NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4 - Determinants

Question 1 (i) Write Minors and Cofactors of the elements of following determinants:

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

Answer:

GIven determinant: $\left|\begin{array}{cc}2& -4\\ 0& 3\end{array}\right|$

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

Therefore we have

$M_{11}$ = 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

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1{)}^{i+j}{M}_{ij}$.

Therefore, we have:

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

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

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

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

Question 1 (ii) Write Minors and Cofactors of the elements of following determinants:

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

Answer:

GIven determinant: $\left|\begin{array}{cc}a& c\\ b& d\end{array}\right|$

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

Therefore we have

$M_{11}$ = 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

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1{)}^{i+j}{M}_{ij}$.

Therefore, we have:

$A_{11}=(-1{)}^{1+1}{M}_{11}=(-1{)}^2(d)=d$

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

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

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

Question 2 (i) Write Minors and Cofactors of the elements of following determinants:

$\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

Answer:

Given determinant : $\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

Finding Minors: by the definition,

$M_{11}=$ minor of $a_{11}=\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1$ $M_{12}=$ minor of $a_{12}=\left|\begin{array}{cc}0& 0\\ 0& 1\end{array}\right|=0$

$M_{13}=$ minor of $a_{13}=\left|\begin{array}{cc}0& 1\\ 0& 0\end{array}\right|=0$ $M_{21}=$ minor of $a_{21}=\left|\begin{array}{cc}0& 0\\ 0& 1\end{array}\right|=0$

$M_{22}=$ minor of $a_{22}=\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1$ $M_{23}=$ minor of $a_{23}=\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|=0$

$M_{31}=$ minor of $a_{31}=\left|\begin{array}{cc}0& 0\\ 1& 0\end{array}\right|=0$ $M_{32}=$ minor of $a_{32}=\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|=0$

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

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11}=(-1{)}^{1+1}{M}_{11}=1$

$A_{12}=$ cofactor of $a_{12}=(-1{)}^{1+2}{M}_{12}=0$

$A_{13}=$ cofactor of $a_{13}=(-1{)}^{1+3}{M}_{13}=0$

$A_{21}=$ cofactor of $a_{21}=(-1{)}^{2+1}{M}_{21}=0$

$A_{22}=$ cofactor of $a_{22}=(-1{)}^{2+2}{M}_{22}=1$

$A_{23}=$ cofactor of $a_{23}=(-1{)}^{2+3}{M}_{23}=0$

$A_{31}=$ cofactor of $a_{31}=(-1{)}^{3+1}{M}_{31}=0$

$A_{32}=$ cofactor of $a_{32}=(-1{)}^{3+2}{M}_{32}=0$

$A_{33}=$ cofactor of $a_{33}=(-1{)}^{3+3}{M}_{33}=1$.

Question:2(ii) Write Minors and Cofactors of the elements of following determinants:

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

Answer:

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

Finding Minors: by the definition,

$M_{11}=$ minor of $a_{11}=\left|\begin{array}{cc}5& -1\\ 1& 2\end{array}\right|=11$ $M_{12}=$ minor of $a_{12}=\left|\begin{array}{cc}3& -1\\ 0& 2\end{array}\right|=6$

$M_{13}=$ minor of $a_{13}=\left|\begin{array}{cc}3& 5\\ 0& 1\end{array}\right|=3$ $M_{21}=$ minor of $a_{21}=\left|\begin{array}{cc}0& 4\\ 1& 2\end{array}\right|=-4$

$M_{22}=$ minor of $a_{22}=\left|\begin{array}{cc}1& 4\\ 0& 2\end{array}\right|=2$ $M_{23}=$ minor of $a_{23}=\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1$

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

$M_{32}=$ minor of $a_{32}=\left|\begin{array}{cc}1& 4\\ 3& -1\end{array}\right|=-1-12=-13$

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

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11}=(-1{)}^{1+1}{M}_{11}=11$

