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

We are given:

$\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix} \cdot A \cdot \begin{bmatrix}2 & 1\\1 & 1\end{bmatrix} = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$

Let $P = \begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}$, $Q = \begin{bmatrix}2 & 1\\1 & 1\end{bmatrix}$, $R = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$

To isolate $A$, multiply both sides by the inverses of $P$ and $Q$:

$A = P^{-1} \cdot R \cdot Q^{-1}$

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For any $2 \times 2$ matrix $\begin{bmatrix}a & b\\c & d\end{bmatrix}$, the inverse is $\dfrac{1}{ad - bc}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

$\det(P) = (3)(1) - (1)(2) = 1$

$P^{-1} = \dfrac{1}{1}\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}$

$= \begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}$

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$\det(Q) = (2)(1) - (1)(1) = 1$

$Q^{-1} = \begin{bmatrix}1 & -1\\-1 & 2\end{bmatrix}$

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$P^{-1} \cdot R = \begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$

$= \begin{bmatrix}(1-0) & (1-1)\\(-2+0) & (-2+3)\end{bmatrix}$

$= \begin{bmatrix}1 & 0\\-2 & 1\end{bmatrix}$

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$A = \begin{bmatrix}1 & 0\\-2 & 1\end{bmatrix}\begin{bmatrix}1 & -1\\-1 & 2\end{bmatrix}$

$= \begin{bmatrix}(1+0) & (-1+0)\\(-2-1) & (2+2)\end{bmatrix}$

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

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