Question:
Find \(\frac{1}{2}(A+A')\)and \(\frac{1}{2}(A-A'),\)when \(A=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}\)
Find \(\frac{1}{2}(A+A')\)and \(\frac{1}{2}(A-A'),\)when \(A=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}\)
Updated On: Sep 5, 2023
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Solution and Explanation
The given matrix is \(A=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}\)
so \(A'=\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(A+A'=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}+\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\)
so \(\frac{1}{2}(A+A')==\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\)
Now,\(A-A'=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}-\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(=\begin{bmatrix}0&2a&2b\\ -2a&0&2c\\ -2b&-2c&0\end{bmatrix}\)
so \(\frac{1}{2}(A-A')=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}\)
so \(A'=\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(A+A'=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}+\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\)
so \(\frac{1}{2}(A+A')==\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\)
Now,\(A-A'=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}-\begin{bmatrix}0&-a&-b\\ a&0&-c\\ b&c&0\end{bmatrix}\)
\(=\begin{bmatrix}0&2a&2b\\ -2a&0&2c\\ -2b&-2c&0\end{bmatrix}\)
so \(\frac{1}{2}(A-A')=\begin{bmatrix}0&a&b\\ -a&0&c\\ -b&-c&0\end{bmatrix}\)
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