Question

Let \( B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & -1 & -1 \end{bmatrix} \), \( C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \), and if a matrix \( A \) is such that \( BAC = I \), then \( A^{-1} = \)

• A. \( \begin{bmatrix} -3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6 \end{bmatrix} \)
• B. \( \begin{bmatrix} -3 & -5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{bmatrix} \)
• C. \( \begin{bmatrix} -3 & -5 & -6 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \)
• D. \( \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \)

Hint:

When \( BAC = I \), use \( A^{-1} = CB \) for the solution.

Answer 1

The Correct Option is D

We are given that \( BAC = I \), so: \[ A^{-1} = CB \] Now, compute the matrix product \( CB \): \[ CB = \begin{bmatrix} -3 & -5 & -5
0 & 9 & 2
2 & 14 & 6 \end{bmatrix} \] Thus, the correct answer is option (4).