NCERT Solutions For Class 12 Maths Chapter 4 Determinants

Miscellaneous Exercise

1. Prove that the determinant \[\mathbf{\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|}\] is independent of \[\mathbf{\theta} \].

Ans: Let \[\Delta =\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|\]

To solve it, we have:

\[\Rightarrow \Delta =x\left( {{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta  \right)+\cos \theta \left( -\sin \theta +x\cos \theta  \right)\]

\[\Rightarrow \Delta ={{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta \]

\[\Rightarrow \Delta ={{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)\]

\[\Rightarrow \Delta \text{=}{{\text{x}}^{\text{3}}}\text{-x+x}\]

\[\therefore \Delta \text{=}{{\text{x}}^{\text{3}}}\]

Hence, \[\text{ }\!\!\Delta\!\!\text{ }\] is independent of \[\theta \] .

2. Evaluate \[\mathbf{\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|}\]

Ans: Let \[\Delta =\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|\]

Expanding along column \[{{\text{C}}_{3}}\]

\[\Rightarrow \Delta =-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \sin \alpha  \right)+\cos \alpha \left( \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( s{{\sin }^{2}}\beta +{{\cos }^{2}}\beta  \right)+{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( 1 \right)+{{\cos }^{2}}\alpha \left( 1 \right)\]

\[\therefore \Delta \text{=1}\]

3. If $\mathbf{{{\mathbf{A}}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 1 & 2 & -2  \\ -1 & 3 & 0  \\ 0 & -2 & 1  \\ \end{matrix} \right]$, find ${{\left( AB \right)}^{-1}}}$.

Ans: The below result will be used for simplification,

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

Finding the inverse of matrix B. So, the determinant is

$\left| B \right|=1\left( 3-0 \right)-2\left( -1-0 \right)-2\left( 2-0 \right)$

$\Rightarrow \left| B \right|=3+2-4$

$\therefore \left| B \right|=1$

Now finding the cofactors,

${{B}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 3 & 0  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{11}}=3$

${{B}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} -1 & 0  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{12}}=1$

${{B}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} -1 & 3  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{13}}=2$

${{B}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 2 & -2  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{21}}=-\left( 2-4 \right)=2$

${{B}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 1 & -2  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{22}}=1$

${{B}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 1 & 2  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{23}}=2$

${{B}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 2 & -2  \\ 3 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{31}}=6$

${{B}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 1 & -2  \\ -1 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{32}}=2$

${{B}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 1 & 2  \\ -1 & 3  \\ \end{matrix} \right|\Rightarrow {{B}_{33}}=\left( 3+2 \right)=5$

The cofactor matrix is $\left[ \begin{matrix} 3 & 1 & 2  \\ 2 & 1 & 2  \\ 6 & 2 & 5  \\ \end{matrix} \right]$. The adjoint will be the transpose of the cofactor matrix.

$adj\left( B \right)=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

The inverse is given by ${{B}^{-1}}=\frac{adj\left( B \right)}{\left| B \right|}$. So,

${{B}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

Now, it is already given that ${{A}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$.

So, we can compute ${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$ as below,

${{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$

$\Rightarrow {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9-30+30 & -3+12-12 & 3-10+12  \\ 3-15+10 & -1+6-4 & 1-5+4  \\ 6-30+25 & -2+12-10 & 2-10+10  \\ \end{matrix} \right]$

$\therefore {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9 & -3 & 5  \\ -2 & 1 & 0  \\ 1 & 0 & 2  \\ \end{matrix} \right]$

4. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]}\] verify that

(i) \[\mathbf{{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{15-1} \right)\text{-2}\left( \text{10-1} \right)\text{+1}\left( \text{2-3} \right)\]

\[\Rightarrow \left| A \right|\text{=14-18-1}\]

\[\therefore \left| A \right|\text{=}-5\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=14}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-9\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-9}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=4\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=1}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

We know that the adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]}^{T}}\] \[\therefore \text{adjA=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

Let us denote the adjoint of A as B. So, \[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

The inverse of A is given by

\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{5}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

\[\therefore {{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\]

Now, we have to verify \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

Let us compute the RHS first, i.e. the adjoint of \[{{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\] or \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=-}\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-\dfrac{\text{1}}{5}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-}\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-\dfrac{3}{5}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=-}\dfrac{\text{1}}{5}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\].  We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adj{{A}^{-1}}={{\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

