Exercise 4.4

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Question 1

Find the adjoint of the matrix

$$ A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}. $$

Solution:

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

Step 1: Find the minors of all the elements.

Minor of the element $1$:

$$ M_{11}= \begin{vmatrix} 4 \end{vmatrix} =4. $$

Minor of the element $2$:

$$ M_{12}= \begin{vmatrix} 3 \end{vmatrix} =3. $$

Minor of the element $3$:

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

Minor of the element $4$:

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

Step 2: Find the cofactors.

Using

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

we get

$$ A_{11}=4, $$ $$ A_{12}=-3, $$ $$ A_{21}=-2, $$ $$ A_{22}=1. $$

Hence, the cofactor matrix is

$$ \begin{bmatrix} 4&-3\\ -2&1 \end{bmatrix}. $$

Step 3: Find the adjoint.

The adjoint of a matrix is the transpose of its cofactor matrix.

$$ \operatorname{adj}(A) = \begin{bmatrix} 4&-2\\ -3&1 \end{bmatrix}. $$

Hence, the adjoint of the given matrix is

$$ \boxed{ \operatorname{adj}(A)= \begin{bmatrix} 4&-2\\ -3&1 \end{bmatrix} } $$

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Question 2

Find the adjoint of the matrix

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

Solution:

We know that the adjoint of a matrix is the transpose of its cofactor matrix.

Step 1: Find the cofactors of all the elements.

Cofactor of the element $1$:

$$ \begin{aligned} A_{11} &=(-1)^{1+1} \begin{vmatrix} 3&5\\ 0&1 \end{vmatrix} \\[2mm] &=3. \end{aligned} $$

Cofactor of the element $-1$:

$$ \begin{aligned} A_{12} &=(-1)^{1+2} \begin{vmatrix} 2&5\\ -2&1 \end{vmatrix} \\[2mm] &=-\left(2+10\right) \\[2mm] &=-12. \end{aligned} $$

Cofactor of the element $2$:

$$ \begin{aligned} A_{13} &=(-1)^{1+3} \begin{vmatrix} 2&3\\ -2&0 \end{vmatrix} \\[2mm] &=0+6 \\[2mm] &=6. \end{aligned} $$

Cofactor of the element $2$:

$$ \begin{aligned} A_{21} &=(-1)^{2+1} \begin{vmatrix} -1&2\\ 0&1 \end{vmatrix} \\[2mm] &=-(-1) \\[2mm] &=1. \end{aligned} $$

Cofactor of the element $3$:

$$ \begin{aligned} A_{22} &=(-1)^{2+2} \begin{vmatrix} 1&2\\ -2&1 \end{vmatrix} \\[2mm] &=1+4 \\[2mm] &=5. \end{aligned} $$

Cofactor of the element $5$:

$$ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 1&-1\\ -2&0 \end{vmatrix} \\[2mm] &=-(0-2) \\[2mm] &=2. \end{aligned} $$

Cofactor of the element $-2$:

$$ \begin{aligned} A_{31} &=(-1)^{3+1} \begin{vmatrix} -1&2\\ 3&5 \end{vmatrix} \\[2mm] &=-5-6 \\[2mm] &=-11. \end{aligned} $$

Cofactor of the element $0$:

$$ \begin{aligned} A_{32} &=(-1)^{3+2} \begin{vmatrix} 1&2\\ 2&5 \end{vmatrix} \\[2mm] &=-(5-4) \\[2mm] &=-1. \end{aligned} $$

Cofactor of the element $1$:

$$ \begin{aligned} A_{33} &=(-1)^{3+3} \begin{vmatrix} 1&-1\\ 2&3 \end{vmatrix} \\[2mm] &=3+2 \\[2mm] &=5. \end{aligned} $$

Step 2: Form the cofactor matrix.

$$ \text{Cofactor Matrix} = \begin{bmatrix} 3&-12&6\\ 1&5&2\\ -11&-1&5 \end{bmatrix}. $$

Step 3: Find the adjoint.

The adjoint is the transpose of the cofactor matrix.

$$ \operatorname{adj}(A) = \begin{bmatrix} 3&1&-11\\ -12&5&-1\\ 6&2&5 \end{bmatrix}. $$

Hence, the adjoint of the given matrix is

$$ \boxed{ \operatorname{adj}(A)= \begin{bmatrix} 3&1&-11\\ -12&5&-1\\ 6&2&5 \end{bmatrix} } $$

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Question 3

Verify that

$$ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I, $$

where

$$ A= \begin{bmatrix} 2&3\\ -4&-6 \end{bmatrix}. $$

Solution:

Step 1: Find the determinant of $A$.

$$ \begin{aligned} |A| &= \begin{vmatrix} 2&3\\ -4&-6 \end{vmatrix} \\[2mm] &=2(-6)-3(-4) \\[2mm] &=-12+12 \\[2mm] &=0. \end{aligned} $$

Hence,

$$ |A|I= 0 \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}. $$

Step 2: Find the adjoint of $A$.

