NCERT Solutions for Class 12 Maths Chapter 4: Determinants - Exercise 4.2

Exercise (4.2)

1. Using the property of determinants and without expanding, prove that:

\[\left| \begin{matrix} \text{x} & \text{a} & \text{x+a} \\ \text{y} & \text{b} & \text{y+b} \\ \text{z} & \text{c} & \text{z+c} \\ \end{matrix} \right|\text{=0}\]. Ans: Given matrix \[\left| \begin{matrix} \text{x} & \text{a} & \text{x+a} \\ \text{y} & \text{b} & \text{y+b} \\ \text{z} & \text{c} & \text{z+c} \\ \end{matrix} \right|\]. Applying the Sum Property of determinants, we have \[\left| \begin{matrix} \text{x} & \text{a} & \text{x+a} \\ \text{y} & \text{b} & \text{y+b} \\ \text{z} & \text{c} & \text{z+c} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{a} & \text{x} \\ \text{y} & \text{b} & \text{y} \\ \text{z} & \text{c} & \text{z} \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{x} & \text{a} & \text{a} \\ \text{y} & \text{b} & \text{b} \\ \text{z} & \text{c} & \text{c} \\ \end{matrix} \right|\] We know, if two rows or columns of a determinant are identical, then the value of the determinant is zero. Since, the two columns in both the determinants are identical, thus its determinant would be zero. \[\Rightarrow \left| \begin{matrix} \text{x} & \text{a} & \text{x+a} \\ \text{y} & \text{b} & \text{y+b} \\ \text{z} & \text{c} & \text{z+c} \\ \end{matrix} \right|=0+0\] \[\therefore \left| \begin{matrix} \text{x} & \text{a} & \text{x+a} \\ \text{y} & \text{b} & \text{y+b} \\ \text{z} & \text{c} & \text{z+c} \\ \end{matrix} \right|=0\]

2. Using the property of determinants and without expanding, prove that:

\[\left| \begin{matrix} \text{a-b} & \text{b-c} & \text{c-a} \\ \text{b-c} & \text{c-a} & \text{a-b} \\ \text{c-a} & \text{a-b} & \text{b-c} \\ \end{matrix} \right|\text{=0}\] Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a} \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b} \\ \text{c}-\text{a} & \text{a}-\text{b} & \text{b}-\text{c} \\ \end{matrix} \right|\] Applying row operation, \[{{\text{R}}_{\text{1}}}\to \text{ }{{\text{R}}_{\text{1}}}\text{+}{{\text{R}}_{\text{2}}}\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a}-c & \text{b}-a & \text{c}-b \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b} \\ c-a & a-b & b-c \\ \end{matrix} \right|\] \[\Rightarrow \Delta =\left| \begin{matrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\ -\left( a-c \right) & -\left( b-a \right) & -\left( c-b \right) \\ \end{matrix} \right|\] Multiplying the third row by $\left( -1 \right)$, we get: \[\Rightarrow \Delta \text{=}-\left| \begin{matrix} \text{a}-c & \text{b}-a & \text{c}-b \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b} \\ \text{a}-\text{c} & \text{b}-\text{a} & \text{c}-\text{b} \\ \end{matrix} \right|\] We know, if two rows or columns of a determinant are identical, then the value of the determinant is zero. Since, the two rows \[{{\text{R}}_{\text{1}}}\] and \[{{\text{R}}_{\text{3}}}\] are identical. \[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\] Hence, \[\left| \begin{matrix} \text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a} \\ \text{b}-\text{c} & \text{c}-\text{a} & \text{a}-\text{b} \\ \text{c}-\text{a} & \text{a}-\text{b} & \text{b}-\text{c} \\ \end{matrix} \right|=0\].

3. Using the property of determinants and without expanding, prove that:

\[\left| \begin{matrix} \text{2} & \text{7} & \text{65} \\ \text{3} & \text{8} & \text{75} \\ \text{5} & \text{9} & \text{86} \\ \end{matrix} \right|\text{=0}\]

Ans: Let $\Delta =\left| \begin{matrix} \text{2} & \text{7} & \text{65} \\ \text{3} & \text{8} & \text{75} \\ \text{5} & \text{9} & \text{86} \\ \end{matrix} \right|$ \[\therefore \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & \text{63+2} \\ \text{3} & \text{8} & \text{72+3} \\ \text{5} & \text{9} & \text{81+5} \\ \end{matrix} \right|\] Applying the Sum Property of determinants, we get \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & \text{63} \\ \text{3} & \text{8} & \text{72} \\ \text{5} & \text{9} & \text{81} \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{2} & \text{7} & \text{2} \\ \text{3} & \text{8} & \text{3} \\ \text{5} & \text{9} & \text{5} \\ \end{matrix} \right|\] The two columns of the second determinant are identical, thus it’s value becomes zero. Hence, \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & 63 \\ \text{3} & \text{8} & 72 \\ \text{5} & \text{9} & 81 \\ \end{matrix} \right|\text{+0}\] \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{2} & \text{7} & 9\left( 7 \right) \\ \text{3} & \text{8} & 9\left( 8 \right) \\ \text{5} & \text{9} & 9\left( 9 \right) \\ \end{matrix} \right|\] Taking $9$ common from the third column, we have \[\Rightarrow \Delta =9\left| \begin{matrix} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \\ \end{matrix} \right|\] Since, the two columns \[C_2\] and \[C_3\] are identical. $\therefore \Delta =0$ Hence, $\left| \begin{matrix} \text{2} & \text{7} & \text{65} \\ \text{3} & \text{8} & \text{75} \\ \text{5} & \text{9} & \text{86} \\ \end{matrix} \right|=0$

4. Using the property of determinants and without expanding, prove that:

\[\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right) \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right) \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right) \\ \end{matrix} \right|\text{=0}\].

