Question

If \( \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix} A = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \), then the matrix A is:

• A. \( \begin{bmatrix} 6 & 2 \\ 11/2 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} -1/5 & 2/5 \\ -3/10 & 1/10 \end{bmatrix} \)
• C. \( \begin{bmatrix} 2 & 6 \\ 11 & 3 \end{bmatrix} \)
• D. \( \begin{bmatrix} 3 & 2 \\ 6 & 5 \end{bmatrix} \)

Hint:

When solving \( MA = B \), ensure you multiply \( M^{-1} \) on the left: \( A = M^{-1}B \), not \( BM^{-1} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let \( M = \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \). We have \( MA = B \), so \( A = M^{-1}B \).

Step 2: Key Formula or Approach:
Calculate \( M^{-1} = \frac{1}{|M|} adj(M) \).
\( |M| = (1)(-2) - (-4)(3) = -2 + 12 = 10 \).
\( adj(M) = \begin{bmatrix} -2 & 4 \\ -3 & 1 \end{bmatrix} \). So, \( M^{-1} = \frac{1}{10} \begin{bmatrix} -2 & 4 \\ -3 & 1 \end{bmatrix} \).

Step 3: Detailed Explanation:
\( A = \frac{1}{10} \begin{bmatrix} -2 & 4 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \)
\( A = \frac{1}{10} \begin{bmatrix} (-2)(-16) + (4)(7) & (-2)(-6) + (4)(2) \\ (-3)(-16) + (1)(7) & (-3)(-6) + (1)(2) \end{bmatrix} \)
\( A = \frac{1}{10} \begin{bmatrix} 32 + 28 & 12 + 8 \\ 48 + 7 & 18 + 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 60 & 20 \\ 55 & 20 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5.5 & 2 \end{bmatrix} \)

Step 4: Final Answer:
The matrix A is \( \begin{bmatrix} 6 & 2 \\ 11/2 & 2 \end{bmatrix} \).