If $A=\left[\begin{array}{ll}4 & 1 \\ 5 & 8\end{array}\right]$ and $A+A^{T}=\left[\begin{array}{ll}8 & a \\ 6 & b\end{array}\right]$ is a symmetric matrix, then a and b are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $a=6, b=16$
$A=\begin{bmatrix}4 & 1 \\ 5 & 8\end{bmatrix}$
$A^T=\begin{bmatrix}4 & 5 \\ 1 & 8\end{bmatrix}$
$A+A^T=\begin{bmatrix}8 & 6 \\ 6 & 16\end{bmatrix}$
$\begin{bmatrix}8 & a \\ 6 & b\end{bmatrix}=\begin{bmatrix}8 & 6 \\ 6 & 16\end{bmatrix}$
$a=6,\;\; b=16$
The values are $a=6,\; b=16$.