Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 \\
x+7 & 2 & 5 \\
\end{vmatrix} \tag{$C_1\rightarrow C_1+C_2+C_3$}$$
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}\tag{1}$$
On opening the first determinant in the last step above we get $2(2-3x)$.
On simplifying the secind determinant we get,
$$(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=(x+5)\begin{vmatrix}
1 & 2 & 3 \\
0 & x-2 & -2 \\
0 & 0 & 2 \\
\end{vmatrix}
(R_2\rightarrow R_2-R_1)
(R_3\rightarrow R_3-R_1)$$
$=2(x+5)(x-2)$
Substituting the values obtained above in $(1)$, we get
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=2(2-3x)+2(x+5)(x-2)=2(2-3x+x^2+3x-10)=2(x^2-8)$$
Now, as $\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$, $\therefore 2(x^2-8)=0\implies x=\pm2\sqrt2$
As you can see there was lot of work in my solution so if anyone can provide me with some techniques to solve it faster, or a technique which includes less amount of pen and more thinking.
Best Answer
Notice that the first two columns are proportional, hence linearly dependent, if
$$\frac x 2 = \frac 4 x$$
which is the same as $x = \pm 2 \sqrt{2}$ after solving. Convince yourself that the third column can never be written as a linear combination of the first two by noticing that the first and second columns have equal components in rows 1 and 3; you can work this into rigorous proof that the above equation gives all solutions.