If the sum of a matrix A and its transpose is \( \begin{bmatrix}-6 & 4\\4 & 10\end{bmatrix} \), then A is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix}-3 & 0\\4 & 5\end{bmatrix} \)
Let \( A=\begin{bmatrix}a & b\\c & d\end{bmatrix} \). Then
\( \displaystyle A+A^T=\begin{bmatrix}2a & b+c\\b+c & 2d\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-6 & 4\\4 & 10\end{bmatrix} \)
\( \displaystyle a=-3 \)
\( \displaystyle d=5 \)
\( \displaystyle b+c=4 \)
The off-diagonal entries fix only the sum b + c, so A is not unique; the options decide it. Only Option (3) has diagonal entries −3 and 5, and its off-diagonal entries 0 and 4 add up to 4. Check:
\( \displaystyle \begin{bmatrix}-3 & 0\\4 & 5\end{bmatrix}+\begin{bmatrix}-3 & 4\\0 & 5\end{bmatrix}=\begin{bmatrix}-6 & 4\\4 & 10\end{bmatrix} \)