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