If \(A=\begin{bmatrix}-2\\-1\\-4\end{bmatrix}, B = \begin{bmatrix}-1&2&3\end{bmatrix}\), then the value of \(A' B'\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}-12\end{bmatrix}\)
The transposes are
\(\displaystyle A' = \begin{bmatrix}-2&-1&-4\end{bmatrix},\qquad B' = \begin{bmatrix}-1\\2\\3\end{bmatrix}\)
A′ is 1 × 3 and B′ is 3 × 1, so A′B′ is a 1 × 1 matrix:
\(\displaystyle A'B' = \begin{bmatrix}(-2)(-1) + (-1)(2) + (-4)(3)\end{bmatrix}\)
\(= \begin{bmatrix}2 - 2 - 12\end{bmatrix} = \begin{bmatrix}-12\end{bmatrix}\)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}2&-4&-6\\1&-2&-3\\4&-8&-12\end{bmatrix}\) — Incorrect. This is AB (3 × 1 times 1 × 3), not A′B′.
Option (2) \(\begin{bmatrix}2&1&4\\-4&-2&-8\\-6&-3&-12\end{bmatrix}\) — Incorrect. This is B′A′ = (AB)′. Since (BA)′ = A′B′, the required product is the transpose of BA, which is 1 × 1.
Option (3) \(\begin{bmatrix}12\end{bmatrix}\) — Incorrect. The sum is 2 − 2 − 12 = −12; the sign is negative.