If $\mathrm{A}=\left[\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right]$, show that $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is symmetric

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If $\mathrm{A}=\left[\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right]$ , show that $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is symmetric

If $\mathrm{A}=\left[\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right]$ , show that $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is symmetric

Solution:

We have $\left(\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right)$ .
The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element $a_{j i}$ shifted to new position $a_{j i}$ .
The symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, $A^{T}=A$ .
The skew-symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, $A^{T}=-A$ .
$A^{T}=\left(\begin{array}{ll} 4 & 5 \\ 1 & 8 \end{array}\right)^{T}=\left(\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right)$
Therefore finally we have
$A+A^{T}=\left(\begin{array}{ll} 4 & 5 \\ 1 & 8 \end{array}\right)+\left(\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right)=\left(\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right)$
$\operatorname{Here}\left(A+A^{T}\right)^{T}=\left(\begin{array}{cc}8 & 6 \\ 6 & 16\end{array}\right)^{T}=\left(\begin{array}{cc}8 & 6 \\ 6 & 16\end{array}\right)=A+A^{T}$
Therefore $A+A^{T}$ is symmetric.