If A, B and C are square matrices of order 2 such that \(AB = \begin{bmatrix}2&-1\\3&0\end{bmatrix}\) and \(AC = \begin{bmatrix}1&4\\-2&5\end{bmatrix}\), then \(A(B - 2C)\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}0&-9\\7&-10\end{bmatrix}\)
Matrix multiplication is distributive over addition, and a scalar can be taken out of a product:
\(A(B - 2C) = AB - 2AC\)
\( \displaystyle 2AC = \begin{bmatrix}2&8\\-4&10\end{bmatrix} \)
\( \displaystyle AB - 2AC = \begin{bmatrix}2 - 2&-1 - 8\\3 + 4&0 - 10\end{bmatrix} = \begin{bmatrix}0&-9\\7&-10\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}4&7\\-1&10\end{bmatrix}\) — Incorrect. This is \(AB + 2AC\); the sign of 2AC is wrong.
Option (3) \(\begin{bmatrix}0&9\\-7&10\end{bmatrix}\) — Incorrect. This is \(2AC - AB\), the negative of the answer.
Option (4) \(\begin{bmatrix}1&-5\\5&-5\end{bmatrix}\) — Incorrect. This is \(AB - AC\); the factor 2 is left out.