Question

If \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \), then \( A^{-1} \) is:

• A. \[ \frac{1}{2} \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} \]
• B. \[ \begin{bmatrix} 1 & -1 & -1 \\ 2 & -3 & 2 \\ -4 & -3 & -1 \end{bmatrix} \]
• C. \[ \frac{1}{2} \begin{bmatrix} 1 & -1 & 5 \\ -6 & 3 & -1 \\ 1 & 2 & -1 \end{bmatrix} \]
• D. \[ \begin{bmatrix} -1 & -1 & -2 \\ 1 & -6 & 3 \\ 5 & -3 & 1 \end{bmatrix} \]

Hint:

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

Answer 1

The Correct Option is A

The inverse of a 3x3 matrix \( A \) is given by the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \] Step 1: Calculate the determinant of \( A \). For the matrix \[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}, \] the determinant is calculated as: \[ \text{det}(A) = 0 \times \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 1 \times \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} + 2 \times \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} \] \[ \text{det}(A) = 0 - 1 \times (-8) + 2 \times (-5) = 8 - 10 = -2 \] Step 2: Calculate the adjoint of \( A \). The adjoint of \( A \) is the transpose of the cofactor matrix. After calculating the cofactors, the adjoint \( \text{adj}(A) \) is: \[ \text{adj}(A) = \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} \] Step 3: Compute the inverse. Now, using the formula for the inverse, we have: \[ A^{-1} = \frac{1}{-2} \times \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} \] Conclusion: The correct inverse matrix is: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} -1 & -1 & -1 \\ -8 & 6 & -1 \\ 5 & -3 & 1 \end{bmatrix} \]