Ans: Let\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right) \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right) \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right) \\ \end{matrix} \right|\]

Applying the column operation, \[{{\text{C}}_{\text{3}}}\to {{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{2}}}\]. \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{bc} & \text{ab+bc+ca} \\ \text{1} & \text{ca} & \text{ab+bc+ca} \\ \text{1} & \text{ab} & \text{ab+bc+ca} \\ \end{matrix} \right|\] Taking $\left( ab+bc+ca \right)$ common from the third column, we get: \[\text{ }\!\!\Delta\!\!\text{ =}\left( ab+bc+ca \right)\left| \begin{matrix} \text{1} & \text{bc} & 1 \\ \text{1} & \text{ca} & 1 \\ \text{1} & \text{ab} & 1 \\ \end{matrix} \right|\] Since, the two columns \[C_1\] and \[C_3\] are identical. \[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\] Hence, \[\left| \begin{matrix} \text{1} & \text{bc} & \text{a}\left( \text{b+c} \right) \\ \text{1} & \text{ca} & \text{b}\left( \text{c+a} \right) \\ \text{1} & \text{ab} & \text{c}\left( \text{a+b} \right) \\ \end{matrix} \right|=0\].

5. Using the property of determinants and without expanding, prove that:

\[\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{a+b} & \text{p+q} & \text{x+y} \\ \end{matrix} \right|\text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\].

Ans: Let\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{a+b} & \text{p+q} & \text{x+y} \\ \end{matrix} \right|\] Applying the Sum Property, we get \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{a} & \text{p} & \text{x} \\ \end{matrix} \right|\text{+}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{b} & \text{q} & \text{y} \\ \end{matrix} \right|\] Suppose \[\Delta \text{=}{\Delta}_\text{1+ }\!\!{\Delta}_\text{ 2}\] ……(1) Now, \[\text{ }\!\!\Delta_\text{1 = }\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{a} & \text{p} & \text{x} \\ \end{matrix} \right|\] Applying the row operation, \[{R_2}\to{R_2}-{R_3}\] \[\Rightarrow \text{ }\!\!\Delta_\text{1 = }\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c} & \text{r} & \text{z} \\ \text{a} & \text{p} & \text{x} \\ \end{matrix} \right|\] Again, applying the row operation, \[{R_1}\to {R_1}-{R_2}\] \[\Rightarrow \text{ }\!\!\Delta_\text{1=}\left| \begin{matrix} \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \text{a} & \text{p} & \text{x} \\ \end{matrix} \right|\] We know that if any two rows or columns of a determinant are interchanged, the value of the determinant is multiplied by $\left( -1 \right)$. Hence, interchanging the rows, \[{R_1}\leftrightarrow {R_3}\] and \[{R_2}\leftrightarrow {R_3}\], we have \[\Rightarrow \text{ }\!\!\Delta_\text{1=}{{\left( \text{-1} \right)}^{\text{2}}}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\] \[\therefore \Delta_1\text{=}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\] ……(2) We have, \[\text{ }\!\!{\Delta}_\text{2=}\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{b} & \text{q} & \text{y} \\ \end{matrix} \right|\] Applying the row operation, \[{R_1}\to{R}1-{R_3}\] \[\Rightarrow \text{ }\!\!\Delta_\text{2}=\left| \begin{matrix} \text{c} & \text{r} & \text{z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{b} & \text{q} & \text{y} \\ \end{matrix} \right|\] Applying the row operation, \[{R_2}\to{R_2}-{R_1}\] \[\Rightarrow \text{ }\!\!\Delta_\text{2=}\left| \begin{matrix} \text{c} & \text{r} & \text{z} \\ \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \end{matrix} \right|\] Interchanging the rows, \[{{\text{R}}_{\text{1}}}\leftrightarrow {{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{\text{2}}}\leftrightarrow {{\text{R}}_{\text{3}}}\] \[\Rightarrow \text{ }\!\!\Delta_\text{2=}{{\left( \text{-1} \right)}^{\text{2}}}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\] \[\therefore \text{ }\!\!\Delta_\text{2=}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\] …… (3) From (2) and (3), we get: \[\Rightarrow \text{ }\!\!\Delta_\text{1}=\text{ }\!\!\Delta_\text{2}=\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\] From (1), we have: \[\Rightarrow \text{ }\!\!\Delta=2\Delta_ \text{1}\] \[\therefore \Delta \text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\]. Hence, \[\left| \begin{matrix} \text{b+c} & \text{q+r} & \text{y+z} \\ \text{c+a} & \text{r+p} & \text{z+x} \\ \text{a+b} & \text{p+q} & \text{x+y} \\ \end{matrix} \right|\text{=2}\left| \begin{matrix} \text{a} & \text{p} & \text{x} \\ \text{b} & \text{q} & \text{y} \\ \text{c} & \text{r} & \text{z} \\ \end{matrix} \right|\].

