HSC 12th Mathematics: Matrices (Exercise 2.2, Q5 & Q6)
Question 5
Find the inverse of the following matrix by the adjoint method:
$$A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$$
Solution:
First, we find the determinant of matrix $A$:
$$|A| = \begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix} = (2)(3) - (-2)(4) = 6 + 8 = 14$$
Since $|A| = 14 \neq 0$, the inverse $A^{-1}$ exists.
Next, we find the cofactors $A_{ij} = (-1)^{i+j} M_{ij}$ of all elements:
$A_{11} = (-1)^{1+1}(3) = 3$
$A_{12} = (-1)^{1+2}(4) = -4$
$A_{21} = (-1)^{2+1}(-2) = 2$
$A_{22} = (-1)^{2+2}(2) = 2$
The cofactor matrix is:
$$C = \begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix}$$
The adjoint of matrix $A$ is the transpose of the cofactor matrix:
$$\text{adj}(A) = C^T = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$$
Using the formula for the inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$$
Question 6
Find the inverse of the following matrix by the adjoint method:
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$$
Solution:
First, find the determinant of matrix $A$:
$$|A| = 1 \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} + 3 \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix}$$
$$|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10$$
Since $|A| = 10 \neq 0$, the inverse $A^{-1}$ exists.
Now, we calculate the cofactors $A_{ij} = (-1)^{i+j} M_{ij}$ for each element:
$A_{11} = (-1)^{1+1}(10 - 0) = 10$
$A_{12} = (-1)^{1+2}(0 - 0) = 0$
$A_{13} = (-1)^{1+3}(0 - 0) = 0$
$A_{21} = (-1)^{2+1}(10 - 0) = -10$
$A_{22} = (-1)^{2+2}(5 - 0) = 5$
$A_{23} = (-1)^{2+3}(0 - 0) = 0$
$A_{31} = (-1)^{3+1}(8 - 6) = 2$
$A_{32} = (-1)^{3+2}(4 - 0) = -4$
$A_{33} = (-1)^{3+3}(2 - 0) = 2$
The cofactor matrix $C$ is:
$$C = \begin{bmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{bmatrix}$$
The adjoint of $A$ is the transpose of the cofactor matrix:
$$\text{adj}(A) = C^T = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$
Finally, finding the inverse using $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$$A^{-1} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$
