# Find the values of $x$, if $\text{: } \begin{vmatrix} 2&3 \\ 4&5 \end{vmatrix} = \begin{vmatrix} x&3 \\ 2x&5 \end{vmatrix}$

Note: This is part 2 of a 2 part question, split as 2 separate questions here.

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

(ii)$\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}=\begin{vmatrix}x &3\\2x & 5\end{vmatrix}$

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

Hence LHS=$2\times 5-3\times 4$

$\qquad\qquad=10-12$

$\qquad\qquad=-12$

RHS=$x\times 5-3\times 2x$

$\qquad=5x-6x=-x$

Now equating the both,

-x=-2.

$\Rightarrow x=2.$

edited Mar 6, 2013