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

Given matrix \( A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 3 \\ 5 & 2 & -1 \end{bmatrix} \), its inverse is to be determined.

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

Step 2: Calculate the adjoint of \( A \), which is the transpose of its cofactor matrix.

Step 3: Apply the formula \( A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \): \[ A^{-1} = \frac{1}{-9} \times \text{adj}(A) \]

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