If \(A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}\) and \(B = \begin{bmatrix}1&2\\-2&3\end{bmatrix}\), then \((A + 2B)'\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}4&-1\\3&10\end{bmatrix}\)
Multiplying B by 2 multiplies every element by 2:
\( \displaystyle 2B = \begin{bmatrix}2&4\\-4&6\end{bmatrix} \)
Adding the corresponding elements:
\( \displaystyle A + 2B = \begin{bmatrix}2 + 2&-1 + 4\\3 - 4&4 + 6\end{bmatrix} = \begin{bmatrix}4&3\\-1&10\end{bmatrix} \)
The transpose turns rows into columns:
\( \displaystyle (A + 2B)' = \begin{bmatrix}4&-1\\3&10\end{bmatrix} \)
Why the other options are incorrect
Option (2) \(\begin{bmatrix}4&3\\-1&10\end{bmatrix}\) — Incorrect. This is \(A + 2B\) itself, before taking the transpose.
Option (3) \(\begin{bmatrix}4&7\\-5&10\end{bmatrix}\) — Incorrect. This is \(A' + 2B\); the transpose was applied to A only. The correct result is \(A' + 2B'\).
Option (4) \(\begin{bmatrix}5&4\\0&11\end{bmatrix}\) — Incorrect. This is \((2A + B)'\); the factor 2 was put on the wrong matrix.