Solve the following matrices

We have

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

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

We know that a matrix is symmetric if $A=A^{\prime}$.

Thus,

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

Now,

$2 b=3$

$\Rightarrow b=\frac{3}{2}$

Also,

$3 a=-2$

$\Rightarrow a=\frac{-2}{3}$

Therefore,

$a=\frac{-2}{3}$ and $b=\frac{3}{2}$