Question

If $\left[\begin{array}{ccc}1& 3& 0\\ 1& 0& -2\\ -4& -4& 4\end{array}\right]=A+B$ where $A$ is symmetric matrix and $B$ is skew-symmetric, then $A-B$ is equal to

• A. $\left[\begin{array}{ccc}1& 1& -4\\ 3& 0& -4\\ 0& -2& 4\end{array}\right]$
• B. $\left[\begin{array}{ccc}2& 1& 3\\ -1& 2& 4\\ 3& -1& 2\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& 1& -1\\ 2& 3& 4\\ -4& 1& 2\end{array}\right]$
• D. $\left[\begin{array}{ccc}2& -3& 0\\ 0& 1& 2\\ 2& 4& 0\end{array}\right]$

Answer 1

Answer: (A)

$\left[\begin{array}{ccc}1& 1& -4\\ 3& 0& -4\\ 0& -2& 4\end{array}\right]$

The symmetric matrix A is given by

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

$A=\frac{1}{2}\left[\begin{array}{ccc}2& 4& -4\\ 4& 0& -6\\ -4& -6& 8\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 2& -2\\ 2& 0& -3\\ -2& -3& 4\end{array}\right]$

and antisymmetric B is

$B=\frac{1}{2}(\left[\begin{array}{ccc}1& 3& 0\\ 1& 0& -2\\ -4& -4& 4\end{array}\right]-\left[\begin{array}{ccc}1& 1& -4\\ 3& 0& -4\\ 0& -2& 4\end{array}\right]$

$B=\frac{1}{2}\left[\begin{array}{ccc}0& 2& 4\\ -2& 0& 2\\ -4& -2& 0\end{array}\right]$

$B=\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 1\\ -2& -1& 0\end{array}\right]$

$A-B$ = $\left[\begin{array}{ccc}1& 2& -2\\ 2& 0& -3\\ -2& -3& 4\end{array}\right]-\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 1\\ -2& -1& 0\end{array}\right]$

$A-B$ = $\left[\begin{array}{ccc}1& 1& -4\\ 3& 0& -4\\ 0& -2& 4\end{array}\right]$

Hence the answer is option A