For a matrix

$$ \begin{bmatrix} a&b\\ c&d \end{bmatrix}, $$

its adjoint is

$$ \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}. $$

Therefore,

$$ \operatorname{adj}(A) = \begin{bmatrix} -6&-3\\ 4&2 \end{bmatrix}. $$

Step 3: Find $A(\operatorname{adj}A)$.

$$ \begin{aligned} A(\operatorname{adj}A) &= \begin{bmatrix} 2&3\\ -4&-6 \end{bmatrix} \begin{bmatrix} -6&-3\\ 4&2 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 2(-6)+3(4) & 2(-3)+3(2)\\ (-4)(-6)+(-6)(4) & (-4)(-3)+(-6)(2) \end{bmatrix} \\[2mm] &= \begin{bmatrix} -12+12 & -6+6\\ 24-24 & 12-12 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}. \end{aligned} $$

Step 4: Find $(\operatorname{adj}A)A$.

$$ \begin{aligned} (\operatorname{adj}A)A &= \begin{bmatrix} -6&-3\\ 4&2 \end{bmatrix} \begin{bmatrix} 2&3\\ -4&-6 \end{bmatrix} \\[2mm] &= \begin{bmatrix} (-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6)\\ 4(2)+2(-4) & 4(3)+2(-6) \end{bmatrix} \\[2mm] &= \begin{bmatrix} -12+12 & -18+18\\ 8-8 & 12-12 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}. \end{aligned} $$

Since

$$ A(\operatorname{adj}A) = (\operatorname{adj}A)A = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} = |A|I, $$

therefore,

$$ \boxed{ A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I. } $$

Hence verified.

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Question 4

Verify that

$$ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I, $$

where

$$ A= \begin{bmatrix} 1&-1&2\\ 3&0&-2\\ 1&0&3 \end{bmatrix}. $$

Solution:

Step 1: Find the determinant of $A$.

Expanding along the first row,

$$ \begin{aligned} |A| &= 1 \begin{vmatrix} 0&-2\\ 0&3 \end{vmatrix} -(-1) \begin{vmatrix} 3&-2\\ 1&3 \end{vmatrix} + 2 \begin{vmatrix} 3&0\\ 1&0 \end{vmatrix} \\[2mm] &= 1(0) +(9+2) +2(0) \\[2mm] &=11. \end{aligned} $$

Hence,

$$ |A|I= 11 \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix}. $$

Step 2: Find the cofactors.

$$ A_{11}=0, \qquad A_{12}=-11, \qquad A_{13}=0, $$ $$ A_{21}=3, \qquad A_{22}=1, \qquad A_{23}=-1, $$ $$ A_{31}=2, \qquad A_{32}=8, \qquad A_{33}=3. $$

Therefore, the cofactor matrix is

$$ \begin{bmatrix} 0&-11&0\\ 3&1&-1\\ 2&8&3 \end{bmatrix}. $$

The adjoint of $A$ is the transpose of the cofactor matrix.

$$ \operatorname{adj}(A) = \begin{bmatrix} 0&3&2\\ -11&1&8\\ 0&-1&3 \end{bmatrix}. $$

Step 3: Find $A(\operatorname{adj}A)$.

$$ \begin{aligned} A(\operatorname{adj}A) &= \begin{bmatrix} 1&-1&2\\ 3&0&-2\\ 1&0&3 \end{bmatrix} \begin{bmatrix} 0&3&2\\ -11&1&8\\ 0&-1&3 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix}. \end{aligned} $$

Step 4: Find $(\operatorname{adj}A)A$.

$$ \begin{aligned} (\operatorname{adj}A)A &= \begin{bmatrix} 0&3&2\\ -11&1&8\\ 0&-1&3 \end{bmatrix} \begin{bmatrix} 1&-1&2\\ 3&0&-2\\ 1&0&3 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix}. \end{aligned} $$

Thus,

$$ A(\operatorname{adj}A) = (\operatorname{adj}A)A = \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix} = |A|I. $$

Hence,

$$ \boxed{ A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I. } $$

Hence verified.

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