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If the sum of a matrix A and its transpose is [−64410]\begin{bmatrix} -6 & 4 \\ 4 & 10 \end{bmatrix}, then A is:

Solution

✅ Correct Option: 3

Diagonal entries double, so 2a11=−62a_{11}=-6 and 2a22=102a_{22}=10, giving a11=−3a_{11}=-3, a22=5a_{22}=5.

Off diagonal: a12+a21=4a_{12}+a_{21}=4. Only option 3 satisfies all: A=[−3045]A=\begin{bmatrix}-3&0\\4&5\end{bmatrix} gives A+AT=[−64410]A+A^T=\begin{bmatrix}-6&4\\4&10\end{bmatrix}.

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