of the matrix and ${{\lambda }}$ is a constant.

Therefore, 

⇒ $\left| {{{3A}}} \right|{{ = \;}}{{{3}}^{{3}}}\left| {{A}} \right|{{ = 27}}\left| {{A}} \right|$

 Hence, If A is a matrix of order, ${{3 \times 3,}}$ then $\left| {{{3A}}} \right|{{ = 27}}\left| {{A}} \right|$.

39. If A is invertible matrix of order $3 \times 3$, then $\left| {{{\mathbf{A}}^{ - 1}}} \right| = $…………….

Ans: Here, A is invertible matrix of order ${{3 \times 3}}$

Therefore, 

$ \Rightarrow {{A}}{{{A}}^{ - 1}} = {{I}}$ 

⇒ $\left| {{{A}}{{{A}}^{ - 1}}} \right| = \left| {{I}} \right|$

⇒ $\left| {{A}} \right|{{.}}\left| {{{{A}}^{{{ - 1}}}}} \right|{{ = 1}}$

⇒ $\left| {{{{A}}^{{{ - 1}}}}} \right|{{ = }}\dfrac{{{1}}}{{\left| {{A}} \right|}}$

Hence, If A is invertible matrix of order ${{3 \times 3}}$, then $\left| {{{{A}}^{ - 1}}} \right| = \dfrac{{{1}}}{{\left| {{A}} \right|}}$ . 

40. If ${\mathbf{x}},\;{\mathbf{y}},\;{\mathbf{z}}\; \in {\mathbf{R}},$ then the value of determinant $\begin{vmatrix} {{{\left( {{2^x} + {2^{ - x}}} \right)}^2}} & {{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} & 1 \\ {{{\left( {{3^x} + {3^{ - x}}} \right)}^2}} & {{{\left( {{3^x} - {3^{ - x}}} \right)}^2}} & 1 \\ {{{\left( {{4^x} + {4^{ - x}}} \right)}^2}} & {{{\left( {{4^x} - {4^{ - x}}} \right)}^2}} & 1 \\ \end{vmatrix}$ is equal to ………. .

Ans: Here, we have $\begin{vmatrix} {{{\left( {{{{2}}^{{x}}}{{ + }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{3}}^{{x}}}{{ + }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{4}}^{{x}}}{{ + }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ \end{vmatrix}$, where ${{x}},{{\;y}},{{\;z\;}} \in {{R}},$

Applying ${{{C}}_1} \to {{{C}}_1} - {{{C}}_2}$,we get

$= \begin{vmatrix} {{{\left( {{{{2}}^{{x}}}{{ + }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}{{ - }}{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{3}}^{{x}}}{{ + }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}{{ - }}{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{4}}^{{x}}}{{ + }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}{{ - }}{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ \end{vmatrix}$

$= \begin{vmatrix} {{{4}}{{.}}{{{2}}^{{x}}}{{.}}{{{2}}^{{{ - x}}}}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{4}}{{.}}{{{3}}^{{x}}}{{.}}{{{3}}^{{{ - x}}}}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{4}}{{.}}{{{4}}^{{x}}}{{.}}{{{4}}^{{{ - x}}}}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ \end{vmatrix}$               [${\left( {{{a}} + {{b}}} \right)^2} - {\left( {{{a}} - {{b}}} \right)^2} = {{4ab}}]$

$= \begin{vmatrix}  {{4}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{4}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{4}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ \end{vmatrix}$

Taking common 4 from ${{{C}}_1}$

$= \begin{vmatrix} {{1}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{1}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{1}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ \end{vmatrix}{{ = 0}}$

Since, a determinant having all elements of any column or row gives the value of determinant is zero.

Hence, then the value of determinant $\begin{vmatrix} {{{\left( {{{{2}}^{{x}}}{{ + }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{2}}^{{x}}}{{ - }}{{{2}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{3}}^{{x}}}{{ + }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{3}}^{{x}}}{{ - }}{{{3}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}} \\ {{{\left( {{{{4}}^{{x}}}{{ + }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{{\left( {{{{4}}^{{x}}}{{ - }}{{{4}}^{{{ - x}}}}} \right)}^{{2}}}} & {{1}}  \\ \end{vmatrix}$ is equal to 0.

41. If ${\mathbf{cos}}\;2{\mathbf{\theta }} = 0$, then ${\begin{vmatrix} 0 & {{\mathbf{cos}}\;{\mathbf{\theta }}} & {{\mathbf{sin}}\;{\mathbf{\theta }}} \\ {{\mathbf{cos}}\;{\mathbf{\theta }}} & {{\mathbf{sin}}\;{\mathbf{\theta }}} & 0 \\ {{\mathbf{sin}}\;{\mathbf{\theta }}} & 0 & {{\mathbf{cos}}\;{\mathbf{\theta }}} \\ \end{vmatrix}^2} = $ …………….

