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MATHEMATICS CLASS- 12

CHAPTER-4
(DETERMINANTS)

CBSEChapter 4EXERCISE 4.3

Exercise 4.3



Question 1

Write the minors and cofactors of the elements of the following determinants.

(i)

Given determinant is

$$ \Delta= \begin{vmatrix} 2&-4\\ 0&3 \end{vmatrix}. $$

Solution:

The minor of an element is obtained by deleting the row and the column containing that element.

The cofactor of an element is given by

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

where $M_{ij}$ is the minor of the element.

Minor and cofactor of each element are as follows:

For the element $2$:

$$ M_{11}= \begin{vmatrix} 3 \end{vmatrix} =3. $$ $$ A_{11}=(-1)^{1+1}M_{11}=3. $$

For the element $-4$:

$$ M_{12}= \begin{vmatrix} 0 \end{vmatrix} =0. $$ $$ A_{12}=(-1)^{1+2}M_{12}=0. $$

For the element $0$:

$$ M_{21}= \begin{vmatrix} -4 \end{vmatrix} =-4. $$ $$ A_{21}=(-1)^{2+1}(-4)=4. $$

For the element $3$:

$$ M_{22}= \begin{vmatrix} 2 \end{vmatrix} =2. $$ $$ A_{22}=(-1)^{2+2}(2)=2. $$

Hence,

Element Minor Cofactor
$2$ $3$ $3$
$-4$ $0$ $0$
$0$ $-4$ $4$
$3$ $2$ $2$


(ii)

Given determinant is

$$ \Delta= \begin{vmatrix} a&c\\ b&d \end{vmatrix}. $$

Solution:

The minor of each element is obtained by deleting the corresponding row and column.

The cofactor is given by

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

Minor and cofactor of each element are as follows:

For the element $a$:

$$ M_{11}=d. $$ $$ A_{11}=d. $$

For the element $c$:

$$ M_{12}=b. $$ $$ A_{12}=-b. $$

For the element $b$:

$$ M_{21}=c. $$ $$ A_{21}=-c. $$

For the element $d$:

$$ M_{22}=a. $$ $$ A_{22}=a. $$

Hence,

Element Minor Cofactor
$a$ $d$ $d$
$c$ $b$ $-b$
$b$ $c$ $-c$
$d$ $a$ $a$


Question 2

Write the minors and cofactors of the elements of the following determinants.

(i)

Given determinant is

$$ \Delta= \begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{vmatrix}. $$

Solution:

The minor of an element is obtained by deleting the row and the column containing that element.

The cofactor of an element is given by

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

where $M_{ij}$ is the minor of the element.

Now, find the minor and cofactor of each element.

For the element $1$ at position $(1,1)$:

$$ M_{11}= \begin{vmatrix} 1&0\\ 0&1 \end{vmatrix} =1. $$ $$ A_{11}=1. $$

For the element $0$ at position $(1,2)$:

$$ M_{12}= \begin{vmatrix} 0&0\\ 0&1 \end{vmatrix} =0. $$ $$ A_{12}=0. $$

For the element $0$ at position $(1,3)$:

$$ M_{13}= \begin{vmatrix} 0&1\\ 0&0 \end{vmatrix} =0. $$ $$ A_{13}=0. $$

For the element $0$ at position $(2,1)$:

$$ M_{21}= \begin{vmatrix} 0&0\\ 0&1 \end{vmatrix} =0. $$ $$ A_{21}=0. $$

For the element $1$ at position $(2,2)$:

$$ M_{22}= \begin{vmatrix} 1&0\\ 0&1 \end{vmatrix} =1. $$ $$ A_{22}=1. $$

For the element $0$ at position $(2,3)$:

$$ M_{23}= \begin{vmatrix} 1&0\\ 0&0 \end{vmatrix} =0. $$ $$ A_{23}=0. $$

For the element $0$ at position $(3,1)$:

$$ M_{31}= \begin{vmatrix} 0&0\\ 1&0 \end{vmatrix} =0. $$ $$ A_{31}=0. $$

For the element $0$ at position $(3,2)$:

$$ M_{32}= \begin{vmatrix} 1&0\\ 0&0 \end{vmatrix} =0. $$ $$ A_{32}=0. $$

For the element $1$ at position $(3,3)$:

$$ M_{33}= \begin{vmatrix} 1&0\\ 0&1 \end{vmatrix} =1. $$ $$ A_{33}=1. $$

Hence, the matrix of minors is

$$ \boxed{ \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} } $$

and the matrix of cofactors is

$$ \boxed{ \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} } $$

(ii)

Given determinant is

$$ \Delta= \begin{vmatrix} 1&0&4\\ 3&5&-1\\ 0&1&2 \end{vmatrix}. $$

Solution:

Find the minor and cofactor of each element.

Element Minor Cofactor
$1$ $ \begin{vmatrix} 5&-1\\ 1&2 \end{vmatrix} =11 $ $11$
$0$ $ \begin{vmatrix} 3&-1\\ 0&2 \end{vmatrix} =6 $ $-6$
$4$ $ \begin{vmatrix} 3&5\\ 0&1 \end{vmatrix} =3 $ $3$
$3$ $ \begin{vmatrix} 0&4\\ 1&2 \end{vmatrix} =-4 $ $4$
$5$ $ \begin{vmatrix} 1&4\\ 0&2 \end{vmatrix} =2 $ $2$
$-1$ $ \begin{vmatrix} 1&0\\ 0&1 \end{vmatrix} =1 $ $-1$
$0$ $ \begin{vmatrix} 0&4\\ 5&-1 \end{vmatrix} =-20 $ $-20$
$1$ $ \begin{vmatrix} 1&4\\ 3&-1 \end{vmatrix} =-13 $ $13$
$2$ $ \begin{vmatrix} 1&0\\ 3&5 \end{vmatrix} =5 $ $5$

Hence, the matrix of minors is

$$ \boxed{ \begin{bmatrix} 11&6&3\\ -4&2&1\\ -20&-13&5 \end{bmatrix} } $$

and the matrix of cofactors is

$$ \boxed{ \begin{bmatrix} 11&-6&3\\ 4&2&-1\\ -20&13&5 \end{bmatrix} } $$

Question 3

Using cofactors of the elements of the second row, evaluate

$$ \Delta= \begin{vmatrix} 5&3&8\\ 2&0&1\\ 1&2&3 \end{vmatrix}. $$

Solution:

Expand the determinant along the second row.

