Let A and B be square matrices of order 3 with \(|A| = 4\) and \(|B| = -2\). Then the value of \(|2A^{-1}B|\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-4\)
Three results are used for matrices of order 3:
\(|kM| = k^{3}|M|\)
\(|MN| = |M|\,|N|\)
\(|A^{-1}| = \displaystyle\frac{1}{|A|}\)
So
\( \displaystyle |2A^{-1}B| = 2^{3} \times \frac{1}{|A|} \times |B| \)
\( \displaystyle |2A^{-1}B| = 8 \times \frac{1}{4} \times (-2) \)
\(|2A^{-1}B| = -4\)
Why the other options are incorrect
Option (1) \(-1\) — Incorrect. This takes the factor 2 only once; it comes out of each of the 3 rows, giving \(2^{3}\).
Option (3) 4 — Incorrect. The sign is lost: \(|B| = -2\) makes the product negative.
Option (4) \(-64\) — Incorrect. This uses \(|A| = 4\) in place of \(|A^{-1}| = \frac{1}{4}\).