$A_{12}=$ cofactor of $a_{12}=(-1{)}^{1+2}{M}_{12}=-6$

$A_{13}=$ cofactor of $a_{13}=(-1{)}^{1+3}{M}_{13}=3$

$A_{21}=$ cofactor of $a_{21}=(-1{)}^{2+1}{M}_{21}=4$

$A_{22}=$ cofactor of $a_{22}=(-1{)}^{2+2}{M}_{22}=2$

$A_{23}=$ cofactor of $a_{23}=(-1{)}^{2+3}{M}_{23}=-1$

$A_{31}=$ cofactor of $a_{31}=(-1{)}^{3+1}{M}_{31}=-20$

$A_{32}=$ cofactor of $a_{32}=(-1{)}^{3+2}{M}_{32}=13$

$A_{33}=$ cofactor of $a_{33}=(-1{)}^{3+3}{M}_{33}=5$.

Question:3 Using Cofactors of elements of second row, evaluate .$\mathrm{\Delta }=\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$

Answer:

Given determinant : $\mathrm{\Delta }=\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$

First finding Minors of the second rows by the definition,

$M_{21}=$ minor of $a_{21}=\left|\begin{array}{cc}3& 8\\ 2& 3\end{array}\right|=9-16=-7$

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

$M_{23}=$ minor of $a_{23}=\left|\begin{array}{cc}5& 3\\ 1& 2\end{array}\right|=10-3=7$

Finding the Cofactors of the second row:

$A_{21}=$ Cofactor of $a_{21}=(-1{)}^{2+1}{M}_{21}=7$

$A_{22}=$ Cofactor of $a_{22}=(-1{)}^{2+2}{M}_{22}=7$

$A_{23}=$ Cofactor of $a_{23}=(-1{)}^{2+3}{M}_{23}=-7$

Therefore we can calculate $\mathrm{△}$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

$\mathrm{△}=a_{21}{A}_{21}+a_{22}{A}_{22}+a_{23}{A}_{23}=2(7)+0(7)+1(-7)=14-7=7$

Question:4 Using Cofactors of elements of third column, evaluate $\mathrm{\Delta }=\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$

Answer:

Given determinant : $\mathrm{\Delta }=\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$

First finding Minors of the third column by the definition,

$M_{13}=$ minor of $a_{13}=\left|\begin{array}{cc}1& y\\ 1& z\end{array}\right|=z-y$

$M_{23}=$ minor of $a_{23}=\left|\begin{array}{cc}1& x\\ 1& z\end{array}\right|=z-x$

$M_{33}=$ minor of $a_{33}=\left|\begin{array}{cc}1& x\\ 1& y\end{array}\right|=y-x$

Finding the Cofactors of the second row:

$A_{13}=$ Cofactor of $a_{13}=(-1{)}^{1+3}{M}_{13}=z-y$

$A_{23}=$ Cofactor of $a_{23}=(-1{)}^{2+3}{M}_{23}=x-z$

$A_{33}=$ Cofactor of $a_{33}=(-1{)}^{3+3}{M}_{33}=y-x$

Therefore we can calculate $\mathrm{△}$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$\mathrm{△}=a_{13}{A}_{13}+a_{23}{A}_{23}+a_{33}{A}_{33}$

$=(z-y)yz+(x-z)zx+(y-x)xy$

$=y{z}^2-y^{2}z+z{x}^2-x{z}^2+x{y}^2-x^{2}y$

$=z(x^2-y^2)+z^2(y-x)+xy(y-x)$

$=(x-y)[zx+zy-z^2-xy]$

$=(x-y)[z(x-z)+y(z-x)]$

$=(x-y)(z-x)[-z+y]$

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

Thus, we have value of $\mathrm{△}=(x-y)(y-z)(z-x)$.

Question 5: If $\mathrm{\Delta }=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$ and $A_{ij}$ is the cofactor of $a_{ij}$, then the value of $\mathrm{\Delta }$ 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) $a_{11}{A}_{11}+a_{21}{A}_{21}+a_{31}{A}_{31}$

Answer: (D) $a_{11}{A}_{11}+a_{21}{A}_{21}+a_{31}{A}_{31}$

By the definition itself, $\mathrm{\Delta }$ is equal to the sum of the products of the elements of any row or column with their corresponding cofactors.