Solution

(i) The given determinants is $\begin{vmatrix}2 & -4 \\ 0 & 3\end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$

$\therefore M _{11}=$ minor of element $a _{11}=3$

$M _{12}=$ minor of element $a _{12}=0$

$M _{21}=$ minor of element $a _{21}=-4$

$M _{22}=$ minor of element $a _{22}=2$

Cofactor of $a _{i j}$ is $A _{i j}=(-1)^{i+j} M _{i j}$.

$\therefore A _{11}=(-1)^{1+1} M _{11}=(-1)^{2}(3)=3$

$A _{12}=(-1)^{1+2} M _{12}=(-1)^{3}(0)=0$

$A _{21}=(-1)^{2+1} M _{21}=(-1)^{3}(-4)=4$

$A _{22}=(-1)^{2+2} M _{22}=(-1)^{4}(2)=2$

(ii) The given determinant is $ \begin{vmatrix} a & c \\ b & d\end{vmatrix} $. Minor of element $a _{i j}$ is $M _{i j}$. $\therefore M _{11}=$ minor of element $a _{11}=d$

$M _{12}=$ minor of element $a _{12}=b$

$M _{21}=$ minor of element $a _{21}=c$

$M _{22}=$ minor of element $a _{22}=a$

Cofactor of $a _{i j}$ is $A _{i j}=(-1)^{i+j} M _{i j}$.

$ \therefore A _{11}=(-1)^{1+1} M _{11}=(-1)^{2}(d)=d $

$A _{12}=(-1)^{1+2} M _{12}=(-1)^{3}(b)=-b$

$A _{21}=(-1)^{2+1} M _{21}=(-1)^{3}(c)=-c$

$A _{22}=(-1)^{2+2} M _{22}=(-1)^{4}(a)=a$

Solution

(i) The given determinant is $ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix} $.

By the definition of minors and cofactors, we have:

$M _{11}=$ minor of $a _{11}= \begin{vmatrix} 1 & 0 \\ 0 & 1\end{vmatrix} =1$

$M _{12}=$ minor of $a _{12}= \begin{vmatrix} 0 & 0 \\ 0 & 1\end{vmatrix} =0$

$ \begin{aligned} & M _{13}=\text{ minor of } a _{13}= \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} =0 \\ & M _{21}=\text{ minor of } a _{21}= \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} =0 \\ & M _{22}=\text{ minor of } a _{22}= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} =1 \\ & M _{23}=\text{ minor of } a _{23}= \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} =0 \\ & M _{31}=\text{ minor of } a _{31}= \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} =0 \\ & M _{32}=\text{ minor of } a _{32}= \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} =0 \\ & M _{33}=\text{ minor of } a _{33}= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} =1 \\ & A _{11}=\text{ cofactor of } a _{11}=(-1)^{1+1} M _{11}=1 \\ & A _{12}=\text{ cofactor of } a _{12}=(-1)^{1+2} M _{12}=0 \\ & A _{13}=\text{ cofactor of } a _{13}=(-1)^{1+3} M _{13}=0 \\ & A _{21}=\text{ cofactor of } a _{21}=(-1)^{2+1} M _{21}=0 \\ & A _{22}=\text{ cofactor of } a _{22}=(-1)^{2+2} M _{22}=1 \\ & A _{23}=\text{ cofactor of } a _{23}=(-1)^{2+3} M _{23}=0 \\ & A _{31}=\text{ cofactor of } a _{31}=(-1)^{3+1} M _{31}=0 \\ & A _{32}=\text{ cofactor of } a _{32}=(-1)^{3+2} M _{32}=0 \\ & A _{33}=\text{ cofactor of } a _{33}=(-1)^{3+3} M _{33}=1 \\ & \text{ (ii) The given determinant is } \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix} \text{. } \end{aligned} $

By definition of minors and cofactors, we have:

$ M _{11}=\text{ minor of } a _{11}= \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} =10+1=11 $

