Equation with one variable unknow how to solve easily

Question

I have

$$\left|\begin{array}{ccc} x & -2 & 3x-6 \\ 2x & \phantom{-}0 & 2-x \\ -x & \phantom{-}5 & x-2 \end{array}\right| = 0$$

How can I solve this with a fast way? I thought I can do $$x-2+3x-6=0 \implies 4x=-8 \implies x=-2$$ and I will continue with the other two, but I don't know if I am right.

Answer 1

Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:

$$0 = \left|\begin{array}{ccc} \phantom{-}x & -2 & 3x-6 \\ 2x & \phantom{-}0 & 2-x \\ -x & \phantom{-}5 & x-2 \end{array}\right| = x(x-2) \left|\begin{array}{ccc} \phantom{-}1 & -2 & \phantom{-}3 \\ \phantom{-}2& \phantom{-}0 & -1 \\ -1& \phantom{-}5 & \phantom{-}1 \end{array}\right| = 37x(x-2) \quad\implies\quad x=0,2$$

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To add a bit of detail ...

Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):

$$\left|\begin{array}{ccc} pa & b & c \\ pd & e & f \\ ph & i & j \end{array}\right| \;=\; p\left|\begin{array}{ccc} a & b & c \\ d & e & f \\ h & i & j \end{array}\right|$$

The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2\times 2$:

$$\left|\begin{array}{cc} pa & b \\ pc & d \end{array}\right| = pa\cdot d -pc\cdot b = p(ad-cb) = p\left|\begin{array}{cc} a & b \\ c & d \end{array}\right|$$ Every term in the expansion includes the multiplier, which can be factored-out.

Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).

Answer 2

This is wrong: $$x-2+3x-6=0 \implies 4x=-8 \implies x=-2 $$

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This is the right way to do it $$\left|\begin{array}{ccc} x & -2 & 3x-6 \\ 2x & \phantom{-}0 & 2-x \\ -x & \phantom{-}5 & x-2 \end{array}\right| = 0$$

Apply $R3 \rightarrow R3+R1$ $$\left|\begin{array}{ccc} x & -2 & 3x-6 \\ 2x & \phantom{-}0 & 2-x \\ 0 & \phantom{-}3 & 4x-8 \end{array}\right| = 0$$

Apply $R2\rightarrow R2-2R1$ $$\left|\begin{array}{ccc} x & -2 & 3x-6 \\ 0 & \phantom{-}4 & 14-7x \\ 0 & \phantom{-}3 & 4x-8 \end{array}\right| = 0$$

$$x\left( 4(4x-8)-3(14-7x)\right)=0$$ $$37x(x-2)=0 \implies x=0,2$$

Answer 3

You can also use the Laplace formula on the second column and get

$$\begin{vmatrix} x & -2 & 3x-6 \\ 2x & 0 & 2-x \\ -x & 5 & x-2 \end{vmatrix} = 2 \begin{vmatrix} 2x & 2-x \\ -x & x-2 \end{vmatrix} - 5 \begin{vmatrix} x & 3x-6 \\ 2x & 2-x \end{vmatrix}=37x(x-2) $$

Hence $x \in \{0,2\}$.