Q8:

Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$

Step 1: Evaluation of the Determinant

Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$, we must first confirm the existence of $A^{-1}$ by calculating the determinant $|A|$. A matrix is invertible if and only if $|A| \neq 0$ [By the Invertibility Criterion].

Expanding along the first row ($R_1$):

$|A| = 1 \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} - 0 \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} + 0 \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix}$

$|A| = 1[(3)(-1) - (0)(2)] - 0 + 0$

$|A| = 1[-3 - 0] = -3$

Since $|A| = -3 \neq 0$, the inverse matrix $A^{-1}$ exists.

Step 2: Calculation of the Adjoint Matrix

The inverse is defined as $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. We compute the cofactor matrix $C_{ij} = (-1)^{i+j} M_{ij}$.

• $C_{11} = + \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = -3$
• $C_{12} = - \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = -(-3) = 3$
• $C_{13} = + \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = (6 - 15) = -9$
• $C_{21} = - \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = 0$
• $C_{22} = + \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = -1$
• $C_{23} = - \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = -(2 - 0) = -2$
• $C_{31} = + \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = 0$
• $C_{32} = - \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = 0$
• $C_{33} = + \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = 3$

The cofactor matrix is $C = \begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}$. The adjoint is the transpose of the cofactor matrix, $\text{adj}(A) = C^T$.

$\text{adj}(A) = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

Step 3: Determination of the Inverse

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{-3}{-3} & \frac{0}{-3} & \frac{0}{-3} \\ \frac{3}{-3} & \frac{-1}{-3} & \frac{0}{-3} \\ \frac{-9}{-3} & \frac{-2}{-3} & \frac{3}{-3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1/3 & 0 \\ 3 & 2/3 & -1 \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}$