6. By using properties of determinants, show that:

\[\left| \begin{matrix} \text{0} & \text{a} & \text{-b} \\ \text{-a} & \text{0} & \text{-c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|\text{=0}\]. Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{a} & -\text{b} \\ -\text{a} & \text{0} & -\text{c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|\]

Applying \[{R_1}\to \text{cR}_1\]: \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{c}}\left| \begin{matrix} \text{0} & \text{ac} & -\text{bc} \\ -\text{a} & \text{0} & -\text{c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|\] Applying \[{R_1}\to R_1-{bR_2}\] \[\text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{c}}\left| \begin{matrix} \text{ab} & \text{ac} & \text{0} \\ -\text{a} & \text{0} & -\text{c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|\] Taking $a$ common from the first row, we have: \[\Rightarrow \Delta \text{=}\dfrac{\text{a}}{\text{c}}\left| \begin{matrix} \text{b} & \text{c} & \text{0} \\ -\text{a} & \text{0} & -\text{c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|\] Since, the two rows \[{{\text{R}}_{\text{1}}}\] and \[{{\text{R}}_{\text{3}}}\] are identical. \[\therefore \text{ }\!\!\Delta\!\!\text{ =0}\] Hence, \[\left| \begin{matrix} \text{0} & \text{a} & -\text{b} \\ -\text{a} & \text{0} & -\text{c} \\ \text{b} & \text{c} & \text{0} \\ \end{matrix} \right|=0\].

7. By using properties of determinants, show that:

\[\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac} \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc} \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\].

Ans: Let\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac} \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc} \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] Taking out $a,b,c$ from $R_1$, $R_2$and $R_3$ respectively, we have: \[\Rightarrow \Delta \text{=abc}\left| \begin{matrix} \text{-a} & \text{b} & \text{c} \\ \text{a} & \text{-b} & \text{c} \\ \text{a} & \text{b} & \text{-c} \\ \end{matrix} \right|\] Similarly, taking out $a,b,c$ from $C_1$, $C_2$ and $C_3$ respectively, we have: \[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left| \begin{matrix} \text{-1} & \text{1} & \text{1} \\ \text{1} & \text{-1} & \text{1} \\ \text{1} & \text{1} & \text{-1} \\ \end{matrix} \right|\] Applying the row operations, \[R_2\to R_2\text{+ }R_1\] and \[R_3\to R_3\text{+}R_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left| \begin{matrix} \text{-1} & \text{1} & \text{1} \\ \text{0} & \text{0} & \text{2} \\ \text{0} & \text{2} & \text{0} \\ \end{matrix} \right|\] Solving it, we get: \[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left( \text{-1} \right)\left| \begin{matrix} \text{0} & \text{2} \\ \text{2} & \text{0} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\left( \text{-1} \right)\left( \text{0-4} \right)\] \[\therefore \Delta \text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\] Hence, \[\left| \begin{matrix} \text{-}{{\text{a}}^{\text{2}}} & \text{ab} & \text{ac} \\ \text{ba} & \text{-}{{\text{b}}^{\text{2}}} & \text{bc} \\ \text{ca} & \text{cb} & \text{-}{{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\text{=4}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}\].

8. By using properties of determinants, show that:

i. \[\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}} \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|=\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\]

Ans: Let\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}} \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] Applying the row operations, \[R_1\to R_1-R_3\] and \[R_2\to R_2-R_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{a-c} & {{\text{a}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}} \\ \text{0} & \text{b-c} & {{\text{b}}^{\text{2}}}\text{-}{{\text{c}}^{\text{2}}} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}\left( \text{c}-\text{a} \right)\left( \text{b}-c \right)\left| \begin{matrix} \text{0} & -\text{1} & -\left( a+c \right) \\ \text{0} & \text{1} & \text{b+c} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] Applying \[R_1\to R_1+R_2\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\left| \begin{matrix} \text{0} & \text{0} & -a+b \\ \text{0} & \text{1} & \text{b+c} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] Taking out \[\left( \text{a}-\text{b} \right)\] common from \[R_1\] \[\Rightarrow \Delta \text{=}\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\left| \begin{matrix} \text{0} & \text{0} & -\text{1} \\ \text{0} & \text{1} & \text{b+c} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|\] Expanding along \[\text{C}_1\], \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{0} & \text{-1} \\ \text{1} & \text{b+c} \\ \end{matrix} \right|\] \[\therefore \Delta \text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\] Hence, \[\left| \begin{matrix} \text{1} & \text{a} & {{\text{a}}^{\text{2}}} \\ \text{1} & \text{b} & {{\text{b}}^{\text{2}}} \\ \text{1} & \text{c} & {{\text{c}}^{\text{2}}} \\ \end{matrix} \right|=\left( \text{a}-\text{b} \right)\left( \text{b}-\text{c} \right)\left( \text{c}-\text{a} \right)\]