Ans: Here, we have ${{cos\;2\theta }} = {{0}}$

⇒ ${{cos\;2\theta }} = \cos {{90}}^\circ $

⇒ ${{\;2\theta }} = {{90}}^\circ $

⇒ ${{\;\theta }} = {{45}}^\circ $…………….. (i)

Also, we have  ${\begin{vmatrix} {{0}} & {{{cos\;\theta }}} & {{{sin\;\theta }}} \\ {{{cos\;\theta }}} & {{{sin\;\theta }}} & {{0}} \\  {{{sin\;\theta }}} & {{0}} & {{{cos\;\theta }}} \\ \end{vmatrix}^2}$

Expanding along ${{{R}}_1}$

$ = $ ${\left[ {{{0 - cos\theta }}\left( {{{co}}{{{s}}^{{2}}}{{\theta  - 0}}} \right){{ + sin\theta \;}}\left( {{{0 - si}}{{{n}}^{{2}}}{{\theta }}} \right)} \right]^2}$

$= {\left[ { - {{co}}{{{s}}^3}{{\theta }} - {{si}}{{{n}}^3}{{\theta }}} \right]^2}$

$= {\left[ {{{co}}{{{s}}^3}{{\theta }} + {{si}}{{{n}}^3}{{\theta }}} \right]^2}$

$= {\left[ {{{\left( {\dfrac{{{1}}}{{\sqrt {{2}} }}} \right)}^{{3}}}{{ + }}{{\left( {\dfrac{{{1}}}{{\sqrt {{2}} }}} \right)}^{{3}}}} \right]^2}$

$= {\left[ {\dfrac{{{1}}}{{{{2}}\sqrt {{2}} }}{{ + }}\dfrac{{{1}}}{{{{2}}\sqrt {{2}} }}} \right]^2}$

$= {\left[ {\dfrac{{{1}}}{{\sqrt {{2}} }}} \right]^2}$

$= \dfrac{{{1}}}{{{2}}}$

Hence, If ${{cos\;2\theta  = 0}}$, then ${\begin{vmatrix} {{0}} & {{{cos\;\theta }}} & {{{sin\;\theta }}} \\ {{{cos\;\theta }}} & {{{sin\;\theta }}} & {{0}} \\ {{{sin\;\theta }}} & {{0}} & {{{cos\;\theta }}} \\ \end{vmatrix}^{{2}}}{{ = }}\dfrac{{{1}}}{{{2}}}$.

42. If A is a matrix of order $3 \times 3$, then ${\left( {{A^2}} \right)^{ - 1}} = $…………..

Ans: If A is a matrix of order ${{3 \times 3}}$, then ${\left( {{{{A}}^2}} \right)^{ - 1}} = $ ${\left( {{{{A}}^{ - 1}}} \right)^{2}}$.

43. If A is a matrix of order $3 \times 3$, then the number of minors in the determinant of A are …………….. .

Ans: If A is a matrix of order $3 \times 3$, then the number of minors in determinant of A are 9. because, in a $3 \times 3$ matrix, there are 9 elements.

44. The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ………….. .

Ans: The sum of the products of elements of any row with the co-factors of corresponding elements is equal to the value of determinant.

Let $\Delta = \begin{vmatrix} {{{{a}}_{11}}} & {{{{a}}_{12}}} & {{{{a}}_{13}}} \\ {{{{a}}_{21}}} & {{{{a}}_{22}}} & {{{{a}}_{23}}} \\ {{{{a}}_{31}}} & {{{{a}}_{32}}} & {{{{a}}_{33}}} \\ \end{vmatrix}$

Expanding along ${{{R}}_1}$

⇒ $\Delta \; = {{{a}}_{11}}\left( {{{{a}}_{22}}.{{\;}}{{{a}}_{33}} - {{{a}}_{23}}.{{{a}}_{32}}} \right) - {{{a}}_{12}}\left( {{{{a}}_{21}}.{{\;}}{{{a}}_{33}} - {{{a}}_{23}}.{{{a}}_{31}}} \right) + {{{a}}_{13}}\left( {{{{a}}_{21}}.{{\;}}{{{a}}_{32}} - {{{a}}_{22}}.{{{a}}_{31}}} \right)$

⇒ $\Delta \; = {{{a}}_{11}}\left( {{{{a}}_{22}}.{{\;}}{{{a}}_{33}} - {{{a}}_{23}}.{{{a}}_{32}}} \right) + {{{a}}_{12}}\left( {{{{a}}_{23}}.{{{a}}_{31}} - {{{a}}_{21}}.{{\;}}{{{a}}_{33}}} \right) + {{{a}}_{13}}\left( {{{{a}}_{21}}.{{\;}}{{{a}}_{32}} - {{{a}}_{22}}.{{{a}}_{31}}} \right)$

⇒ $\Delta \; = {{{a}}_{11}}{{{A}}_{11}} + {{{a}}_{12}}{{{A}}_{12}} + {{{a}}_{13}}{{{A}}_{13}}$

⇒ $\Delta \; = $ sum of the products of elements of ${{{R}}_1}$ with their corresponding cofactors.

