Question

The inverse of the matrix \[ \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 3 \\ 5 & 2 & -1 \end{bmatrix} \] is:

• A. \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 2 & -3 \end{bmatrix} \]
• B. \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]
• C. \[ \frac{-1}{3} \begin{bmatrix} 3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]
• D. \[ \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]

Hint:

To find the inverse of a matrix, first calculate its determinant, then find the adjoint, and apply the formula \( A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \).

Answer 1

The Correct Option is B

We are given the matrix \[ A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 3 \\ 5 & 2 & -1 \end{bmatrix} \] and we need to find its inverse. 

Step 1: First, calculate the determinant of matrix \( A \). \[ \text{det}(A) = 1 \times \begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} - 0 \times \begin{vmatrix} 3 & 3 \\ 5 & -1 \end{vmatrix} + 0 \times \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} \] \[ \text{det}(A) = 1 \times ((3 \times (-1)) - (3 \times 2)) = 1 \times (-3 - 6) = -9 \] 

Step 2: Now, calculate the adjoint of matrix \( A \). The adjoint is the transpose of the cofactor matrix.

 Step 3: Using the formula for the inverse of a matrix \( A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \), we find: \[ A^{-1} = \frac{1}{-9} \times \text{adj}(A) \]

 Conclusion: The correct inverse matrix is: \[ A^{-1} = \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]