If $A =\begin{bmatrix}1&1&1\\1&0&2\\x&1&1\end{bmatrix}$ and $A^{-1}=\frac{1}{4}\begin{bmatrix}-2&0&y\\5&-2&-1\\1&2&-1\end{bmatrix}$, then values of $x$ and $y$, are:

$x = 3, y=2$

The correct answer is Option (1) → $x = 3, y=2$

$A=\begin{pmatrix}1&1&1\\[4pt]1&0&2\\[4pt]x&1&1\end{pmatrix}$

$A^{-1}=\frac{1}{4}\begin{pmatrix}-2&0&y\\[4pt]5&-2&-1\\[4pt]1&2&-1\end{pmatrix}$

Using the identity $A\cdot A^{-1}=I$.

$A\cdot A^{-1}=\frac{1}{4}\begin{pmatrix} 1&1&1\\ 1&0&2\\ x&1&1 \end{pmatrix} \begin{pmatrix} -2&0&y\\ 5&-2&-1\\ 1&2&-1 \end{pmatrix}$

Compute the $(3,1)$ entry of the product because it contains $x$:

$x(-2)+1(5)+1(1)=0$

$-2x+6=0$

$x=3$

Now use the $(1,3)$ entry to find $y$:

$1\cdot y + 1(-1) + 1(-1)=0$

$y - 2 = 0$

$y = 2$

The values of $x$ and $y$ are $x=3,\; y=2$.