45. If ${{x =  - 9}}$ is a root of $\begin{vmatrix} {{x}} & {{3}} & {{7}} \\ {{2}} & {{x}} & {{2}} \\ {{7}} & {{6}} & {{x}} \\ \end{vmatrix}= 0$, then the other two roots are ……..

Ans: Here, we have ${{x =  - 9}}$ is a root of $\begin{vmatrix} {{x}} & {{3}} & {{7}} \\ {{2}} & {{x}} & {{2}} \\ {{7}} & {{6}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

⇒ $\begin{vmatrix} {{x}} & {{3}} & {{7}} \\ {{2}} & {{x}} & {{2}} \\ {{7}} & {{6}} & {{x}} \\ \end{vmatrix}{{ = 0}}$

Expanding along ${{{R}}_1}$, we get

⇒ ${{x}}\left( {{{{x}}^{{2}}}{{ - 12}}} \right){{ - 3}}\left( {{{2x - 14}}} \right){{ + 7}}\left( {{{12 - 7x}}} \right){{ = 0}}$

⇒ ${{{x}}^{{3}}}{{ - 12x - 6x + 42 + 84 - 49x = 0}}$

⇒ ${{{x}}^{{3}}}{{ - 12x - 6x + 42 + 84 - 49x = 0}}$

⇒ ${{{x}}^{{3}}}{{ - 67x + 126 = 0}}$

⇒ ${{{x}}^{{3}}}{{ + 9}}{{{x}}^{{2}}}{{ - 9}}{{{x}}^{{2}}}{{ - 81x + 14x + 126 = 0}}$

⇒ ${{{x}}^{{2}}}\left( {{{x + 9}}} \right){{ - 9x}}\left( {{{x + 9}}} \right){{ + 14}}\left( {{{x + 9}}} \right){{ = 0}}$

⇒ $\left( {{{x + 9}}} \right){{\;}}\left( {{{{x}}^{{2}}}{{ - 9x + 14}}} \right){{ = 0}}$

⇒ $\left( {{{x + 9}}} \right)\left( {{{x - 7}}} \right)\left( {{{x - 2}}} \right){{ = 0}}$

⇒ ${{x =  - 9,\;7,\;2}}$

Hence, other two roots of $\begin{vmatrix} {{x}} & {{3}} & {{7}} \\ {{2}} & {{x}} & {{2}} \\ {{7}} & {{6}} & {{x}} \\ \end{vmatrix}{{ = 0\;}}$ are 7 and 2.

46. $\begin{vmatrix} 0 & {xyz} & {x - z} \\ {y - x} & 0 & {y - z} \\ {z - x} & {z - y} & 0 \\ \end{vmatrix} = $ ………….

Ans: Here, we have $\begin{vmatrix} {{0}} & {{{xyz}}} & {{{x - z}}} \\  {{{y - x}}} & {{0}} & {{{y - z}}} \\  {{{z - x}}} & {{{z - y}}} & {{0}} \\ \end{vmatrix}$

Applying ${{{C}}_1} \to {{{C}}_1} - {{{C}}_3}$, we get

$ = \begin{vmatrix} {{{z - x}}} & {{{xyz}}} & {{{x - z}}} \\ {{{z - x}}} & {{0}} & {{{y - z}}} \\ {{{z - x}}} & {{{z - y}}} & {{0}}  \\ \end{vmatrix}$ 

Taking common $\left( {{{z}} - {{x}}} \right)$ from ${{{C}}_1}$

$ = \left( {{{z - x}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{xyz}}} & {{{x - z}}} \\ {{1}} & {{0}} & {{{y - z}}} \\ {{1}} & {{{z - y}}} & {{0}} \\ \end{vmatrix}$ 

Applying $[{{{R}}_2} \to {{{R}}_2} - {{{R}}_1}{{\;and\;\;}}{{{R}}_3} \to {{{R}}_3} - {{{R}}_1}]$, we get

$ = \left( {{{z - x}}} \right){{\;}}\begin{vmatrix} {{1}} & {{{xyz}}} & {{{x - z}}} \\ {{0}} & {{{ - xyz}}} & {{{y - x}}} \\ {{0}} & {{{z - y - xyz}}} & {{{z - x}}} \\ \end{vmatrix}$ 

Expanding along ${{{C}}_1}$

$ = \left( {{{z}} - {{x}}} \right)\left[ { - {{xyz}}\left( {{{z}} - {{x}}} \right) - \left( {{{y}} - {{x}}} \right)\left( {{{z}} - {{y}} - {{xyz}}} \right)} \right]$ 

$ = \left( {{{z}} - {{x}}} \right)\left[ { - {{xy}}{{{z}}^2} + {{{x}}^2}{{yz}} - \left( {{{\;yz}} - {{{y}}^2} - {{x}}{{{y}}^2}{{z}} - {{xz}} + {{xy}} + {{{x}}^2}{{yz}}} \right)} \right]$ 

$ = \left( {{{z}} - {{x}}} \right)\left[ { - {{xy}}{{{z}}^2} - {{\;yz}} + {{{y}}^2} + {{x}}{{{y}}^2}{{z}} + {{xz}} - {{xy}}} \right]$ 

