Question

If $A=\begin{bmatrix}\sqrt{3} & 2 & 3 \\ 2 & -1.5 & -1 \\ 3 & -1 & 1\end{bmatrix}$, then

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

Answer 1

Answer: (B)

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