Question 1

Write Minors and Cofactors of the elements of following determinants:

(i) $\left|\begin{array}{cc}2 & -4 \ 0 & 3\end{array}ight|$

(ii) $\left|\begin{array}{ll}a & c \ b & d\end{array}ight|$

Accepted Answer

(i) The given determinant is $\left|\begin{array}{cc}2 & -4 \ 0 & 3\end{array}ight|$

Minor of element $a_{i j}$ is $M_{i j}$.

$M_{11}=$ 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 $A_{i j}=(-1)^{i+j} M_{i j}$

$A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(3)=3$

$A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(0)=0$

$A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(-4)=4$

$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(2)=2$

(ii) The given determinant is $\left|\begin{array}{ll}a & c \ b & d\end{array}ight|$

Minor of element $a_{i j}$ is $M_{i j}$.

$M_{11}=$ 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 $A_{i j}=(-1)^{i+j} M_{i j}$

$$

\begin{aligned}

& 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

\end{aligned}

$$

Question 2

Write Minors and Cofactors of the elements of following determinants:

(i) $\left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight|$

(ii) $\left|\begin{array}{ccc}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}ight|$

Accepted Answer

(i) The given determinant is $\left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight|$

Minor of element $a_{i j}$ is $M_{i j}$.

$$

\begin{aligned}

& M_{11}=	ext { minor of element } \quad a_{11}=\left|\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight|=1 \

& M_{12}=	ext { minor of element } \quad a_{12}=\left|\begin{array}{ll}

0 & 0 \

0 & 1

\end{array}ight|=0 \

& M_{13}=	ext { minor of element } \quad a_{13}=\left|\begin{array}{ll}

0 & 1 \

0 & 0

\end{array}ight|=0 \

& M_{21}=	ext { minor of element } \quad a_{21}=\left|\begin{array}{ll}

0 & 0 \

0 & 1

\end{array}ight|=0 \

& M_{22}=	ext { minor of element } \quad a_{22}=\left|\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight|=1 \

& M_{23}=	ext { minor of element } \quad a_{23}=\left|\begin{array}{ll}

1 & 0 \

0 & 0

\end{array}ight|=0

\end{aligned}

$$

$M_{31}=$ minor of element $a_{31}=\left|\begin{array}{ll}0 & 0 \ 1 & 0\end{array}ight|=0$

$M_{32}=$ minor of element $a_{32}=\left|\begin{array}{ll}1 & 0 \ 0 & 0\end{array}ight|=0$

$M_{33}=$ minor of element $a_{33}=\left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight|=1$

Cofactor of $a_{i j}$ is $A_{i j}=(-1)^{i+j} M_{i j}$

$A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(1)=1$

$A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(0)=0$

$A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(0)=0$

$A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(0)=0$

$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(1)=1$

$A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(0)=0$

$A_{31}=(-1)^{3+1} M_{31}=(-1)^{4}(0)=0$

$A_{32}=(-1)^{3+2} M_{32}=(-1)^{5}(0)=0$

$A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(1)=1$

(ii) The given determinant is $\left|\begin{array}{ccc}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}ight|$

Minor of element $a_{i j}$ is $M_{i j}$.

$M_{11}=$ minor of element $a_{11}=\left|\begin{array}{cc}5 & -1 \ 1 & 2\end{array}ight|=11$

$M_{12}=$ minor of element $a_{12}=\left|\begin{array}{cc}3 & -1 \ 0 & 2\end{array}ight|=6$

$M_{13}=$ minor of element $a_{13}=\left|\begin{array}{ll}3 & 5 \ 0 & 1\end{array}ight|=3$

$M_{21}=$ minor of element $a_{21}=\left|\begin{array}{ll}0 & 4 \ 1 & 2\end{array}ight|=-4$

$M_{22}=$ minor of element $a_{22}=\left|\begin{array}{ll}1 & 4 \ 0 & 2\end{array}ight|=2$

$M_{23}=$ minor of element $a_{23}=\left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight|=1$

$M_{31}=$ minor of element $a_{31}=\left|\begin{array}{cc}0 & 4 \ 5 & -1\end{array}ight|=-20$

$M_{32}=$ minor of element $a_{32}=\left|\begin{array}{cc}1 & 4 \ 3 & -1\end{array}ight|=-13$

$M_{33}=$ minor of element $a_{33}=\left|\begin{array}{ll}1 & 0 \ 3 & 5\end{array}ight|=5$

Cofactor of $a_{i j}$ is $A_{i j}=(-1)^{i+j} M_{i j}$

$A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(11)=11$

$A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(6)=-6$

$A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(3)=3$

$A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(-4)=4$

$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(2)=2$

$A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(1)=-1$

$A_{31}=(-1)^{3+1} M_{31}=(-1)^{4}(-20)=-20$

$A_{32}=(-1)^{3+2} M_{32}=(-1)^{5}(-13)=13$

$A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(5)=5$

Question 3

Using Cofactors of elements of second row, evaluate $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}ight|$

Accepted Answer

The given determinant is $\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}ight|$

$M_{21}=$ minor of element $a_{21}=\left|\begin{array}{ll}3 & 8 \ 2 & 3\end{array}ight|=-7$

$A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(-7)=7$

$M_{22}=$ minor of element $a_{22}=\left|\begin{array}{ll}5 & 8 \ 1 & 3\end{array}ight|=15-8=7$

$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(7)=7$

$M_{23}=$ minor of element $a_{23}=\left|\begin{array}{ll}5 & 3 \ 1 & 2\end{array}ight|=7$

$A_{23}=(-1)^{2+3} M_{21}=(-1)^{5}(7)=-7$

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

Therefore,

$$

\begin{aligned}

\Delta & =a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23} \

& =2(7)+0(7)+1(-7) \

& =14-7 \

& =7

\end{aligned}

$$

Question 4

Using Cofactors of elements of third column, evaluate $\Delta=\left|\begin{array}{ccc}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}ight|$

Accepted Answer

The given determinant is $\left|\begin{array}{ccc}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}ight|$

Therefore,

$$

\begin{aligned}

M_{13} & =\left|\begin{array}{ll}

1 & y \

1 & z

\end{array}ight|=z-y \

M_{23} & =\left|\begin{array}{ll}

1 & x \

1 & z

\end{array}ight|=z-x \

M_{33} & =\left|\begin{array}{ll}

1 & x \

1 & y

\end{array}ight|=y-x

\end{aligned}

$$

\begin{aligned}

& A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(z-y)=z-y \

& A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(z-x)=-(z-x)=x-z \

& A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(y-x)=y-x

\end{aligned}

$$

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

Therefore,

$$

\begin{aligned}

\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 \

& =\left(x^{2} z-y^{2} zight)+\left(y z^{2}-x z^{2}ight)+\left(x y^{2}-x^{2} yight) \

& =z\left(x^{2}-y^{2}ight)+z^{2}(y-x)+x y(y-x) \

& =z(x-y)(x+y)+z^{2}(y-x)+x y(y-x) \

& =(x-y)\left[z x+z y-z^{2}-x yight] \

& =(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)

$$

Question 5

$\Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33}\end{array}ight|$ and $A_{i j}$ is the cofactor of $a_{i j}$, then the value of $\Delta$ is given by:

A. $a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}$

B. $a_{11} A_{11}+a_{12} A_{21}+a_{13} A_{31}$

C. $a_{21} A_{11}+a_{22} A_{12}+a_{23} A_{13}$

D. $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$

Accepted Answer

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

$$

\Delta=a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}

$$

Thus, the correct option is D.

\section*{EXERCISE 4.5}