$ = \left( {{{z}} - {{x}}} \right)\left[ {{{{y}}^2} - {{\;yz}} + {{x}}{{{y}}^2}{{z}} - {{xy}}{{{z}}^2} + {{xz}} - {{xy}}} \right]$ 

$ = \left( {{{z}} - {{x}}} \right)\left[ {{{y}}\left( {{{y}} - {{z}}} \right) + {{xyz}}\left( {{{y}} - {{z}}} \right) - {{x}}\left( {{{y}} - {{z}}} \right)} \right]$ 

$ = \left( {{{z}} - {{x}}} \right)\left( {{{y}} - {{z}}} \right)\left( {{{y}} + {{xyz}} - {{x}}} \right)$ 

Hence, $\begin{vmatrix} {{0}} & {{{xyz}}} & {{{x - z}}} \\ {{{y - x}}} & {{0}} & {{{y - z}}} \\ {{{z - x}}} & {{{z - y}}} & {{0}} \\ \end{vmatrix}{{ = }}\left( {{{z - x}}} \right)\left( {{{y - z}}} \right)\left( {{{y + xyz - x}}} \right){{.}}$

47. If ${\mathbf{f}}\left( {\mathbf{x}} \right)$ =$\begin{vmatrix} {{{\left( {1 + {\mathbf{x}}} \right)}^{17}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{19}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{23}}} \\ {{{\left( {1 + {\mathbf{x}}} \right)}^{23}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{29}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{34}}} \\ {{{\left( {1 + {\mathbf{x}}} \right)}^{41}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{43}}} & {{{\left( {1 + {\mathbf{x}}} \right)}^{47}}}\\ \end{vmatrix}$ = $A+Bx+C{x}^{2}$ …, then A = ………….

Ans: Here, we have ${{f}}\left( {{x}} \right){{ = }}\begin{vmatrix} {{{\left( {{{1 + x}}} \right)}^{{{17}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{19}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{23}}}}} \\ {{{\left( {{{1 + x}}} \right)}^{{{23}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{29}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{34}}}}} \\  {{{\left( {{{1 + x}}} \right)}^{{{41}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{43}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{47}}}}}  \\ \end{vmatrix}$

Taking common ${\left( {{{1 + x}}} \right)^{{{17}}}}{{,\;}}{\left( {{{1 + x}}} \right)^{{{23}}}}{{,\;}}{\left( {{{1 + x}}} \right)^{{{41}}}}$ from ${{R}_{1}}, {{R}_{2}} \text{ and } {{R}_{3}}$ respectively, we get

${{ = }}{\left( {{{1 + x}}} \right)^{{{17}}}}{{.\;}}{\left( {{{1 + x}}} \right)^{{{23}}}}{{.}}{\left( {{{1 + x}}} \right)^{{{41}}}}\begin{vmatrix} {{1}} & {{{\left( {{{1 + x}}} \right)}^{{2}}}} & {{{\left( {{{1 + x}}} \right)}^{{6}}}} \\  {{1}} & {{{\left( {{{1 + x}}} \right)}^{{6}}}} & {{{\left( {{{1 + x}}} \right)}^{{{11}}}}} \\ {{1}} & {{{\left( {{{1 + x}}} \right)}^{{2}}}} & {{{\left( {{{1 + x}}} \right)}^{{6}}}}  \\ \end{vmatrix}$ 

= 0.

Here, ${{R}_{1}} \& {{R}_{3}}$ are identical, 

Thus, ${{f}}\left( {{x}} \right){{ = }}\begin{vmatrix} {{{\left( {{{1 + x}}} \right)}^{{{17}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{19}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{23}}}}} \\  {{{\left( {{{1 + x}}} \right)}^{{{23}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{29}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{34}}}}} \\ {{{\left( {{{1 + x}}} \right)}^{{{41}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{43}}}}} & {{{\left( {{{1 + x}}} \right)}^{{{47}}}}} \\ \end{vmatrix}{{ = 0}}{{.}}$

Hence, the value of A is 0.

State True or False for the statements of the following Exercises:

48. ${\left( {{{\mathbf{A}}^3}} \right)^{ - 1}} = {\left( {{{\mathbf{A}}^{ - 1}}} \right)^3}$, where A is square matrix and $\left| {\mathbf{A}} \right| \ne 0.$

Ans: Here, the given statement is true.

Since, ${\left( {{{{A}}^{{n}}}} \right)^{ - 1}} = {\left( {{{{A}}^{ - 1}}} \right)^{{n}}}$, where ${{n}} \in {{N}}.$

49. ${\left( {{\mathbf{aA}}} \right)^{ - 1}} = \dfrac{1}{{\mathbf{a}}}\;{{\mathbf{A}}^{ - 1}}$, where ${\mathbf{a}}$ is any real number and A is a square matrix.

