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.
