# Find the value of $x$, if $\begin{vmatrix} 2&4 \\ 5&1 \end{vmatrix} = \begin{vmatrix} 2x&4 \\ 6&x \end{vmatrix}$

Toolbox:
• A determinant of order $2\times 2$ can be evaluated as $\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$
• $\mid A\mid=a_{11}a_{22}-a_{21}a_{12}$
Find the values of x if

(i)$\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x &4\\6 & x\end{vmatrix}$

We know the value of determinant of order $2\times 2$ is $(a_{11}a_{22}-a_{21}a_{12})$

Hence LHS=$2\times 1-5\times 4$

$\qquad\qquad=2-20$

$\qquad\qquad=-18$

RHS=$2x\times x-6\times 4$

$\qquad=2x^2-24$

Now equating the both,

$2x^2-24=-18$

$\Rightarrow 2x^2=24-18$

$\;\;2x^2=6$

$\Rightarrow x^2=\frac{6}{2}=3$

$x=\pm\sqrt 3$.

edited Feb 20, 2013