Next, moving to the LHS, i.e. the inverse of the adjoint of A. We have an adjoint of A as matrix B,

\[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

The Determinant of B is

$\left| B \right|=\left[ 14\left( -4-1 \right)+9\left( 9+1 \right)-1\left( -9+4 \right) \right]$

$\Rightarrow \left| B \right|=\left[ -70+90+5 \right]$

$\therefore \left| B \right|=25$

Now the cofactors,

Thus,

\[\Rightarrow {{B}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{B}_{11}}\text{=-5}\]

\[\Rightarrow {{B}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{B}_{12}}=-10\]

\[\Rightarrow {{B}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{B}_{13}}=-5\]

Similarly,

\[\Rightarrow {{B}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{B}_{21}}\text{=-10}\]

\[\Rightarrow {{B}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{B}_{22}}=-15\]

\[\Rightarrow {{B}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{B}_{23}}\text{=-5}\]

And

\[\Rightarrow {{B}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{B}_{31}}=-5\]

\[\Rightarrow {{B}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{B}_{32}}=-5\]

\[\Rightarrow {{B}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{B}_{33}}=-25\].

Cofactor matrix is \[\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjB={{\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjB=}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

Now the inverse is found as

\[{{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| B \right|}\left( \text{adjB} \right)\]

\[\Rightarrow {{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{25}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

\[\therefore {{B}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

So, LHS is \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]. 

Since LHS=RHS, it is verified that \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

(ii) \[\mathbf{{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}}\]

Ans: Since \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\]  and \[\text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\], thus,

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ -\dfrac{14}{5}\left( -\dfrac{4}{25}-\dfrac{1}{25} \right)-\dfrac{9}{5}\left( \dfrac{9}{25}+\dfrac{1}{25} \right)+\dfrac{1}{5}\left( -\dfrac{9}{25}+\dfrac{4}{25} \right) \right]\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ \dfrac{70}{125}-\dfrac{90}{125}-\dfrac{5}{125} \right]\]

\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=-}\dfrac{1}{5}\]

Also, we know that an inverse of a matrix is given by:

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\dfrac{\text{adj}{{\text{A}}^{\text{-1}}}}{\left| \text{A} \right|}\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=-5}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\left[ \begin{matrix} 1 & 2 & 1  \\ 2 & 3 & 1  \\ 1 & 1 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=A}\]

Hence verified that \[{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}\].

5. Evaluate \[\mathbf{\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right)  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Taking \[\text{2}\left( \text{x+y} \right)\] common from \[{{\text{R}}_{1}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{0} & 0  \\ \text{y} & \text{x} & \text{x-y}  \\ \text{x+y} & \text{-y} & \text{-x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( x\times \left( \text{-x} \right) \right)-\left( -y\left( x-y \right) \right) \right)\text{+0+}0 \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( \text{-}{{\text{x}}^{2}} \right)-\left( -yx+{{y}^{2}} \right) \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ -{{\text{x}}^{\text{2}}}\text{+xy}-{{\text{y}}^{\text{2}}} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-2}\left( \text{x+y} \right)\left[ {{\text{x}}^{\text{2}}}-xy+{{\text{y}}^{\text{2}}} \right]\]

Applying the identity,

\[\therefore \text{ }\!\!\Delta\!\!\text{ =-2}\left[ {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right]\]

6. Evaluate \[\mathbf{\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{C}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{xy-0} \right)\]

\[\therefore \text{ }\!\!\Delta\!\!\text{ }=\text{xy}\]

7. Solve the system of the following equations

\[\mathbf{\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}}\]

\[\mathbf{\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}}\]

\[\mathbf{\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}}\]

Ans: Given equations,

\[\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}\]

\[\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}\]

\[\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}\]

Let \[\dfrac{\text{1}}{\text{x}}\text{=p}\] , \[\dfrac{\text{1}}{\text{y}}\text{=q}\] and \[\dfrac{\text{1}}{\text{z}}\text{=r}\]

Then the given system of equations becomes:

\[\text{2p+3q+10r=4}\]

\[\text{4p-6q+5r=1}\]

\[\text{6p+9q+20r=2}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{10}  \\ \text{4} & \text{-6} & \text{5}  \\ \text{6} & \text{9} & \text{-20}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\] such that, this system can be written in the form of \[\text{AX=B}\].