Using the expansion formula,

$$ \Delta=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}. $$

Here,

$$ a_{21}=2,\qquad a_{22}=0,\qquad a_{23}=1. $$

Now find the cofactors.

Cofactor of $a_{21}$:

$$ \begin{aligned} A_{21} &=(-1)^{2+1} \begin{vmatrix} 3&8\\ 2&3 \end{vmatrix} \\[2mm] &=- (3\times3-8\times2) \\[2mm] &=-(9-16) \\[2mm] &=7. \end{aligned} $$

Cofactor of $a_{22}$:

$$ \begin{aligned} A_{22} &=(-1)^{2+2} \begin{vmatrix} 5&8\\ 1&3 \end{vmatrix} \\[2mm] &=15-8 \\[2mm] &=7. \end{aligned} $$

Cofactor of $a_{23}$:

$$ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 5&3\\ 1&2 \end{vmatrix} \\[2mm] &=- (10-3) \\[2mm] &=-7. \end{aligned} $$

Substituting these values in the expansion,

$$ \begin{aligned} \Delta &=2(7)+0(7)+1(-7) \\[2mm] &=14+0-7 \\[2mm] &=7. \end{aligned} $$

Hence,

$$ \boxed{ \Delta=7. } $$

Question 4

Using cofactors of the elements of the third column, evaluate

$$ \Delta= \begin{vmatrix} 1&x&yz\\ 1&y&zx\\ 1&z&xy \end{vmatrix}. $$

Solution:

Expand the determinant along the third column.

Using the expansion formula,

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

Here,

$$ a_{13}=yz,\qquad a_{23}=zx,\qquad a_{33}=xy. $$

Now find the cofactors.

Cofactor of $a_{13}$:

$$ \begin{aligned} A_{13} &=(-1)^{1+3} \begin{vmatrix} 1&y\\ 1&z \end{vmatrix} \\[2mm] &= 1(z)-1(y) \\[2mm] &=z-y. \end{aligned} $$

Cofactor of $a_{23}$:

$$ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 1&x\\ 1&z \end{vmatrix} \\[2mm] &=- (z-x) \\[2mm] &=x-z. \end{aligned} $$

Cofactor of $a_{33}$:

$$ \begin{aligned} A_{33} &=(-1)^{3+3} \begin{vmatrix} 1&x\\ 1&y \end{vmatrix} \\[2mm] &= y-x. \end{aligned} $$

Substituting these values,

$$ \begin{aligned} \Delta &=yz(z-y)+zx(x-z)+xy(y-x). \end{aligned} $$

Expanding the terms,

$$ \begin{aligned} \Delta &=yz^2-y^2z+x^2z-xz^2+xy^2-x^2y. \end{aligned} $$

Rearranging the terms,

$$ \begin{aligned} \Delta &=(x^2z-x^2y)+(xy^2-y^2z)+(yz^2-xz^2) \\[2mm] &=x^2(z-y)+y^2(x-z)+z^2(y-x). \end{aligned} $$

Using the identity

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

we obtain

$$ \boxed{ \Delta=(x-y)(y-z)(z-x). } $$

Hence, the value of the determinant is

$$ \boxed{(x-y)(y-z)(z-x).} $$

Question 5

If

$$ A= \begin{bmatrix} 1&2&4\\ 2&-1&3\\ 0&5&2 \end{bmatrix}, $$

then $A_{21}+A_{32}$ is equal to

(A) $0$
(B) $7$
(C) $14$
(D) $-7$

Solution:

We know that the cofactor of an element is given by

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

where $M_{ij}$ is the minor of the element.

Step 1: Find $A_{21}$.

Deleting the second row and first column, we get the minor

$$ M_{21}= \begin{vmatrix} 2&4\\ 5&2 \end{vmatrix}. $$

Evaluating the determinant,

$$ \begin{aligned} M_{21} &=2(2)-4(5) \\[2mm] &=4-20 \\[2mm] &=-16. \end{aligned} $$

Therefore,

$$ \begin{aligned} A_{21} &=(-1)^{2+1}M_{21} \\[2mm] &=-(-16) \\[2mm] &=16. \end{aligned} $$

Step 2: Find $A_{32}$.

Deleting the third row and second column, we get the minor

$$ M_{32}= \begin{vmatrix} 1&4\\ 2&3 \end{vmatrix}. $$

Evaluating the determinant,

$$ \begin{aligned} M_{32} &=1(3)-4(2) \\[2mm] &=3-8 \\[2mm] &=-5. \end{aligned} $$

Therefore,

$$ \begin{aligned} A_{32} &=(-1)^{3+2}M_{32} \\[2mm] &=-(-5) \\[2mm] &=5. \end{aligned} $$

Step 3: Find $A_{21}+A_{32}$.

$$ \begin{aligned} A_{21}+A_{32} &=16+5 \\[2mm] &=21. \end{aligned} $$

Thus,

$$ \boxed{A_{21}+A_{32}=21.} $$