Ans: Since, we know that, if A is a non- singular square matrix, then for any scalar a (non – zero), aA is invertible such that 

⇒ $\left( {{{aA}}} \right){{ = }}\left( {\dfrac{{{1}}}{{{a}}}{{\;}}{{{A}}^{{{ - 1}}}}} \right){{ = }}\left( {{{a}}{{.}}\dfrac{{{1}}}{{{a}}}{{\;}}} \right)\left( {{{A}}{{{A}}^{{{ - 1}}}}} \right)$ i.e., (aA) is inverse of $\left( {\dfrac{{{1}}}{{{a}}}{{\;}}{{{A}}^{{{ - 1}}}}} \right)$ or ${\left( {{{aA}}} \right)^{{{ - 1}}}}{{ = }}\dfrac{{{1}}}{{{a}}}{{\;}}{{{A}}^{{{ - 1}}}}$, where a is any non -zero scalar false. 

In the above statement, a is any real number. So, we can conclude that above statement is false.

50. $\left| {{A^{ - 1}}} \right| \ne {\left| A \right|^{ - 1}},$ where A is non – singular matrix.

Ans: Since, $\left| {{{{A}}^{ - 1}}} \right| = {\left| {{A}} \right|^{ - 1}}$, where A is non – singular matrix.

Hence, the given statement is false.

51. If A and B are the matrices of order 3 and |A|=5, |B|=3, then |3AB| $= 27 \times 5 \times 3 = 405.$

Ans: Here, A and B are the matrices of order 3 and $\left| {{A}} \right|{{ = 5,\;}}\left| {{B}} \right|{{ = 3}}$

⇒ $\left| {{{3AB}}} \right|{{ = \;}}{{{3}}^{{3}}}{{ \times }}\left| {{A}} \right|{{ \times }}\left| {{B}} \right|$

⇒ $\left| {{{3AB}}} \right|{{ = \;27 \times 5 \times 3}}$

⇒ $\left| {{{3AB}}} \right|{{ = \;405}}$

Hence, the given statement is true.

52. If the value of a third order determinant is 12, then the value of the determinants formed by replacing each element by its co – factor will be 144. 

Ans: let A be the determinant, $\left| {{A}} \right|{{ = 12}}$

Also, we know that, if A is the square matrix of order n, 

then $\left| {{{Adj\;A}}} \right| = {\left| {{A}} \right|^{{{n}} - 1}}$

⇒ $\left| {{{Adj\;A}}} \right|{{ = }}{\left( {{{12}}} \right)^{{{3 - 1}}}}$

⇒ $\left| {{{Adj\;A}}} \right|{{ = }}{\left( {{{12}}} \right)^{{2}}}$

⇒ $\left| {{{Adj\;A}}} \right|{{ = 144}}$

Hence, the given statement is true.

53. $\begin{vmatrix} {{\mathbf{x}} + 1} & {{\mathbf{x}} + 2} & {{\mathbf{x}} + {\mathbf{a}}} \\ {{\mathbf{x}} + 2} & {{\mathbf{x}} + 3} & {{\mathbf{x}} + {\mathbf{b}}} \\ {{\mathbf{x}} + 3} & {{\mathbf{x}} + 4} & {{\mathbf{x}} + {\mathbf{c}}} \\ \end{vmatrix} = 0$, where ${\mathbf{a}},\;{\mathbf{b}},\;{\mathbf{c}}$ are in ${\mathbf{A}}.{\mathbf{P}}.$

Ans: Here, we have $\begin{vmatrix} {{{x + 1}}} & {{{x + 2}}} & {{{x + a}}} \\  {{{x + 2}}} & {{{x + 3}}} & {{{x + b}}} \\  {{{x + 3}}} & {{{x + 4}}} & {{{x + c}}} \\ \end{vmatrix}$

Applying ${{{C}}_1} \to {{{C}}_1} - {{{C}}_2}{{\;}}$, we get

${{ = }}\begin{vmatrix} {{{ - 1}}} & {{{x + 2}}} & {{{x + a}}} \\ {{{ - 1}}} & {{{x + 3}}} & {{{x + b}}} \\  {{{ - 1}}} & {{{x + 4}}} & {{{x + c}}} \\ \end{vmatrix}$ 

Applying \[R_{2} \to R_{2} ~R_{1}~ and ~R_{3} ~\to ~R_{3} ~R_{1}\] , we get

$= \begin{vmatrix} {{{ - 1}}} & {{{x + 2}}} & {{{x + a}}} \\ {{0}} & {{1}} & {{{b - a}}} \\ {{0}} & {{2}} & {{{c - a}}} \\ \end{vmatrix}$ 

Expanding along ${{{C}}_1}$

${{ =  - 1}}\left[ {\left( {{{c - a}}} \right){{ - 2}}\left( {{{b - a}}} \right)} \right]$ 

${{ =  - 1}}\left( {{{c - a - 2b + 2a}}} \right]$ 

${{ =  - 1}}\left( {{{c - 2b + a}}} \right]$ 

Since, ${{a}},{{\;b}}$ and ${{c}}$ are in A.P. 

Therefore, ${{2b}} = {{a}} + {{c}}$

${{ =  - 1}}\left( {{{2b - 2b}}} \right)$ 

= 0.

