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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.

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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.

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    $\begingroup$ Another way of seeing that those are all solutions is to note that $\det A$ is a second-degree polynomial in $x$, so it has only two roots. $\endgroup$ – Robin Ekman Aug 13 '16 at 1:57
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Subtract the first row from the last and use Laplace's formula for the third row: $$\begin{vmatrix} x & 2 & 3 \\ 4 & x & 1 \\ x & 2 & 5 \\ \end{vmatrix}= \begin{vmatrix} x & 2 & 3 \\ 4 & x & 1 \\ 0 & 0 & 2 \\ \end{vmatrix} = 2(-1)^{3+3}\begin{vmatrix} x & 2\\ 4 & x \end{vmatrix} = 2(x^2-8) = 2(x-2\sqrt{2})(x+2\sqrt{2})=0.$$ Now the roots are $x = \pm 2\sqrt{2}$.

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Just expand the determinant! We want $$ 5x^2+2x+24-2x-40-3x^2=0 $$ which simplifies to $2x^2-16=0$.

(In general, finding determinants via row/column relations is faster when a matrix is large. But $3 \times 3$ matrices aren't that large yet...)

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    $\begingroup$ +1 You use Sarrus rule to calculate the determinant. This is the fastetst and simplest way. But it seems that more complicated solutions that took longer are prefered by the readers. $\endgroup$ – miracle173 Aug 12 '16 at 18:34
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    $\begingroup$ @miracle173: In all fairness, they posted their solutions many hours before I did... $\endgroup$ – Micah Aug 12 '16 at 19:02

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