Browse Questions

# If $A = \begin{bmatrix} 4 & 1 \\ 5 & 8 \end{bmatrix},$ show that A + A' is symmetric matrix.

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• A square matrix A = $[a_{ij}]$ is said to be symmetric matrix if $a_{ij}=a_{ji}$ for all possible value of i and j
Step1:
Given:
$A=\begin{bmatrix}4 & 1\\5 & 8\end{bmatrix}$
The transpose of a matrix can be obtained by changing the rows into column.
$A'=\begin{bmatrix}4 & 5\\1 & 8\end{bmatrix}$
Step2:
$A+A'=\begin{bmatrix}4 & 1\\5 & 8\end{bmatrix}+\begin{bmatrix}4 & 5\\1 & 8\end{bmatrix}$
$\qquad\;\;\;=\begin{bmatrix}8 & 6\\6 & 16\end{bmatrix}$
Step3:
Here the element is symmetric about the main (principal) diagonal
Here $a_{21}=6=a_{12}$
$\Rightarrow A+A'$ is a symmetric matrix.