Hence, the given statement is true.

54. $\left| {{\mathbf{adj}}.\;{\mathbf{A}}} \right| = {\left| {\mathbf{A}} \right|^2}$, where A is a square matrix of order two.

Ans: Here, A is square matrix of order two

We know that, 

$ \Rightarrow {{A}}.{{\;Adj\;A}} = {{\;}}\left| {{A}} \right|{{\;I\;}}$ 

$ \Rightarrow \left| {{{A}}.{{\;Adj\;A}}} \right| = {{\;}}\left| {\left| {{A}} \right|{{\;I}}} \right|$ 

$ \Rightarrow \left| {{A}} \right|\left| {{{\;Adj\;A}}} \right| = {{\;}}{\left| {{A}} \right|^{{n}}}$ 

$ \Rightarrow \left| {{{\;Adj\;A}}} \right| = \dfrac{{{{\left| {{A}} \right|}^{{n}}}}}{{\left| {{A}} \right|}}$ 

$ \Rightarrow \left| {{{\;Adj\;A}}} \right| = {\left| {{A}} \right|^{{{n}} - 1}}$  

Here, ${{n}} = {{2}}$

$ \Rightarrow \left| {{{\;Adj\;A}}} \right| = {\left| {{A}} \right|^{2 - 1}}$ 

$ \Rightarrow \left| {{{\;Adj\;A}}} \right| = \left| {{A}} \right|$ 

Hence, the given statement is false.

55. The determinant $\begin{vmatrix} {\sin {\mathbf{A}}} & {\cos {\mathbf{A}}} & {\sin {\mathbf{A}} + \cos {\mathbf{B}}} \\ {\sin {\mathbf{B}}} & {\cos {\mathbf{A}}} & {\sin {\mathbf{B}} + \cos {\mathbf{B}}} \\ {\sin {\mathbf{C}}} & {\cos {\mathbf{A}}} & {\sin {\mathbf{C}} + \cos {\mathbf{B}}} \\ \end{vmatrix}$ is equal to zero.

Ans: Here, we have $\begin{vmatrix} {\sin {{A}}} & {\cos {{A}}} & {\sin {{A}} + \cos {{B}}} \\ {\sin {{B}}} & {\cos {{A}}} & {\sin {{B}} + \cos {{B}}} \\ {\sin {{C}}} & {\cos {{A}}} & {\sin {{C}} + \cos {{B}}} \\ \end{vmatrix}$

$= \begin{vmatrix} {\sin {{A}}} & {\cos {{A}}} & {\sin {{A}}} \\ {\sin {{B}}} & {\cos {{A}}} & {\sin {{B}}} \\  {\sin {{C}}} & {\cos {{A}}} & {\sin {{C}}} \\ \end{vmatrix}$ + $\begin{vmatrix} {\sin {{A}}} & {\cos {{A}}} & {\cos {{B}}} \\  {\sin {{B}}} & {\cos {{A}}} & {\cos {{B}}} \\  {\sin {{C}}} & {\cos {{A}}} & {\cos {{B}}}  \\ \end{vmatrix}$ 

Since, the value of the determinant having two identical rows and columns is zero. Therefore, 

$ = {{0}} + \begin{vmatrix} {\sin {{A}}} & {\cos {{A}}} & {\cos {{B}}} \\  {\sin {{B}}} & {\cos {{A}}} & {\cos {{B}}} \\ {\sin {{C}}} & {\cos {{A}}} & {\cos {{B}}} \\ \end{vmatrix}$ 

$= \begin{vmatrix} {\sin {{A}}} & {\cos {{A}}} & {\cos {{B}}} \\ {\sin {{B}}} & {\cos {{A}}} & {\cos {{B}}} \\ {\sin {{C}}} & {\cos {{A}}} & {\cos {{B}}} \\ \end{vmatrix}$ 

Taking common $\cos {{A}}$ and $\cos {{B}}$ from ${{{C}}_2}$ and ${{{C}}_3}$

$ = \cos {{A}}.\cos {{B\;}}\begin{vmatrix} {{{sinA}}} & {{1}} & {{1}} \\ {{{sinB}}} & {{1}} & {{1}} \\  {{{sinC}}} & {{1}} & {{1}} \\ \end{vmatrix}$ 

${{ = }}\,{{0}}$ 

Since, the value of the determinant having two identical rows and columns is zero. 

Hence, the given statement is true.

