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Mathematics Question on Matrices

Find 12(A+A′)\frac{1}{2}(A+A')and 12(A−A′),\frac{1}{2}(A-A'),when A=[0ab −a0c −b−c0]A=\begin{bmatrix}0&a&b\\\ -a&0&c\\\ -b&-c&0\end{bmatrix}

Answer

The given matrix is A=[0ab −a0c −b−c0]A=\begin{bmatrix}0&a&b\\\ -a&0&c\\\ -b&-c&0\end{bmatrix}
so A′=[0−a−b a0−c bc0]A'=\begin{bmatrix}0&-a&-b\\\ a&0&-c\\\ b&c&0\end{bmatrix}
A+A′=[0ab −a0c −b−c0]+[0−a−b a0−c bc0]A+A'=\begin{bmatrix}0&a&b\\\ -a&0&c\\\ -b&-c&0\end{bmatrix}+\begin{bmatrix}0&-a&-b\\\ a&0&-c\\\ b&c&0\end{bmatrix}
=[000 000 000]=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}
so 12(A+A′)==[000 000 000]\frac{1}{2}(A+A')==\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}
Now,A−A′=[0ab −a0c −b−c0]−[0−a−b a0−c bc0]A-A'=\begin{bmatrix}0&a&b\\\ -a&0&c\\\ -b&-c&0\end{bmatrix}-\begin{bmatrix}0&-a&-b\\\ a&0&-c\\\ b&c&0\end{bmatrix}
=[02a2b −2a02c −2b−2c0]=\begin{bmatrix}0&2a&2b\\\ -2a&0&2c\\\ -2b&-2c&0\end{bmatrix}
so 12(A−A′)=[0ab −a0c −b−c0]\frac{1}{2}(A-A')=\begin{bmatrix}0&a&b\\\ -a&0&c\\\ -b&-c&0\end{bmatrix}