Question

If $A=\left[\begin{array}{ccc}\sqrt{3}& 2& 3\\ 2& -1.5& -1\\ 3& -1& 1\end{array}\right]$, then

• A. $A^{\prime }=-A$
• B. $A^{\prime }=A$
• C. $A^{\prime }=2A$
• D. $A^{\prime }=-2A$

Answer 1

Answer: (B)

We have, $A=\left[\begin{array}{ccc}\sqrt{3}& 2& 3\\ 2& -1.5& -1\\ 3& -1& 1\end{array}\right]$
$A^{\prime }=\left[\begin{array}{ccc}\sqrt{3}& 2& 3\\ 2& -1\cdot b& -1\\ 3& -1& 1\end{array}\right]=A$
$A^{\prime }=A$
Thus, $A$ is a symmetric matrix.