Adjoint of a Matrix (Practice Problems)

The adjoint (or adjugate) of a matrix is the transpose matrix of the cofactor of the given matrix. For any square matrix A, to calculate its adjoint matrix, we have to first calculate the cofactor matrix of the given matrix and then find its determinant.

Solved Examples

Q1. A=\begin{bmatrix} 5 & 3\\ -2 & 4 \end{bmatrix}

Solution:

For A=\begin{bmatrix} a & c\\ b & d \end{bmatrix} ,

adj(A)= \begin{bmatrix} d & -c\\ -b & a \end{bmatrix}

Therefore:

Adj (A) = \begin{bmatrix} 4 & -3\\ 2 & 5 \end{bmatrix}

Q2. A=\begin{bmatrix} 2 & 1 & 4\\ 1 & 0 & 2\\ 3 & 2 & 1 \end{bmatrix}A=\begin{bmatrix} 4 & 3\\ 2 & 5 \end{bmatrix}

Solution:

The cofactor matrix is : \begin{bmatrix} -4 & 7 & 2\\ 5 & -10 & 0\\ 2 & -1 & -1 \end{bmatrix}

Therefore,

adj(A)=C^T

\begin{bmatrix} -4 & 5 & 2\\ 7 & -10 & -1\\ 2 & 0 & -1 \end{bmatrix}

Also,

det(A)=1.

Q3. A=\begin{bmatrix} k & 3\\ 2 & 5 \end{bmatrix}

Given:

det(A)=14.

Solution: Therefore,

5k−(2)(3)=14

5k−6=14

5k=20

k=4

Thus,

A=\begin{bmatrix} 4 & 3\\ 2 & 5 \end{bmatrix}

\text{Hence,}\quad \operatorname{adj}(A)= \begin{bmatrix} 5 & -2\\ -3 & 4 \end{bmatrix}

Q4. Find adj⁡(A), where A=\begin{pmatrix} 2 & 0 & 1\\ -1 & 4 & 2\\ 3 & 5 & -2 \end{pmatrix}

Solution:

First, find the matrix of cofactors.

C11 ​=​ \begin{pmatrix} 4 & 5\\ 2 & -2 \end{pmatrix}=−8 − 10 = −18

C12​ = - \begin{pmatrix} 0 & 1\\ 5 & -2 \end{pmatrix} = - (-5) = 5

C13 ​= \begin{pmatrix} 0 & 1\\ 4 & 2 \end{pmatrix}

Similarly,

C21​ = −\begin{pmatrix} -1 & 2\\ 3 & -2 \end{pmatrix} = − (−4−6) = 10

C22 ​= \begin{pmatrix} 2 & 1\\ 3 & -2 \end{pmatrix} = − 4 − 3 = −7

C23​ = − \begin{pmatrix} 2 & 1\\ -1 & 2 \end{pmatrix} = − 5

C31 ​= \begin{pmatrix} -1 & 4\\ 3 & 5 \end{pmatrix} = − 5 − 12 = −17

C32​ = − \begin{pmatrix} 2 & 0\\ 3 & 5 \end{pmatrix} = −10

C33​ = \begin{pmatrix} 2 & 0\\ -1 & 4 \end{pmatrix} = 8

Thus, the cofactor matrix is : C= \begin{pmatrix} -18 & 10 & -17\\ 5 & -7 & -10\\ -4 & -5 & 8 \end{pmatrix}

Q 5. Find the Adjoint of the given matrix A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}.