$ \begin{aligned} & M _{12}=\text{ minor of } a _{12}= \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} =6-0=6 \\ & M _{13}=\text{ minor of } a _{13}= \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} =3-0=3 \\ & M _{21}=\text{ minor of } a _{21}= \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} =0-4=-4 \\ & M _{22}=\text{ minor of } a _{22}= \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} =2-0=2 \\ & M _{23}=\text{ minor of } a _{23}= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} =1-0=1 \\ & M _{31}=\text{ minor of } a _{31}= \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} =0-20=-20 \\ & M _{32}=\text{ minor of } a _{32}= \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} =-1-12=-13 \\ & A _{33}= \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} =5-0=5 \\ & A _{32}=\text{ cofactor of } a _{32}=(-1)^{3+2} M _{32}=13 \\ & A _{11}=\text{ cofactor of } a _{33}=(-1)^{3+3} M _{33}=5 \\ & A _{12}=\text{ cofactor of } a _{12}=(-1)^{1+2} M _{12}=-6 \\ & A _{13}=\text{ cofactor of } a _{13}=(-1)^{1+3} M _{13}=3 \\ & A _{21}=\text{ cofactor of } a _{21}=(-1)^{2+1} M _{21}=4 \\ & A _{22}=\text{ cofactor of } a _{22}=(-1)^{2+2} M _{22}=2 \\ & A _{23}=\text{ cofactor of } a _{23}=(-1)^{2+3} M _{23}=-1 \\ & M _{11}=11 \end{aligned} $

Solution

We have:

$M _{21}= \begin{vmatrix} 3 & 8 \\ 2 & 3\end{vmatrix} =9-16=-7$

$\therefore A _{21}=$ cofactor of $a _{21}=(-1)^{2+1} M _{21}=7$

$M _{22}= \begin{vmatrix} 5 & 8 \\ 1 & 3\end{vmatrix} =15-8=7$

$\therefore A _{22}=$ cofactor of $a _{22}=(-1)^{2+2} M _{22}=7$

$M _{23}= \begin{vmatrix} 5 & 3 \\ 1 & 2\end{vmatrix} =10-3=7$

$\therefore A _{23}=$ cofactor of $a _{23}=(-1)^{2+3} M _{23}=-7$

We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors. $\therefore \Delta=a _{21} A _{21}+a _{22} A _{22}+a _{23} A _{23}=2(7)+0(7)+1(-7)=14-7=7$

Solution

The given determinant is $ \begin{vmatrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{vmatrix} $.

We have:

$ \begin{aligned} & M _{13}= \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} =z-y \\ & M _{23}= \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} =z-x \\ & M _{33}= \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} =y-x \\ & \therefore A _{13}=\text{ cofactor of } a _{13}=(-1)^{1+3} M _{13}=(z-y) \end{aligned} $

$A _{23}=$ cofactor of $a _{23}=(-1)^{2+3} M _{23}=-(z-x)=(x-z)$

$A _{33}=$ cofactor of $a _{33}=(-1)^{3+3} M _{33}=(y-x)$

We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

$ \begin{aligned} \therefore \Delta & =a _{13} A _{13}+a _{23} A _{23}+a _{33} A _{33} \\ & =y z(z-y)+z x(x-z)+x y(y-x) \\ & =y z^{2}-y^{2} z+x^{2} z-x z^{2}+x y^{2}-x^{2} y \\ & =(x^{2} z-y^{2} z)+(y z^{2}-x z^{2})+(x y^{2}-x^{2} y) \\ & =z(x^{2}-y^{2})+z^{2}(y-x)+x y(y-x) \\ & =z(x-y)(x+y)+z^{2}(y-x)+x y(y-x) \\ & =(x-y)[z x+z y-z^{2}-x y] \\ & =(x-y)[z(x-z)+y(z-x)] \\ & =(x-y)(z-x)[-z+y] \\ & =(x-y)(y-z)(z-x) \end{aligned} $

Hence, $\Delta=(x-y)(y-z)(z-x)$.

Solution

Given,

$\begin{aligned} & \Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \\ & =a_{11}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|-a_{12}\left|\begin{array}{ll} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right|+a_{13}\left|\begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right| \\ & =a_{11}(-1)^{1+1}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|+a_{12}(-1)^{1+2}\left|\begin{array}{ll} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right|+a_{13}(-1)^{1+3} & \left|\begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right| \\ & =a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31} \end{aligned}$

So, the correct option is D.