If \(A =\begin{bmatrix}1&1&1\\1&0&2\\x&1&1\end{bmatrix}\) and \(A^{-1}=\displaystyle\frac{1}{4}\begin{bmatrix}-2&0&y\\5&-2&-1\\1&2&-1\end{bmatrix}\), then values of \(x\) and \(y\), are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x = 3, y=2\)
Let \(M = 4A^{-1}\), the matrix inside the bracket. Since \(AA^{-1} = I\),
\(AM = 4I\)
So each row of A times a column of M gives 4 on the diagonal and 0 elsewhere.
Row 1 of A with column 3 of M (an off-diagonal entry):
\(1 \cdot y + 1 \cdot (-1) + 1 \cdot (-1) = 0\)
\(y - 2 = 0,\quad y = 2\)
Row 3 of A with column 1 of M (an off-diagonal entry):
\(x \cdot (-2) + 1 \cdot 5 + 1 \cdot 1 = 0\)
\(-2x + 6 = 0,\quad x = 3\)
Check with row 3 of A and column 3 of M (a diagonal entry):
\(3 \cdot 2 + 1 \cdot (-1) + 1 \cdot (-1) = 4\)
So x = 3 and y = 2.