Solution:

Step 1: To find the cofactor of each element
To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of elements at A[1, 1] = 1: +\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix}          = +(4 x 9 - 8 x 5) = -4
Cofactor of elements at A[1, 2] = 2: -\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix}          = -(7 x 9 - 6 x 5) = -33
Cofactor of elements at A[1, 3] = 3: +\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}          = +(7 x 8 - 6 x 4) = 32
Cofactor of elements at A[2, 1] = 7: -\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix}          = -(2 x 9 - 8 x 3) = 6
Cofactor of elements at A[2, 2] = 4: +\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix}          = +(1 x 9 - 6 x 3) = -9
Cofactor of elements at A[2, 3] = 5: -\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix}          = -(1 x 8 - 6 x 2) = 4
Cofactor of elements at A[3, 1] = 6: +\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}          = +(2 x 5 - 4 x 3) = -2
Cofactor of elements at A[3, 2] = 8: -\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix}          = -(1 x 5 - 7 x 3) = 16
Cofactor of elements at A[3, 3] = 9: +\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix}          = +(1 x 4 - 7 x 2) = -10

The matrix looks like with the cofactors:

A =\begin{bmatrix} +\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix} & -\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix} & +\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}\\ \\ -\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix} & +\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix} & -\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix} \\ \\         +\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} & -\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix} & +\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix} \end{bmatrix}

The final cofactor matrix:
A =\begin{bmatrix} -4 & -33 & 32\\ 6 & -9 & 4 \\ -2 & 16 & -10 \end{bmatrix}

Step 2: Find the transpose of the matrix obtained in step 1
adj(A) =\begin{bmatrix} -4 & 6 & -2\\ -33 & -9 & 16 \\ 32 & 4 & -10 \end{bmatrix}

This is the Adjoint of the matrix.

Q 6. Find the Adjoint of the given matrix A =\begin{bmatrix} -1 & -2 & -2\\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}.

Solution:

Step 1: To find the cofactor of each element

To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of element at A[1, 1] = -1: +\begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}          = +(1 x 1 - (-2) x (-2)) = -3
Cofactor of elements at A[1, 2] = -2: -\begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix}          = -(2 x 1 - 2 x (-2)) = -6
Cofactor of elements at A[1, 3] = -2: +\begin{bmatrix} 2 & 1 \\ 2 & -2 \end{bmatrix}          = +(2 x (-2) - 2 x 1) = -6
Cofactor of elements at A[2, 1] = 2: -\begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}          = -((-2) x 1 - (-2) x (-2)) = 6
Cofactor of elements at A[2, 2] = 1:  +\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}           = +((-1) x 1 - 2 x (-2)) = 3
Cofactor of elements at A[2, 3] = -2: -\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}           = -((-1) x (-2) - 2 x (-2)) = -6
Cofactor of elements at A[3, 1] = 2: +\begin{bmatrix} -2 & -2 \\ 1 & -2 \end{bmatrix}          = +((-2) x (-2) - 1 x (-2)) = 6
Cofactor of elements at A[3, 2] = -2: -\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}           = -((-1) x (-2) - 2 x (-2)) = -6
Cofactor of elements at A[3, 3] = 1: +\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}          = +((-1) x (-1)- 2 x (-2)) = 3

The final cofactor matrix:
A =\begin{bmatrix} -3 & -6 & -6\\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}

Step 2: Find the transpose of the matrix obtained in Step 1
adj(A) =\begin{bmatrix} -3 & 6 & 6\\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}

This is the Adjoint of the matrix.

Practice Questions

Q1. A = \begin{bmatrix} 4 & 2\\ 1 & 3 \end{bmatrix}

Q2. A = \begin{bmatrix} 3 & 5\\ 2 & 4 \end{bmatrix}

Q3. A=\begin{bmatrix} 1 & 2 & 3\\ 2 & 1 & 0\\ 3 & 0 & 1 \end{bmatrix}

Q4. Find the adjoint of : A= \begin{pmatrix} 2 & 0 & 5\\ 1 & 4 & 2\\ 3 & -1 & 1 \end{pmatrix}

Q5. A= \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 2\\ 3 & 2 & 1 \end{pmatrix} find adj⁡(A)\operatorname{adj}(A) and verify that:

Q6. Find x if A= \begin{pmatrix} x & 1 & 1\\ 1 & x & 1\\ 1 & 1 & x \end{pmatrix} is singular. Then find adj⁡(A)\operatorname{adj}(A) for those values of x.