Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right] \\ \Rightarrow A^T=\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right] \\ \mathrm{Let}X=\frac{1}{2}\left(A+A^T\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]+\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right]\right)=\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{5}{2}\\ \frac{3}{2}& 4& 4\\ \frac{5}{2}& 4& 8\end{array}\right] \\ \mathrm{Let}Y=\frac{1}{2}\left(A-A^T\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]-\left[\begin{array}{ccc}3& 1& -2\\ 2& 4& 5\\ 7& 3& 8\end{array}\right]\right)=\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{9}{2}\\ \frac{-1}{2}& 0& -1\\ \frac{-9}{2}& 1& 0\end{array}\right] \\ {X}^T={\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{5}{2}\\ \frac{3}{2}& 4& 4\\ \frac{5}{2}& 4& 8\end{array}\right]}^T={\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{5}{2}\\ \frac{3}{2}& 4& 4\\ \frac{5}{2}& 4& 8\end{array}\right]}^T=X \\ {Y}^T={\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{9}{2}\\ \frac{-1}{2}& 0& -1\\ \frac{-9}{2}& 1& 0\end{array}\right]}^T=\left[\begin{array}{ccc}0& \frac{-1}{2}& \frac{-9}{2}\\ \frac{1}{2}& 0& 1\\ \frac{9}{2}& -1& 0\end{array}\right]=-\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{9}{2}\\ \frac{-1}{2}& 0& -1\\ \frac{-9}{2}& 1& 0\end{array}\right]=Y \\ \mathrm{Thus},X\mathrm{is}\mathrm{a}\mathrm{symmetric}\mathrm{matrix}\mathrm{and}Y\mathrm{is}\mathrm{skew}-\mathrm{symmetric}\mathrm{matrix}. \\ \mathrm{Now}, \\ X+Y=\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{5}{2}\\ \frac{3}{2}& 4& 4\\ \frac{5}{2}& 4& 8\end{array}\right]+\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{9}{2}\\ \frac{-1}{2}& 0& -1\\ \frac{-9}{2}& 1& 0\end{array}\right]=\left[\begin{array}{ccc}3& 2& 7\\ 1& 4& 3\\ -2& 5& 8\end{array}\right]=A \\ \therefore X=\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{5}{2}\\ \frac{3}{2}& 4& 4\\ \frac{5}{2}& 4& 8\end{array}\right]\mathrm{and}Y=\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{9}{2}\\ \frac{-1}{2}& 0& -1\\ \frac{-9}{2}& 1& 0\end{array}\right] \end{gathered}$$