If \(A=\begin{bmatrix}0&1&3\\1&2&x\\2&3&1\end{bmatrix}\) and \(A^{-1}=\begin{bmatrix}\displaystyle\frac{1}{2}&-4&\displaystyle\frac{5}{2}\\-\displaystyle\frac{1}{2}&3&-\displaystyle\frac{3}{2}\\\displaystyle\frac{1}{2}&y&\displaystyle\frac{1}{2}\end{bmatrix}\), then value of \(8x + 5y\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
Since \(A^{-1}\) is the inverse of A, \(AA^{-1} = I\). Each entry of \(AA^{-1}\) is a row of A times a column of \(A^{-1}\), and every off-diagonal entry of I is 0.
Row 2 of A times column 3 of \(A^{-1}\) gives the (2, 3) entry, which is 0:
\( \displaystyle 1 \times \frac{5}{2} + 2 \times \left(-\frac{3}{2}\right) + x \times \frac{1}{2} = 0 \)
\( \displaystyle \frac{5}{2} - 3 + \frac{x}{2} = 0 \)
\(x = 1\)
Row 1 of A times column 2 of \(A^{-1}\) gives the (1, 2) entry, which is 0:
\(0 \times (-4) + 1 \times 3 + 3 \times y = 0\)
\(3 + 3y = 0\)
\(y = -1\)
Hence
\(8x + 5y = 8 - 5 = 3\)