# How to solve this determinant equation in a simpler way

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.

• use Sarrus rule Commented Aug 12, 2016 at 18:36

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.

• 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. Commented Aug 13, 2016 at 1:57

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 $x = \pm 2\sqrt{2}$.

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...)

• +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. Commented Aug 12, 2016 at 18:34
• @miracle173: In all fairness, they posted their solutions many hours before I did... Commented Aug 12, 2016 at 19:02