Question

If $A=\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]$, then find matrices $X$ and $Y$, such that $X+Y=A$, where $X$ is a symmetric.

Answer 1

Given $A=\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]$

$⟹A^T=\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right]$

Let $X=\frac{1}{2}(A+A^T)=\frac{1}{2}$ $\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]$ $+\frac{1}{2}\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right]$ $=\frac{1}{2}\left[\begin{array}{ccc}6& 3& 5\\ 3& 8& 8\\ 5& 8& 16\end{array}\right]$ $=\left[\begin{array}{ccc}3& 3/2& 5/2\\ 3/2& 4& 4\\ 5/2& 4& 8\end{array}\right]$

$X^T=\left[\begin{array}{ccc}3& 3/2& 5/2\\ 3/2& 4& 4\\ 5/2& 4& 8\end{array}\right]=X$

So $X$ is a symmetric matrix

Let $Y=\frac{1}{2}(A-A^T)=\frac{1}{2}$ $\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]$ $-\frac{1}{2}\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right]$ $=\frac{1}{2}\left[\begin{array}{ccc}0& 1& 9\\ -1& 0& -2\\ -9& 2& 0\end{array}\right]$ $=\left[\begin{array}{ccc}0& 1/2& 9/2\\ -1/2& 0& -1\\ -9/2& 1& 0\end{array}\right]$

$Y^T=\left[\begin{array}{ccc}0& -1/2& -9/2\\ 1/2& 0& 1\\ 9/2& -1& 0\end{array}\right]=-Y$

So $Y$ is a skew symmetric matrix

$X+Y=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=A$

Hence $X=\left[\begin{array}{ccc}3& 3/2& 5/2\\ 3/2& 4& 4\\ 5/2& 4& 8\end{array}\right]$ and $Y=\left[\begin{array}{ccc}0& 1/2& 9/2\\ -1/2& 0& -1\\ -9/2& 1& 0\end{array}\right]$