Question

If $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& -1& 4\end{array}\right],A=B+C,B=B^T,C=-C^T$, then $C$ is equal to

• A. $\left[\begin{array}{ccc}0& 0.5& 0\\ -0.5& 0& 0\\ 0& 0& 0\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0.5\\ 0& -0.5& 0\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& -0.5& 0.5\\ 0.5& 0& 0.5\\ -05& 0& 0\end{array}\right]$
• D. $\left[\begin{array}{ccc}0& 0.5& 0\\ -0.5& 0& 0.5\\ 0& -0.5& 0\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0.5\\ 0& -0.5& 0\end{array}\right]$

Given, $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& -1& 4\end{array}\right]\text{Since}A=B+C\text{and}B=B^T,C=-C^T$
A can be represented as
$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)\therefore \frac{1}{2}(A+A^T)=B\text{and}\frac{1}{2}(A-A^T)=C2C=\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& -1& 4\end{array}\right]-\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& -1\\ 1& 0& 4\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right]C=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0.5\\ 0& -0.5& 0\end{array}\right]$