HSC 12th Mathematics: Matrices (Exercise 2.2, Q3)
Question 3 (i)
Find the adjoint of the following matrix:
$$A = \begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$$
Solution:
Let $A = \begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$
We first find the cofactors of all elements. The cofactor is given by $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of element $a_{ij}$.
$A_{11} = (-1)^{1+1} (5) = 1(5) = 5$
$A_{12} = (-1)^{1+2} (3) = -1(3) = -3$
$A_{21} = (-1)^{2+1} (-3) = -1(-3) = 3$
$A_{22} = (-1)^{2+2} (2) = 1(2) = 2$
The matrix of cofactors, let's call it $C$, is:
$$C = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} 5 & -3 \\ 3 & 2 \end{bmatrix}$$
The adjoint of matrix $A$ is the transpose of the cofactor matrix ($C^T$):
$$\text{adj}(A) = C^T = \begin{bmatrix} 5 & 3 \\ -3 & 2 \end{bmatrix}$$
Question 3 (ii)
Find the adjoint of the following matrix:
$$A = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{bmatrix}$$
Solution:
Let $A = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{bmatrix}$
We will calculate the cofactors $A_{ij} = (-1)^{i+j} M_{ij}$ for each element in the matrix.
For the first row:
$A_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 5 \\ 0 & -1 \end{vmatrix} = 1[(3)(-1) - (5)(0)] = -3 - 0 = -3$
$A_{12} = (-1)^{1+2} \begin{vmatrix} -2 & 5 \\ -2 & -1 \end{vmatrix} = -1[(-2)(-1) - (5)(-2)] = -(2 - (-10)) = -12$
$A_{13} = (-1)^{1+3} \begin{vmatrix} -2 & 3 \\ -2 & 0 \end{vmatrix} = 1[(-2)(0) - (3)(-2)] = 0 - (-6) = 6$
For the second row:
$A_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & -1 \end{vmatrix} = -1[(-1)(-1) - (2)(0)] = -(1 - 0) = -1$
$A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = 1[(1)(-1) - (2)(-2)] = -1 - (-4) = -1 + 4 = 3$
$A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -1[(1)(0) - (-1)(-2)] = -(0 - 2) = 2$
For the third row:
$A_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = 1[(-1)(5) - (2)(3)] = -5 - 6 = -11$
$A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ -2 & 5 \end{vmatrix} = -1[(1)(5) - (2)(-2)] = -(5 - (-4)) = -(5 + 4) = -9$
$A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ -2 & 3 \end{vmatrix} = 1[(1)(3) - (-1)(-2)] = 3 - 2 = 1$
The cofactor matrix $C$ is:
$$C = \begin{bmatrix} -3 & -12 & 6 \\ -1 & 3 & 2 \\ -11 & -9 & 1 \end{bmatrix}$$
The adjoint of matrix $A$ is the transpose of the cofactor matrix ($C^T$):
$$\text{adj}(A) = C^T = \begin{bmatrix} -3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{bmatrix}$$