ii.\[\left| \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{a} & \text{b} & \text{c} \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{a} & \text{b} & \text{c} \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Applying the column operations, \[\text{C}_1\to \text{C}_1-\text{C}_3\] and \[C_2\to C_2-\text{C}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ \text{a-c} & \text{b-c} & \text{c} \\ {{\text{a}}^{\text{3}}}\text{-}{{\text{c}}^{\text{3}}} & {{\text{b}}^{\text{3}}}\text{-}{{\text{c}}^{\text{3}}} & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ \text{a-c} & \text{b-c} & \text{c} \\ \left( \text{a-c} \right)\left( {{\text{a}}^{\text{2}}}\text{+ac+}{{\text{c}}^{\text{2}}} \right) & \left( \text{b-c} \right)\left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Taking out \[\left( c-a \right)\] and \[\left( b-c \right)\] common from \[C_1\] and \[C_2\] respectively, \[\Rightarrow \Delta \text{=}\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ -1 & \text{1} & \text{c} \\ -\left( {{\text{a}}^{\text{2}}}\text{+ac+}{{\text{c}}^{\text{2}}} \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Applying \[C_1\to C_1+C_2\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ 0 & \text{1} & \text{c} \\ \left( {{b}^{2}}-{{a}^{2}} \right)+c\left( b-a \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Taking out $\left( a-b \right)$ common from \[C_1\], we get: \[\Rightarrow \Delta \text{=}\left( a-b \right)\left( c-a \right)\left( b-c \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{c} \\ \text{-}\left( \text{a+b+c} \right) & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Again taking out $\left( a+b+c \right)$ common from \[C_1\], we get: \[\Rightarrow \Delta \text{=}\left( a-b \right)\left( b-c \right)\left( c-a \right)\left| \begin{matrix} \text{0} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{c} \\ -1 & \left( {{\text{b}}^{\text{2}}}\text{+bc+}{{\text{c}}^{\text{2}}} \right) & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\] Expanding along \[C_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( \text{a+b+c} \right)\left( \text{-1} \right)\left| \begin{matrix} \text{0} & \text{1} \\ \text{1} & \text{c} \\ \end{matrix} \right|\] \[\therefore \Delta \text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)\] Hence, \[\left| \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{a} & \text{b} & \text{c} \\ {{\text{a}}^{\text{3}}} & {{\text{b}}^{\text{3}}} & {{\text{c}}^{\text{3}}} \\ \end{matrix} \right|\text{=}\left( \text{a-b} \right)\left( \text{b-c} \right)\left( \text{c-a} \right)\left( \text{a+b+c} \right)\]

9. By using properties of determinants, show that:

\[\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx} \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy} \\ \end{matrix} \right|\text{=}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\left( \text{xy+yz+zx} \right)\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx} \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy} \\ \end{matrix} \right|\] Applying \[\text{R}_2\to \text{R}_2-\text{R}_1\] and \[\text{R}_3\to \text{R}_3-\text{R}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{y-x} & {{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}} & \text{zx-yz} \\ \text{z-x} & {{\text{z}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}} & \text{xy-yz} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{-}\left( \text{x-y} \right) & \text{-}\left( \text{x-y} \right)\left( \text{x+y} \right) & \text{z}\left( \text{x-y} \right) \\ \left( \text{z-x} \right) & \left( \text{z-x} \right)\left( \text{z+x} \right) & \text{-y}\left( \text{z-x} \right) \\ \end{matrix} \right|\] Taking out \[\left( x-y \right)\] and \[\left( z-x \right)\] common from \[R_2\] and \[R_3\] respectively \[\Rightarrow \Delta \text{=}\left( x-y \right)\left( z-x \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ -1 & -x-y & \text{z} \\ \text{1} & z+x & -y \\ \end{matrix} \right|\] Applying \[R_3\to R_3+R_2\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( x-y \right)\left( z-x \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{-1} & \text{-x-y} & \text{z} \\ 0 & z-y & z-y \\ \end{matrix} \right|\] Taking out \[\left( z-y \right)\] common from \[R_3\], we get: \[\Rightarrow \Delta \text{=}\left( x-y \right)\left( z-x \right)\left( z-y \right)\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ -1 & -x-y & \text{z} \\ \text{0} & \text{1} & \text{1} \\ \end{matrix} \right|\] Expanding along \[R_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left[ \left( x-y \right)\left( z-x \right)\left( z-y \right) \right]\left[ \left( -1 \right)\left| \begin{matrix} \text{x} & \text{yz} \\ -1 & \text{z} \\ \end{matrix} \right|\text{+1}\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} \\ -1 & -x-y \\ \end{matrix} \right| \right]\] \[\Rightarrow \Delta \text{=}\left( \text{x-y} \right)\left( \text{z-x} \right)\left( \text{z-y} \right)\left[ \left( \text{-xz-yz} \right)\text{+}\left( \text{-}{{\text{x}}^{\text{2}}}\text{-xy+}{{\text{x}}^{\text{2}}} \right) \right]\] \[\Rightarrow \Delta \text{=}-\left( x-y \right)\left( z-x \right)\left( z-y \right)\left( \text{xy+yz+zx} \right)\] \[\therefore \Delta \text{=}\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( \text{xy+yz+zx} \right)\] Hence, \[\left| \begin{matrix} \text{x} & {{\text{x}}^{\text{2}}} & \text{yz} \\ \text{y} & {{\text{y}}^{\text{2}}} & \text{zx} \\ \text{z} & {{\text{z}}^{\text{2}}} & \text{xy} \\ \end{matrix} \right|\text{=}\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( \text{xy+yz+zx} \right)\]

10. By using properties of determinants, show that:

i. \[\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x} \\ \text{2x} & \text{x+4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x+4} \\ \end{matrix} \right|\text{=}\left( \text{5x+4} \right){{\left( \text{4-x} \right)}^{\text{2}}}\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x} \\ \text{2x} & \text{x+4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x+4} \\ \end{matrix} \right|\]

