If \(A = \begin{bmatrix}1&3\\-2&0\end{bmatrix}\) and \((A + B)' = \begin{bmatrix}4&1\\5&-2\end{bmatrix}\), then B is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}3&2\\3&-2\end{bmatrix}\)
Since \(\left((A + B)'\right)' = A + B\), taking the transpose of the given matrix gives A + B:
\( \displaystyle A + B = \begin{bmatrix}4&5\\1&-2\end{bmatrix} \)
Subtracting A element by element:
\( \displaystyle B = \begin{bmatrix}4 - 1&5 - 3\\1 + 2&-2 - 0\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}3&2\\3&-2\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}3&-2\\7&-2\end{bmatrix}\) — Incorrect. This subtracts A from \((A + B)'\) itself, without first taking the transpose back.
Option (2) \(\begin{bmatrix}3&3\\2&-2\end{bmatrix}\) — Incorrect. This is \(B'\), not B.
Option (3) \(\begin{bmatrix}5&8\\-1&-2\end{bmatrix}\) — Incorrect. This adds A to \(A + B\) instead of subtracting it.