If \(A'=\begin{bmatrix}3 & 5\\ -6 & 5\end{bmatrix}\) and \(B=\begin{bmatrix}4 & 7\\ -1 & 6\end{bmatrix}\) then (AB)′ =
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}18 & 15\\ -15 & 65\end{bmatrix}\)
By the reversal law of transposes, (AB)′ = B′A′, so A itself is not needed.
\( \displaystyle B' = \begin{bmatrix}4 & -1\\ 7 & 6\end{bmatrix} \)
\( \displaystyle B'A' = \begin{bmatrix}4 & -1\\ 7 & 6\end{bmatrix}\begin{bmatrix}3 & 5\\ -6 & 5\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}12 + 6 & 20 - 5\\ 21 - 36 & 35 + 30\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}18 & 15\\ -15 & 65\end{bmatrix} \)
Tip: (AB)′ is B′A′, in reverse order. Multiplying in the given order, A′B′, gives the matrix in Option (3), a common trap.