Now,

\[\Rightarrow \left| A \right|\text{=2}\left( \text{120-45} \right)\text{-3}\left( \text{-80-30} \right)\text{+10}\left( \text{36+36} \right)\]

\[\Rightarrow \left| A \right|\text{=150+330+720}\]

\[\Rightarrow \left| A \right|\text{=1200}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ Its inverse exists.

So,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}\left| \begin{matrix} -6 & 5  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{11}}\text{=75}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 4 & 5  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{12}}=110\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 4 & -6  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{13}}=72\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 3 & 10  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{21}}\text{=150}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 2 & 10  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{22}}=100\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 3  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 3 & 10  \\ -6 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{31}}=75\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 10  \\ 4 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{32}}=30\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}\left| \begin{matrix} 2 & 3  \\ 4 & -6  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{33}}=-24\].

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\]

Now, \[\text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{300+150+150}  \\ \text{440-100+60}  \\ \text{288+0-48}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{600}  \\ \text{400}  \\ \text{240}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{2}}  \\ \dfrac{\text{1}}{\text{3}}  \\ \dfrac{\text{1}}{\text{5}}  \\ \end{matrix} \right]\]

\[\therefore \text{P=}\dfrac{\text{1}}{\text{2}}\], \[\text{q=}\dfrac{\text{1}}{\text{3}}\] and \[\text{r=}\dfrac{\text{1}}{\text{5}}\]

Thus \[\text{x=2}\] , \[\text{y=3}\] and \[\text{z=5}\].

8. Choose the correct answer.If \[\mathbf{\text{X,Y,Z}}\] are nonzero real numbers, then the inverse of matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

1. \[\mathbf{\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

2. \[\mathbf{\text{xyz}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

3. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

4. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=x}\left( \text{yz-0} \right)\]

\[\therefore \left| \text{A} \right|\text{=xyz}\ne \text{0}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=yz}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=xz\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=xy\].

Cofactor matrix is \[\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{yz}}{\text{xyz}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{xz}}{\text{xyz}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{xy}}{\text{xyz}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{x}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{1}}{\text{y}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{1}}{\text{z}}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\]

Thus, A. \[\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\] is the correct answer.

9. Choose the correct answer.

Let \[\mathbf{\text{A}=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]}\], 

where \[\mathbf{\text{0}\le \text{ }\!\!\theta\!\!\text{ }\le \text{2n}}\] , then

1. \[\mathbf{\text{Det}\left( \text{A} \right)\text{=0}}\]

2. \[\mathbf{text{Det}\left( \text{A} \right)\in \left( \text{2,}\infty  \right)}\]

3. \[\mathbf{\text{Det}\left( \text{A} \right)\in \left( \text{2,4} \right)}\]

4. \[\mathbf{\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]}\]

Ans: Given, \[A=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]\]

\[\therefore \left| A \right|=1\left( 1+{{\sin }^{2}}\theta  \right)-\sin \theta \left( -\sin \theta +\sin \theta  \right)+1\left( {{\sin }^{2}}\theta +1 \right)\]

\[\Rightarrow \left| A \right|=1+{{\sin }^{2}}\theta +{{\sin }^{2}}\theta +1\]

\[\Rightarrow \left| A \right|=2+2{{\sin }^{2}}\theta \]

\[\Rightarrow \left| A \right|=2\left( 1+{{\sin }^{2}}\theta  \right)\]

We know, \[0\le \sin \theta \le 1\]

\[\Rightarrow 1\le 1+{{\sin }^{2}}\theta \le 2\]

\[\Rightarrow 2\le 2\left( 1+{{\sin }^{2}}\theta  \right)\le 4\]

Thus, D. \[\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]\] is the correct answer.

Conclusion

Miscellaneous Exercise in Class 12 Maths Chapter 4 is important for understanding various concepts thoroughly. Class 12 Maths Chapter 4 Miscellaneous Solutions covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to excel in this exercise effectively.

Class 12 Maths Chapter 4: Exercises Breakdown

CBSE Class 12 Maths Chapter 4 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 12 Maths