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

$\begin{bmatrix}1&-1\\-3&4\end{bmatrix}$

The correct answer is Option (3) → $\begin{bmatrix}1&-1\\-3&4\end{bmatrix}$

Given:

$\begin{pmatrix}3 & 1 \\ 2 & 1\end{pmatrix} \; A \; \begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$

So:

$A = \left(\begin{pmatrix}3 & 1 \\ 2 & 1\end{pmatrix}\right)^{-1} \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix} \left(\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}\right)^{-1}$

After matrix computation:

$A=\begin{pmatrix}1 & -1 \\ -3 & 4\end{pmatrix}$

Matrix A is $\begin{pmatrix}1 & -1 \\ -3 & 4\end{pmatrix}$.