Let \(A\) be a square matrix of order 2 such that \(\begin{bmatrix}2&1\\3&2\end{bmatrix} A \begin{bmatrix}-3&2\\5&-3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}\), then \(A\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\)
Let \(P = \begin{bmatrix}2&1\\3&2\end{bmatrix}\) and \(Q = \begin{bmatrix}-3&2\\5&-3\end{bmatrix}\), so that PAQ = I. Multiplying on the left by \(P^{-1}\) and on the right by \(Q^{-1}\):
\(A = P^{-1}Q^{-1}\)
\(|P| = 4 - 3 = 1,\quad |Q| = 9 - 10 = -1\)
Using \(M^{-1} = \frac{1}{|M|}\,\mathrm{adj}\,M\):
\( \displaystyle P^{-1} = \frac{1}{1}\begin{bmatrix}2&-1\\-3&2\end{bmatrix} = \begin{bmatrix}2&-1\\-3&2\end{bmatrix} \)
\( \displaystyle Q^{-1} = \frac{1}{-1}\begin{bmatrix}-3&-2\\-5&-3\end{bmatrix} = \begin{bmatrix}3&2\\5&3\end{bmatrix} \)
\( \displaystyle A = \begin{bmatrix}2&-1\\-3&2\end{bmatrix}\begin{bmatrix}3&2\\5&3\end{bmatrix} \)
\( \displaystyle A = \begin{bmatrix}6 - 5&4 - 3\\-9 + 10&-6 + 6\end{bmatrix} = \begin{bmatrix}1&1\\1&0\end{bmatrix} \)
Shortcut: PAQ = I means \(PA = Q^{-1}\). Multiply P by the option:
\( \displaystyle \begin{bmatrix}2&1\\3&2\end{bmatrix}\begin{bmatrix}1&1\\1&0\end{bmatrix} = \begin{bmatrix}3&2\\5&3\end{bmatrix} = Q^{-1} \)