If \(A = \begin{bmatrix}3 & -4\\ 2 & x\end{bmatrix}\) and \(A^{-1} = A\), then the value of x is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-3\)
\(A^{-1} = A\) means \(A \times A = I\).
\( \displaystyle A^2 = \begin{bmatrix}3 & -4\\ 2 & x\end{bmatrix}\begin{bmatrix}3 & -4\\ 2 & x\end{bmatrix} = \begin{bmatrix}9 - 8 & -12 - 4x\\ 6 + 2x & -8 + x^2\end{bmatrix} \)
Comparing with \(I = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}\):
\( -12 - 4x = 0 \Rightarrow x = -3 \)
\( 6 + 2x = 0 \Rightarrow x = -3 \)
\( -8 + x^2 = -8 + 9 = 1 \)
All four entries agree, so x = −3.
Why the other options are incorrect
Option (1) 3 — Incorrect. The (1, 2) entry of \(A^2\) becomes \(-12 - 12 = -24\), not 0.
Option (2) \(-\frac{7}{3}\) — Incorrect. It makes \(|A| = 3x + 8 = 1\), which is only a necessary condition; the (1, 2) entry of \(A^2\) is then \(-12 + \frac{28}{3} \ne 0\).
Option (3) \(-2\) — Incorrect. The (1, 2) entry of \(A^2\) becomes \(-12 + 8 = -4\), not 0.