For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$,verify that $(A - A^{\prime})$ is a skew-symmetric matrix.

(N/A) Given $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$.
The transpose of $A$ is $A^{\prime} = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.
Now,calculate $A - A^{\prime}$:
$A - A^{\prime} = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Let $B = A - A^{\prime} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Now,find the transpose of $B$:
$B^{\prime} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.
Observe that $B^{\prime} = -\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -B$.
Since $(A - A^{\prime})^{\prime} = -(A - A^{\prime})$,it is verified that $(A - A^{\prime})$ is a skew-symmetric matrix.