Solve the following$\left| {\begin{array}{*{20}{c}}
x&2&0 \\
2&{ - x}&0 \\
0&3&x
\end{array}} \right| = 0$

Hint: Expanding the determinant yields an equation in terms of $x$, which then can be solved to determine the value of $x$.

The given determinant is $\left| {\begin{array}{*{20}{c}}
x&2&0 \\
2&{ - x}&0 \\
0&3&x
\end{array}} \right| = 0$
We need to expand the determinant to get an equation.
While expanding a determinant, the signs are as follows,
$\left| {\begin{array}{*{20}{c}}
+ & - & + \\
- & + & - \\
+ & - & +
\end{array}} \right|$
Choose a row or a column and start taking out one term after another and multiplying it with the
minors along with the corresponding sign.
When a term is taken out, the determinant leaving out the row and column of the term taken out
is called the minor.
Let us expand $\left| {\begin{array}{*{20}{c}}
x&2&0 \\
2&{ - x}&0 \\

0&3&x
\end{array}} \right| = 0$
$ + x\left| {\begin{array}{*{20}{c}}
{ - x}&0 \\
3&x
\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}
2&0 \\
0&x
\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}
2&{ - x} \\
0&3
\end{array}} \right| = 0$
Now we need to expand the $2 \times 2$determinant,
$x( - x(x) - 0(3)) - 2(2(x) - 0(0)) + 0 = 0$
$\begin{gathered}
x( - {x^2} - 0) - 2(2x - 0) + 0 = 0 \\
- {x^3} - 4x = 0 \\
{x^3} + 4x = 0 \\
x({x^2} + 4) = 0 \\
x = 0,{x^2} = - 4 \\
\end{gathered} $
${x^2} = - 4$ is not possible because it yields imaginary roots.
$x = 0$ is the correct answer.

Note: Expanding the determinant yields an equation in terms of $x$, which then can be solved to determine the value of $x$. The signs, while expanding the determinant can be tricky.