If \(\begin{bmatrix}3&1\\2&1\end{bmatrix}A\begin{bmatrix}2&1\\1&1\end{bmatrix}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), then matrix 'A' is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}1&-1\\-3&4\end{bmatrix}\)
Let \(P = \begin{bmatrix}3&1\\2&1\end{bmatrix}\), \(Q = \begin{bmatrix}2&1\\1&1\end{bmatrix}\) and \(R = \begin{bmatrix}1&1\\0&1\end{bmatrix}\), so that PAQ = R.
Multiplying by \(P^{-1}\) on the left and by \(Q^{-1}\) on the right:
\(A = P^{-1}RQ^{-1}\)
Both determinants are \(3 - 2 = 1\) and \(2 - 1 = 1\). For a 2 × 2 matrix, the inverse swaps the diagonal entries, changes the signs of the other two and divides by the determinant:
\( \displaystyle P^{-1} = \begin{bmatrix}1&-1\\-2&3\end{bmatrix} \)
\( \displaystyle Q^{-1} = \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \)
Then
\( \displaystyle P^{-1}R = \begin{bmatrix}1&-1\\-2&3\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix} = \begin{bmatrix}1&0\\-2&1\end{bmatrix} \)
\( \displaystyle A = \begin{bmatrix}1&0\\-2&1\end{bmatrix}\begin{bmatrix}1&-1\\-1&2\end{bmatrix} = \begin{bmatrix}1&-1\\-3&4\end{bmatrix} \)