Applying the row operation,\[R_1\to R_1\text{+}R_2\text{+}R_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{5x+4} & \text{5x+4} & \text{5x+4} \\ \text{2x} & \text{x+4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x+4} \\ \end{matrix} \right|\] Taking out $\left( 5x+4 \right)$ common from \[R_1\] \[\Rightarrow \Delta \text{=}\left( \text{5x+4} \right)\left| \begin{matrix} \text{1} & 1 & 1 \\ \text{2x} & \text{x+4} & 2x \\ \text{2x} & \text{2x} & \text{x+4} \\ \end{matrix} \right|\] Applying the column operations, \[C_2\to C_2-C_1\] and \[\text{C}_3\to C_3-\text{C}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{5x+4} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{2x} & 4-x & \text{0} \\ \text{2x} & \text{0} & 4-x \\ \end{matrix} \right|\] Taking out $\left( 4-x \right)$ common from \[C_2\] and \[C_3\] respectively, \[\Rightarrow \Delta \text{=}\left( \text{5x+4} \right)\left( \text{4-x} \right)\left( \text{4-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{2x} & \text{1} & \text{0} \\ \text{2x} & \text{0} & \text{1} \\ \end{matrix} \right|\] Expanding along \[C_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{5x+4} \right){{\left( 4-x \right)}^{\text{2}}}\left| \begin{matrix} \text{1} & \text{0} \\ \text{2x} & \text{1} \\ \end{matrix} \right|\] \[\therefore \Delta \text{=}\left( \text{5x+4} \right){{\left( 4-x \right)}^{\text{2}}}\] Hence, \[\left| \begin{matrix} \text{x+4} & \text{2x} & \text{2x} \\ \text{2x} & \text{x+4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x+4} \\ \end{matrix} \right|\text{=}\left( \text{5x+4} \right){{\left( \text{4-x} \right)}^{\text{2}}}\]

ii. \[\left| \begin{matrix} \text{y+k} & \text{y} & \text{y} \\ \text{y} & \text{y+k} & \text{y} \\ \text{y} & \text{y} & \text{y+k} \\ \end{matrix} \right|\text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\]

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{y+k} & \text{y} & \text{y} \\ \text{y} & \text{y+k} & \text{y} \\ \text{y} & \text{y} & \text{y+k} \\ \end{matrix} \right|\] Applying the row operation, \[\text{R}_1\to \text{R}_1\text{+R}_2\text{+R}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{3y+k} & \text{3y+k} & \text{3y+k} \\ \text{y} & \text{y+k} & \text{y} \\ \text{y} & \text{y} & \text{y+k} \\ \end{matrix} \right|\] Taking out $\left( 3y+k \right)$ common from \[R_1\] \[\Rightarrow \Delta \text{=}\left( \text{3y+k} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{y} & \text{y+k} & \text{y} \\ \text{y} & \text{y} & \text{y+k} \\ \end{matrix} \right|\] Applying \[C_2\to C_2-\text{C}_1\] and \[C_3\to C_3-\text{C}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{3y+k} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{y} & \text{k} & \text{0} \\ \text{y} & \text{0} & \text{k} \\ \end{matrix} \right|\] Taking out $\left( k \right)$ common from \[C_2\] and \[C_3\] respectively, \[\Rightarrow \Delta \text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{y} & \text{1} & \text{0} \\ \text{y} & \text{0} & \text{1} \\ \end{matrix} \right|\] Expanding along \[C_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\left| \begin{matrix} \text{1} & \text{0} \\ \text{y} & \text{1} \\ \end{matrix} \right|\] \[\therefore \Delta \text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\] Hence, \[\left| \begin{matrix} \text{y+k} & \text{y} & \text{y} \\ \text{y} & \text{y+k} & \text{y} \\ \text{y} & \text{y} & \text{y+k} \\ \end{matrix} \right|\text{=}{{\text{k}}^{\text{2}}}\left( \text{3x+k} \right)\].

11. By using properties of determinants, show that:

i. \[\left| \begin{matrix} \text{a-b-c} & \text{2a} & \text{2a}\\ \text{2b} & \text{b-c-a} & \text{2b} \\ \text{2c} & \text{2c} & \text{c-a-b} \\ \end{matrix} \right|\text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\]

Ans: Let\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a+b+c} & \text{a+b+c} & \text{a+b+c} \\ \text{2b} & \text{b-c-a} & \text{2b} \\ \text{2c} & \text{2c} & \text{c-a-b} \\ \end{matrix} \right|\] Applying the row operation, \[\text{R}_1\to \text{R}_1\text{+ R}_2\text{+R}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{a+b+c} & \text{a+b+c} & \text{a+b+c} \\ \text{2b} & \text{b-c-a} & \text{2b} \\ \text{2c} & \text{2c} & \text{c-a-b} \\ \end{matrix} \right|\] Taking out $\left( a+b+c \right)$ common from \[R_1\] \[\Rightarrow \Delta \text{=}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{2b} & \text{b-c-a} & \text{2b} \\ \text{2c} & \text{2c} & \text{c-a-b} \\ \end{matrix} \right|\] Applying the column operations, \[C_2\to C_2-\text{C}_1\] and \[\text{C}_3\to \text{C}_3-\text{C}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{a+b+c} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{2b} & \text{-}\left( \text{a+b+c} \right) & \text{0} \\ \text{2c} & \text{0} & \text{-}\left( \text{a+b+c} \right) \\ \end{matrix} \right|\] Taking out $\left( a+b+c \right)$ common from \[C_2\] and \[C_3\] respectively, \[\Rightarrow \Delta \text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{2b} & \text{-1} & \text{0} \\ \text{2c} & \text{0} & \text{-1} \\ \end{matrix} \right|\] Expanding along \[\text{C}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{a+b+c} \right)}^{\text{3}}}\left( \text{-1} \right)\left( \text{-1} \right)\] \[\therefore \Delta \text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\] Hence, \[\left| \begin{matrix} \text{a-b-c} & \text{2a} & \text{2a} \\ \text{2b} & \text{b-c-a} & \text{2b} \\ \text{2c} & \text{2c} & \text{c-a-b} \\ \end{matrix} \right|\text{=}{{\left( \text{a+b+c} \right)}^{\text{3}}}\].

