If \( \begin{bmatrix}1 & -4\\3 & -2\end{bmatrix}A=\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix} \), then the matrix \( A \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \begin{bmatrix}6 & 2\\11/2 & 2\end{bmatrix} \)
Let M be the matrix on the left and B the matrix on the right. Since MA = B, \( A=M^{-1}B \).
\( \displaystyle |M|=-2+12=10 \)
\( \displaystyle M^{-1}=\frac{1}{10}\begin{bmatrix}-2 & 4\\-3 & 1\end{bmatrix} \)
Then
\( \displaystyle A=\frac{1}{10}\begin{bmatrix}-2 & 4\\-3 & 1\end{bmatrix}\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix} \)
\( \displaystyle =\frac{1}{10}\begin{bmatrix}60 & 20\\55 & 20\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}6 & 2\\\frac{11}{2} & 2\end{bmatrix} \)
Why the other options are incorrect
Option (2) \( \begin{bmatrix}-1/5 & 2/5\\-3/10 & 1/10\end{bmatrix} \) — Incorrect. This is \( M^{-1} \) itself; it still has to be multiplied by the matrix on the right.
Option (3) \( \begin{bmatrix}2 & 6\\11 & 3\end{bmatrix} \) — Incorrect. The first row of A is (6, 2), not (2, 6).
Option (4) \( \begin{bmatrix}3 & 2\\6 & 5\end{bmatrix} \) — Incorrect. The first row of A is (6, 2), not (3, 2).