EXERCISE 4.4 (Determinants)

Write Minors and Cofactors of the elements of following determinants:(Class 12 ncert solution math exercise 4.4)

Question 1: (i) $\left|\begin{array}{cc}2 & -4 \\ 0 & 3\end{array}\right|$
(ii) $\left|\begin{array}{ll}a & c \\ b & d\end{array}\right|$

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

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 $\left|\begin{array}{ll}a & c \\ b & d\end{array}\right|$

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

$A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(d)=d$
$A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(b)=-b$
$A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(c)=-c$
$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(a)=a$

Question 2: (i) $\left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|$

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

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

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

$M_{11}=\text { minor of element } a_{11}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1$

$M_{12}=\text { minor of element } a_{12}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0$

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

$M_{21}=\text { minor of element } a_{21}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0$

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

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

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

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

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

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 $\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

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

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

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

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

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

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

$M_{23}=\text { minor of element } a_{23}=\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right|=1$

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

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

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

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

$\Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|$

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

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

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

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

$A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(7)=7$
$M_{23}=\text { minor of element } a_{23}=\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=7$

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

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

Therefore,

$\Delta =a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23}$
$=2(7)+0(7)+1(-7)$
$=14-7$
$=7$

Question 4: Using Cofactors of elements of second row, evaluate

$\Delta=\left|\begin{array}{ccc} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right|$

Solution: The given determinant is $\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|$

Therefore,

$M_{13}=\left|\begin{array}{ll} 1 & y \\ 1 & z \end{array}\right|=z-y$

$M_{23}=\left|\begin{array}{ll} 1 & x \\ 1 & z \end{array}\right|=z-x$

$M_{33}=\left|\begin{array}{ll} 1 & x \\ 1 & y \end{array}\right|=y-x$

$A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(z-y)=z-y$
$A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(z-x)=-(z-x)=x-z$
$A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(y-x)=y-x$

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

Therefore,

$\Delta &=a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33}$

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

Hence,

$\Delta=(x-y)(y-z)(z-x)$

Question 5: If $\Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|$ and $A_{i j}$ is Cofactors of $a_{ij}$ , then value of Δ 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. $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$