ii. \[\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y} \\ \text{z} & \text{y+z+2x} & \text{y} \\ \text{z} & \text{x} & \text{z+x+2y} \\ \end{matrix} \right|\text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\]

Ans:  Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y} \\ \text{z} & \text{y+z+2z} & \text{y} \\ \text{z} & \text{x} & \text{z+x+2y} \\ \end{matrix} \right|\] Applying the column operation, \[C_1\to C_1\text{+}C_2\text{+}C_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y+z} \right) & \text{x} & \text{y} \\ \text{2}\left( \text{x+y+z} \right) & \text{y+z+2x} & \text{y} \\ \text{2}\left( \text{x+y+z} \right) & \text{x} & \text{z+x+2y} \\ \end{matrix} \right|\] Taking out $2\left( x+y+z \right)$ common from \[C_1\] \[\Rightarrow \Delta \text{=2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{1} & \text{y+z+2x} & \text{y} \\ \text{1} & \text{x} & \text{z+x+2y} \\ \end{matrix} \right|\] Applying \[R_2\to R_2-R_1\] and \[R_3\to R_3-R_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{0} & \text{x+y+z} & \text{0} \\ \text{0} & \text{0} & \text{x+y+z} \\ \end{matrix} \right|\] Taking out $\left( x+y+z \right)$ common from \[R_2\] and \[R_3\] respectively, \[\Rightarrow \Delta \text{=2}\left( \text{x+y+z} \right)\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right|\] Expanding along \[R_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\left( \text{1} \right)\left( \text{1-0} \right)\] \[\therefore \Delta \text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\] Hence, \[\left| \begin{matrix} \text{x+y+2z} & \text{x} & \text{y} \\ \text{z} & \text{y+z+2x} & \text{y} \\ \text{z} & \text{x} & \text{z+x+2y} \\ \end{matrix} \right|\text{=2}{{\left( \text{x+y+z} \right)}^{\text{3}}}\].

12. By using properties of determinants, show that:

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

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & {{\text{x}}^{\text{2}}} \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x} \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1} \\ \end{matrix} \right|\] Applying \[R_1\to R_1\text{+}R_2+R_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+x+}{{\text{x}}^{\text{2}}} & \text{1+x+}{{\text{x}}^{\text{2}}} & \text{1+x+}{{\text{x}}^{\text{2}}} \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x} \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1} \\ \end{matrix} \right|\] Taking out $\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)$ common from \[R_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left| \begin{matrix} 1 & 1 & 1 \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x} \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1} \\ \end{matrix} \right|\] Applying \[C_2\to C_2-\text{C}_1\] and \[\text{C}_3\to C_3-\text{C}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ {{\text{x}}^{\text{2}}} & \text{1-}{{\text{x}}^{\text{2}}} & \text{x-}{{\text{x}}^{\text{2}}} \\ \text{x} & {{\text{x}}^{\text{2}}}\text{-x} & \text{1-x} \\ \end{matrix} \right|\] Taking out $\left( 1-x \right)$ common from \[C_2\] and \[C_3\] respectively, \[\Rightarrow \Delta \text{=}\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\left( \text{1-x} \right)\left( \text{1-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ {{\text{x}}^{\text{2}}} & \text{1+x} & \text{x} \\ \text{x} & \text{-x} & \text{1} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-x} \right)\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ {{\text{x}}^{\text{2}}} & \text{1+x} & \text{x} \\ \text{x} & \text{-x} & \text{1} \\ \end{matrix} \right|\] Expanding along \[\text{R}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( 1-x \right)\left( \text{1} \right)\left| \begin{matrix} \text{1+x} & \text{x} \\ -x & \text{1} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-x} \right)\left( \text{1+x+}{{\text{x}}^{\text{2}}} \right)\] \[\Rightarrow \Delta \text{=}\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\left( \text{1-}{{\text{x}}^{\text{3}}} \right)\] \[\therefore \Delta \text{=}{{\left( \text{1-}{{\text{x}}^{\text{3}}} \right)}^{\text{2}}}\] Hence, \[\left| \begin{matrix} \text{1} & \text{x} & {{\text{x}}^{\text{2}}} \\ {{\text{x}}^{\text{2}}} & \text{1} & \text{x} \\ \text{x} & {{\text{x}}^{\text{2}}} & \text{1} \\ \end{matrix} \right|\text{=}{{\left( \text{1-}{{\text{x}}^{\text{3}}} \right)}^{\text{2}}}\].

