# Solve this

Question:

Express the matrix $A=\left[\begin{array}{rrr}4 & 2 & -1 \\ 3 & 5 & 7 \\ 1 & -2 & 1\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.

Solution:

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

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

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

$X^{T}=\left[\begin{array}{ccc}4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1\end{array}\right]^{T}=\left[\begin{array}{ccc}4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1\end{array}\right]=X$

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

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

Thus, $X$ is a symmetric matrix and $Y$ is a skew - symmetric matrix.

Now,

$X+Y=\left[\begin{array}{ccc}4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1\end{array}\right]+\left[\begin{array}{ccc}0 & \frac{-1}{2} & -1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & \frac{-9}{2} & 0\end{array}\right]=\left[\begin{array}{ccc}4 & 2 & -1 \\ 3 & 5 & 7 \\ 1 & -2 & 1\end{array}\right]=A$