If matrix \(A=\begin{bmatrix}p & -3\\-4 & p\end{bmatrix}\) and \(|A^3| = 64\), then the value of p is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\pm 4\)
Since \(|AB| = |A||B|\), we have \(|A^3| = |A|^3\). Also
\(|A| = p \cdot p - (-3)(-4) = p^2 - 12\)
So
\((p^2 - 12)^3 = 64 = 4^3\)
The only real cube root of 64 is 4:
\(p^2 - 12 = 4\)
\(p^2 = 16\)
\(p = \pm 4\)