Question

$A=\left[\begin{array}{ccc}6& 8& 5\\ 4& 2& 3\\ 9& 7& 1\end{array}\right]$ is sum of symmetric matrix $B$ and skew symmetric matrix $C$ then $B$ is

• A. $4\left[\begin{array}{ccc}6& 6& 7\\ 6& 2& 5\\ 7& 5& 1\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& 2& -2\\ -2& 5& -2\\ 2& 2& 0\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& 6& 7\\ -6& 2& -5\\ -7& 5& 1\end{array}\right]$
• D. $\left[\begin{array}{ccc}0& 6& -2\\ 2& 0& -2\\ -2& -2& 0\end{array}\right]$

Answer 1

The correct option is A $4\left[\begin{array}{ccc}6& 6& 7\\ 6& 2& 5\\ 7& 5& 1\end{array}\right]$

if we have a matrix $A$,then if

$A^T=A$

$\Rightarrow A$ is symmetric

and if

$A^T=-A$

$\Rightarrow A$ is skew symmetric .

Also, by a theorem for any square for any square matrix $A$ with real entries $A+A^{{}^{\prime }}$ is a skew symmetric matrix.

And by another theorem any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.

$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=B+CB=\frac{1}{2}(A+A^T),C=\frac{1}{2}(A-A^T)\therefore A=\left[\begin{array}{ccc}6& 8& 5\\ 4& 2& 3\\ 9& 7& 1\end{array}\right]A^T=\left[\begin{array}{ccc}6& 4& 9\\ 8& 2& 7\\ 5& 3& 1\end{array}\right]$

So , $B=\frac{1}{2}(A+A^T)=\frac{1}{2}\{\left[\begin{array}{ccc}6& 8& 5\\ 4& 2& 3\\ 9& 7& 1\end{array}\right]+\left[\begin{array}{ccc}6& 4& 9\\ 8& 2& 7\\ 5& 3& 1\end{array}\right]\}$

$=\frac{1}{2}\left[\begin{array}{ccc}12& 12& 14\\ 12& 4& 10\\ 14& 10& 2\end{array}\right]=4\left[\begin{array}{ccc}6& 6& 7\\ 6& 2& 5\\ 7& 5& 1\end{array}\right]$