If A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right], show that \left(A+A^{\prime}\right) is skew-symmetric.
If A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right], show that \left(A+A^{\prime}\right) is skew-symmetric.

Solution:

We have A=\left(\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right)
The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element a_{i j} shifted to new position a_{j i}
The skew-symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, A^{T}=-A.

(i) The transpose of A are given by A^{T} which can be written as
A^{T}=\left(\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right)^{T}=\left(\begin{array}{cc} 3 & 1 \\ -4 & -1 \end{array}\right)

(ii) Next the subtraction of A and A^{T} can be processed as
A-A^{T}=\left(\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right)-\left(\begin{array}{cc} 3 & 1 \\ -4 & -1 \end{array}\right)=\left(\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right)

(iii) To check symmetric nature of A+A^{T} as follow:
\left(A-A^{T}\right)^{T}=\left(\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right)^{T}=\left(\begin{array}{cc} 0 & 5 \\ -5 & 0 \end{array}\right)=-\left(\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right)=-\left(A-A^{T}\right)
Thus it can be said that A-A^{T} is skew-symmetric.