56. If the determinant $\begin{vmatrix} {{\mathbf{x}} + {\mathbf{a}}} & {{\mathbf{p}} + {\mathbf{u}}} & {{\mathbf{l}} + {\mathbf{f}}} \\ {{\mathbf{y}} + {\mathbf{b}}} & {{\mathbf{q}} + {\mathbf{v}}} & {{\mathbf{m}} + {\mathbf{g}}} \\ {{\mathbf{z}} + {\mathbf{c}}} & {{\mathbf{r}} + {\mathbf{w}}} & {{\mathbf{n}} + {\mathbf{h}}} \\ \end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

Ans: Here, we have $\begin{vmatrix} {{{x}} + {{a}}} & {{{p}} + {{u}}} & {{{l}} + {{f}}} \\ {{{y}} + {{b}}} & {{{q}} + {{v}}} & {{{m}} + {{g}}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix}$

Now, splitting first row in the sum of two determinant

$ =$\begin{vmatrix} {{x}} & {{p}} & {{l}} \\  {{{y}} + {{b}}} & {{{q}} + {{v}}} & {{{m}} + {{g}}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix} +$\begin{vmatrix} {{a}} & {{u}} & {{f}} \\ {{{y}} + {{b}}} & {{{q}} + {{v}}} & {{{m}} + {{g}}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix}$ 

Now, splitting second row in the sum of two determinants $= \begin{vmatrix} {{x}} & {{p}} & {{l}} \\ {{y}} & {{q}} & {{m}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix}$ + $\begin{vmatrix} {{x}} & {{p}} & {{l}} \\ {{b}} & {{v}} & {{g}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix}$ + $\begin{vmatrix} {{a}} & {{u}} & {{f}} \\ {{y}} & {{q}} & {{m}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}} \\ \end{vmatrix}$ + $\begin{vmatrix} {{a}} & {{u}} & {{f}} \\ {{b}} & {{v}} & {{g}} \\ {{{z}} + {{c}}} & {{{r}} + {{w}}} & {{{n}} + {{h}}}  \\ \end{vmatrix}$ 

Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants.

Hence, the given statement is true.

57. Let \[\Delta \; = \;\begin{vmatrix} {\mathbf{a}} & {\mathbf{p}} & {\mathbf{x}} \\  {\mathbf{b}} & {\mathbf{q}} & {\mathbf{y}} \\  {\mathbf{c}} & {\mathbf{r}} & {\mathbf{z}} \\ \end{vmatrix} = 16,\] then ${\Delta _1}\; = \;\begin{vmatrix}{{\mathbf{p}} + {\mathbf{x}}} & {{\mathbf{a}} + {\mathbf{x}}} & {{\mathbf{a}} + {\mathbf{p}}} \\ {{\mathbf{q}} + {\mathbf{y}}} & {{\mathbf{b}} + {\mathbf{y}}} & {{\mathbf{b}} + {\mathbf{q}}} \\ {{\mathbf{r}} + {\mathbf{z}}} & {{\mathbf{c}} + {\mathbf{z}}} & {{\mathbf{c}} + {\mathbf{r}}} \\ \end{vmatrix} = 32.$

Ans: Here, we have $\Delta {{\;}} = {{\;}}\begin{vmatrix} {{a}} & {{p}} & {{x}} \\  {{b}} & {{q}} & {{y}} \\  {{c}} & {{r}} & {{z}} \\ \end{vmatrix} = {{16}}$ …….. (i)

Now, \[{\Delta _1}{{\;}} = {{\;}}\begin{vmatrix} {{{p}} + {{x}}} & {{{a}} + {{x}}} & {{{a}} + {{p}}} \\  {{{q}} + {{y}}} & {{{b}} + {{y}}} & {{{b}} + {{q}}} \\ {{{r}} + {{z}}} & {{{c}} + {{z}}} & {{{c}} + {{r}}}  \\ \end{vmatrix}\]

Applying ${{C}_{1}} \to {{C}_{1}} + {{C}_{2}} + {{C}_3}$, we get

⇒ ${\Delta _1}{{\;}} = {{\;}}\begin{vmatrix} {{{2}}\left( {{{p + x + a}}} \right)} & {{{a + x}}} & {{{a + p}}} \\ {{{2}}\left( {{{q + y + b}}} \right)} & {{{b + y}}} & {{{b + q}}} \\ {{{2}}\left( {{{r + z + c}}} \right)} & {{{c + z}}} & {{{c + r}}} \\ \end{vmatrix}$

Taking common 2 from ${{{C}}_1}$

⇒ ${\Delta _1}{{\;}} = {{\;2}}\begin{vmatrix} {\left( {{{p}} + {{x}} + {{a}}} \right)} & {{{a}} + {{x}}} & {{{a}} + {{p}}} \\ {\left( {{{q}} + {{y}} + {{b}}} \right)} & {{{b}} + {{y}}} & {{{b}} + {{q}}} \\ {\left( {{{r}} + {{z}} + {{c}}} \right)} & {{{c}} + {{z}}} & {{{c}} + {{r}}}  \\ \end{vmatrix}$

Applying ${{{C}}_1} \to {{{C}}_1} - {{{C}}_2}$ ,${{\;}}$and ${{{C}}_2} \to {{{C}}_2} - {{{C}}_3}{{\;}}$we get

⇒ ${\Delta _1}{{\;}} = {{\;2}}\begin{vmatrix} {{p}} & {{{x}} - {{p}}} & {{{a}} + {{p}}} \\ {{q}} & {{{y}} - {{q}}} & {{{b}} + {{q}}} \\ {{r}} & {{{z}} - {{r}}} & {{{c}} + {{r}}} \\ \end{vmatrix}$

