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:
Answer & explanation
Correct answer: option 1
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 $x=3,\; y=2$.