13. By using properties of determinants, show that:

\[\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b} \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a} \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|\text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b} \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a} \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|\] Applying the row operations, \[\text{R}_1\to \text{R}_1\text{+bR}_3\] and \[\text{R}_2\to \text{R}_2-\text{aR}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{0} & \text{-b}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right) \\ \text{0} & \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{a}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right) \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|\] Taking out $\left( \text{1+}{{\text{a}}^{2}}\text{+}{{\text{b}}^{\text{2}}} \right)$ common from \[R_1\] and \[R_2\] respectively, \[\Rightarrow \Delta \text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left| \begin{matrix} \text{1} & \text{0} & \text{-b} \\ \text{0} & \text{1} & \text{a} \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|\] Expanding along \[R_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)\left[ \left( \text{1} \right)\left| \begin{matrix} \text{1} & \text{a} \\ -\text{2a} & \text{1}-{{\text{a}}^{\text{2}}}-{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|-b\left| \begin{matrix} \text{0} & \text{1} \\ \text{2b} & -2a \\ \end{matrix} \right| \right]\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left[ \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{+2}{{\text{a}}^{\text{2}}}\text{-b}\left( \text{-2b} \right) \right]\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{2}}\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)\] \[\therefore \text{ }\!\!\Delta\!\!\text{ =}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}\] Hence, \[\left| \begin{matrix} \text{1+}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} & \text{2ab} & \text{-2b} \\ \text{2ab} & \text{1-}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} & \text{2a} \\ \text{2b} & \text{-2a} & \text{1-}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}} \\ \end{matrix} \right|\text{=}{{\left( \text{1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \right)}^{\text{3}}}\].

14. By using properties of determinants, show that:

\[\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac} \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc} \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1} \\ \end{matrix} \right|\text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\].

Ans: Let \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac} \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc} \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1} \\ \end{matrix} \right|\] Taking out \[\text{a,b}\] and \[\text{c}\] from \[R_1\] , \[R_2\] and \[R_3\] respectively \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a+}\dfrac{\text{1}}{\text{a}} & \text{b} & \text{c} \\ \text{a} & \text{b+}\dfrac{\text{1}}{\text{b}} & \text{c} \\ \text{a} & \text{b} & \text{c+}\dfrac{\text{1}}{\text{c}} \\ \end{matrix} \right|\] Applying \[\text{R}_2\to \text{R}_2-\text{R}_1\] and \[\text{R}_3\to \text{R}_3-\text{R}_1\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc}\left| \begin{matrix} \text{a+}\dfrac{\text{1}}{\text{a}} & \text{b} & \text{c} \\ \text{-}\dfrac{\text{1}}{\text{a}} & \dfrac{\text{1}}{\text{b}} & \text{0} \\ \text{-}\dfrac{\text{1}}{\text{a}} & \text{0} & \dfrac{\text{1}}{\text{c}} \\ \end{matrix} \right|\] Applying \[\text{C}_1\to \text{aC}_1\] , \[C_2\to \text{bC}_2\] and \[\text{C}_3\to \text{cC}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =abc }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{abc}}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & {{\text{b}}^{\text{2}}} & {{\text{c}}^{\text{2}}} \\ \text{-1} & \text{1} & \text{0} \\ \text{-1} & \text{0} & \text{1} \\ \end{matrix} \right|\] Expanding along \[\text{R}_3\] \[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-1}\left| \begin{matrix} {{\text{b}}^{\text{2}}} & {{\text{c}}^{\text{2}}} \\ \text{1} & \text{0} \\ \end{matrix} \right|\text{+1}\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & {{\text{b}}^{\text{2}}} \\ \text{-1} & \text{1} \\ \end{matrix} \right|\] \[\Rightarrow \Delta \text{=1}\left( \text{-}{{\text{c}}^{\text{2}}} \right)\text{+}\left( {{\text{a}}^{\text{2}}}\text{+1+}{{\text{b}}^{\text{2}}} \right)\] \[\therefore \Delta \text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\] Hence, \[\left| \begin{matrix} {{\text{a}}^{\text{2}}}\text{+1} & \text{ab} & \text{ac} \\ \text{ab} & {{\text{b}}^{\text{2}}}\text{+1} & \text{bc} \\ \text{ca} & \text{cb} & {{\text{c}}^{\text{2}}}\text{+1} \\ \end{matrix} \right|\text{=1+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\].

15. Choose the correct answer. Let \[\text{A}\] be a square matrix of order \[\text{3 }\!\!\times\!\!\text{ 3}\] , then \[\left| \text{kA} \right|\] is equal to  \[\text{k}\left| \text{A} \right|\]

1. \[{{\text{k}}^{\text{2}}}\left| \text{A} \right|\]

2. \[{{\text{k}}^{\text{3}}}\left| \text{A} \right|\]

3. \[\text{3k}\left| \text{A} \right|\]

Ans: Since, \[\text{A}\] is a square matrix of order \[\text{3 }\!\!\times\!\!\text{ 3}\] . Let us suppose \[\text{A=}\left[ \begin{matrix} \text{a}_1 & \text{b}_1 & \text{c}_1 \\ \text{a}_2 & \text{b}_2 & \text{c}_2 \\ \text{a}_3 & \text{b}_3 & \text{c}_3 \\ \end{matrix} \right]\] Thus, \[\text{kA=}\left[ \begin{matrix} \text{ka}_1 & \text{kb}_1 & \text{kc}_1 \\ \text{ka}_2 & \text{kb}_2 & \text{kc}_2 \\ \text{ka}_3 & \text{kb}_3 & \text{kc}_3 \\ \end{matrix} \right]\] \[\therefore \left| \text{kA} \right|\text{=}\left| \begin{matrix} \text{ka}_1 & \text{kb}_1 & \text{kc}1 \\ \text{ka}_2 & \text{kb}_2 & \text{kc}2 \\ \text{ka}_3 & \text{kb}_3 & \text{kc}3 \\ \end{matrix} \right|\] Taking out $\left( k \right)$ common from each row, we have: \[\Rightarrow \left| kA \right|\text{=}{{\text{k}}^{\text{3}}}\left| \begin{matrix} \text{a}_1 & \text{b}_1 & \text{c}_1 \\ \text{a}_2 & \text{b}_2 & \text{c}_2 \\ \text{a}_3 & \text{b}_3 & \text{c}_3 \\ \end{matrix} \right|\] \[\therefore \left| \text{kA} \right|\text{=}{{\text{k}}^{\text{3}}}\left| \text{A} \right|\] Hence, B. \[{{\text{k}}^{\text{3}}}\left| \text{A} \right|\] is the correct option.