Applying ${{{C}}_2} \to {{{C}}_1} + {{{C}}_2}$

⇒ ${{{\Delta }}_{{1}}}{{\; = \;2}}\begin{vmatrix} {{p}} & {{x}} & {{{a + p}}} \\  {{q}} & {{y}} & {{{b + q}}} \\  {{r}} & {{z}} & {{{c + r}}} \\ \end{vmatrix}$

⇒ $\dfrac{{{{{\Delta }}_{{1}}}}}{{{2}}}{{\; = \;}}\begin{vmatrix} {{p}} & {{x}} & {{a}} \\ {{q}} & {{y}} & {{b}} \\ {{r}} & {{z}} & {{c}} \\ \end{vmatrix}{{ + }}\begin{vmatrix} {{p}} & {{x}} & {{p}} \\ {{q}} & {{y}} & {{q}} \\ {{r}} & {{z}} & {{r}} \\ \end{vmatrix}$

⇒ $\dfrac{{{{{\Delta }}_{{1}}}}}{{{2}}}{{\; = \;}}\begin{vmatrix} {{p}} & {{x}} & {{a}} \\ {{q}} & {{y}} & {{b}} \\ {{r}} & {{z}} & {{c}} \\ \end{vmatrix}{{ + 0}}$

Since, the value of the determinant having two identical rows and columns is zero. 

⇒ ${{{\Delta }}_{{1}}}{{\; = \;2}}\begin{vmatrix} {{p}} & {{x}} & {{a}} \\  {{q}} & {{y}} & {{b}} \\  {{r}} & {{z}} & {{c}} \\ \end{vmatrix}$

⇒ ${{{\Delta }}_{{1}}}{{\; = \;2}}{{.}}{\left( {{{ - 1}}} \right)^{{2}}}\begin{vmatrix} {{a}} & {{p}} & {{x}} \\ {{b}} & {{q}} & {{y}} \\  {{c}} & {{r}} & {{z}} \\ \end{vmatrix}$

⇒ ${{{\Delta }}_{{1}}}{{\; = \;2 \times 16 = 32}}$                              (from eq (i))

Hence, the given statement is true.

58. The maximum value of $\begin{vmatrix} 1 & 1 & 1 \\ 1 & {1 + {\mathbf{sin\theta }}} & 1 \\ 1 & 1 & {1 + {\mathbf{cos\theta }}} \\ \end{vmatrix}$ is $\dfrac{1}{2}\;.$

Ans: Here, we have ${{\Delta \; = \;}}\begin{vmatrix} {{1}} & {{1}} & {{1}} \\ {{1}} & {{{1 + sin\theta }}} & {{1}} \\ {{1}} & {{1}} & {{{1 + cos\theta }}} \\ \end{vmatrix}$

Applying $\left[{{C}_{1}} \to {C}_{1}-{C}_{3}\right] \text{ and }\left[{{C}_{2}} \to {C}_{2}-{C}_{3} \right]$, we get

⇒ ${{\Delta \; = \;}}\begin{vmatrix} {{0}} & {{0}} & {{1}} \\ {{0}} & {{{sin\theta }}} & {{1}} \\ {{{ - cos\theta }}} & {{{ - cos\theta }}} & {{1}} \\ \end{vmatrix}$

Now, expanding along ${{{R}}_1}$

⇒ ${{\Delta \; = \;1}}\left( {{{0 + sin\theta }}{{.cos\theta }}} \right)$

⇒ $\Delta {{\;}} = {{\;}}\sin {{\theta }}.\cos {{\theta }}$

⇒ ${{\Delta \; = \;}}\dfrac{{{2}}}{{{2}}}{{sin\theta }}{{.cos\theta }}$

⇒ ${{\Delta \; = \;}}\dfrac{{{1}}}{{{2}}}{{sin2\theta }}$

Also, we know that, \[ - {{1}} \leqslant \sin {{2\theta }} \leqslant {{1}}\]

⇒$-\dfrac{{{1}}}{{{2}}} \leqslant \dfrac{{{1}}}{{{2}}}\sin {{2\theta }} \leqslant \dfrac{{{1}}}{{{2}}}$

⇒ $ - \dfrac{{{1}}}{{{2}}}\leqslant \Delta {{\;}} \leqslant \dfrac{{{1}}}{{{2}}}$

Thus, maximum value of given determinant is $\dfrac{{{1}}}{{{2}}}$

Hence, the given statement is true.

Determinants

Determinants are the fourth Chapter of NCERT Exemplar Solutions for Class 12 Mathematics. Introduction to Determinants, then a Determinant Matrix of order one, two, or three, properties of the Determinant, minors and cofactor, adjoint and inverse of a Matrix, applications of Matrices, and solution of a System of Linear Equations using the inverse of a Matrix are all covered in this Chapter.

Students can also utilize Vedantu NCERT Exemplar for Class 12 Maths Chapter 4 - Determinants (Book Solutions)  PDF from the links on the Vedantu site to learn and comprehend all of the fundamentals of Determinants in a simple way.