A and B are square matrices of order 3 such that \(|A| = -3\) and \(|A^2B| = 36\). Then \(|B|\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4\)
By the product rule \(|PQ| = |P||Q|\), applied twice:
\(|A^2B| = |A|\,|A|\,|B| = |A|^2\,|B|\)
\((-3)^2\,|B| = 36\)
\(9\,|B| = 36\)
\(|B| = 4\)
Why the other options are incorrect
Option (1) \(-4\) — Incorrect. This takes \(|A|^2 = -9\), but the square of −3 is \(+9\).
Option (3) \(-12\) — Incorrect. This uses \(|A|\) instead of \(|A|^2\): \(-3\,|B| = 36\).
Option (4) \(12\) — Incorrect. This divides 36 by 3 only once; the factor is \(|A|^2 = 9\).