Question

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

Answer 1

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 $\left|\begin{array}{ccc}x& 2& 0\\ 2& -x& 0\\ 0& 3& x\end{array}\right|=0$
We need to expand the determinant to get an equation.
While expanding a determinant, the signs are as follows,
$\left|\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right|$
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 $\left|\begin{array}{ccc}x& 2& 0\\ 2& -x& 0\\ 0& 3& x\end{array}\right|=0$
$+x\left|\begin{array}{cc}-x& 0\\ 3& x\end{array}\right|-2\left|\begin{array}{cc}2& 0\\ 0& x\end{array}\right|+0\left|\begin{array}{cc}2& -x\\ 0& 3\end{array}\right|=0$
Now we need to expand the $2\times 2$determinant,
$x(-x(x)-0(3))-2(2(x)-0(0))+0=0$
$\begin{array}{c}x(-x^2-0)-2(2x-0)+0=0\\ -x^3-4x=0\\ x^3+4x=0\\ x(x^2+4)=0\\ x=0,x^2=-4\end{array}$
$x^2=-4$ 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.