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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.

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.

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