16. Which of the following is correct?

1. Determinant is a square matrix.

2. Determinant is a number associated to a matrix.

3. Determinant is a number associated to a square matrix.

4. None of these.

Ans: For every square matrix, \[\text{A=}\left[ \text{aij} \right]\] of order \[\text{n}\], we can determine or associate a value which is termed as determinant of square matrix A, where \[\text{aij=}{{\left( \text{i,j} \right)}^{\text{th}}}\] element of \[\text{A}\] .

Thus, the determinant is a number associated with a square matrix.

Hence, C. Determinant is a number associated to a square matrix is the correct option.

NCERT Solutions For Class 12 Math Chapter 4 Exercise 4.2

NCERT Solutions for Class 12 Math Chapter 4 Determinants exercise 4.2 contains solutions for all questions covered in the exercise in the pdf format. NCERT Solutions for Class 12 Math Chapter 4 Determinants exercise 4.2 care solved by subject experts at Vedantu in a simple way.. Solving through these questions repeatedly can give students the confidence to write the exams better. Students can download the NCERT Solutions for Class 12 Math  Chapter 4 Exercise 4.2 and practice offline anytime and from anywhere. The questions given in Chapter 4 Exercise 4.2 is based on the topic “Different Properties of Determinants”.

What are Determinants?

In Math, determinants are determined as the scalar quantities. It is obtained by the sum of the elements of a square matrix and their cofactors according to the prescribed rule.

What are the Different Properties of Determinants?

The different properties of Determinants includes:

• Reflection Property

• All Zero Property

• Proportionality Property

• Switching Property

• Scalar Multiple Property

• Factor Property

• Sum Property

• Triangle Property

• Property of Invariance

• Determinants of Cofactor Matrix

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solutions Class 12 Maths Chapter 4 – Other Exercises

The NCERT Solutions Class 12 Chapter 4 Exercise 4.2 Determinants will help students to get a better understanding of the types of questions that are commonly asked from this topic in the exams. Students should download and refer to the Class 12 Maths Chapter 4 Exercise 4.2 for getting a clear idea of the concepts of Determinants. 

All the students can refer to these NCERT Solutions to perform well in their exams and score high marks. If you have any doubts while studying math or while solving the questions given in the exercise in the Maths books, you can refer to the solutions on Vedantu for a detailed stepwise analysis of the sums. Hence, to ensure a better learning experience, download  NCERT Solutions for Maths Class 12 Chapter 4 Exercise 4.2 Determinant.

Properties of Determinants

To solve the sums given in the ex 4.2 class 12 maths solutions, students should know the properties of determinants thoroughly. As mentioned before, a determinant is a special number that can be found by using a square matrix.

The determinant of a matrix, also known as P, can be denoted by det(P) or det P. Determinants have important properties that play a role in permitting an individual to get the same results with simpler and different configurations of entries or elements.

Students should know about these properties if they want to find class 12 maths NCERT solutions. To help students out, we have created a list of those properties. And that list is mentioned below.

• Reflection Property

According to this property, the determinant remains unaltered if the rows are changed into columns. For this property to be effective, the columns should also turn into rows.

• All-Zero Property

If all the elements of a row or column are zero, the value of the determinant is also zero.

• Proportionality or Repetition Property

This property is important for writing NCERT class 12 maths exercise 4.2 solutions. According to this property, if all the elements of a column or row are identical or proportional to the elements of some other column or row, then the value of the determinant is zero.

• Switching Property

The interchange of any two columns or rows of the determinant results in a change in the sign.

• Scalar Multiple Property

If all the elements of a column or row of a determinant are multiplied by a non-zero constant, then the determinant will get multiplied by the same constant.

There are also other properties that students should know about if they want to solve all class 12 maths ch 4 ex 4.2 questions.

Class 12 Determinants Exercise 4.2

Class 12th is an important time for a student. And if a student wants to get the most out of that time, then he or she should focus on getting the best marks in his or her final examination. To achieve this aim in mathematics, students should solve all questions of chapter 4 exercise 4.2.

But what if you need some extra help? In that case, we at Vedantu are always here for all students. We at Vedantu provide students with NCERT solutions class 12 maths pdf files. Students can download these files for free from Vedantu.

Apart from NCERT solutions for class 12 maths exercise 4.2, we also provide solutions for other exercises of the same chapter. There are exercises 4.1, exercise 4.3, exercise 4.4, exercise 4.5, and exercise 4.6 in this chapter. All of these exercises have roughly five questions each.

Why Should You Download Exercise 4.2 Class 12 Maths Answers From Vedantu?

There are a lot of benefits with the NCERT Maths Solutions For Class 12 Chapter 4 Exercise 4.2 Determinants. Some of the benefits of the solutions are: 

• All answers are completely accurate and reliable

• Answers are written by the most talented subject matter experts in India

• We also provide live classes and 24x7 academic assistance to students

• The answers are present in a pdf file, which is very easy to download

• All answers are accompanied by an explanation section. This section provides extra clarity to the answer, and it also highlights the process that the expert followed to arrive at the answer