# If $$A = \begin{bmatrix} 6 & -5 \\ 2 & 3 \end{bmatrix}$$, then show that A-A' is skew 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.
• A square matrix A=[a$_{ij}$] is said to be skew symmetric if A'=-A that is $[a_{ij}]= -[a_{ji}]$ for all possible value of i and j.
• 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.
• From the property of the skew symmetric matrix we have the diagonal of the matrix to be zero.
Step1:
Given:
$A=\begin{bmatrix}6 &-5\\2 & 3\end{bmatrix}$
Transpose can be obtained by changing the rows and column.
$A'=\begin{bmatrix}6 &2\\-5 & 3\end{bmatrix}$
Step2:
A-A'=$\begin{bmatrix}6 &-5\\2 & 3\end{bmatrix}+(-1)\begin{bmatrix}6 &2\\-5& 3\end{bmatrix}$
$\;\;\;=\begin{bmatrix}6 &-5\\2 & 3\end{bmatrix}+\begin{bmatrix}-6 &-2\\5& -3\end{bmatrix}$
$\;\;\;=\begin{bmatrix}0 & -7\\7 & 0\end{bmatrix}$
Since the diagonal of the matrix is zero,we can say that A-A'-skew symmetric matrix.