Let A and B be square matrices of order 3 with \(|A| = 3\) and \(|B| = -2\). Then the value of \(|2A^{-1}B'|\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{16}{3}\)
Three results are used: for order 3, \(|kM| = k^3|M|\); the product rule \(|MN| = |M||N|\); and
\( \displaystyle |A^{-1}| = \frac{1}{|A|}, \quad |B'| = |B| \)
So
\(|2A^{-1}B'| = 2^3\,|A^{-1}|\,|B'|\)
\( \displaystyle |2A^{-1}B'| = 8 \times \frac{1}{3} \times (-2) \)
\( \displaystyle |2A^{-1}B'| = -\frac{16}{3} \)
Why the other options are incorrect
Option (1) \(-\displaystyle\frac{4}{3}\) — Incorrect. This takes the factor 2 outside as 2 instead of \(2^3 = 8\).
Option (2) \(\displaystyle\frac{16}{3}\) — Incorrect. The sign is wrong: \(|B'| = |B| = -2\) is negative.
Option (4) \(-48\) — Incorrect. This uses \(|A| = 3\) in place of \(|A^{-1}| = \displaystyle\frac{1}{3}\).