# $\text{If } \begin{vmatrix} 2 x&5 \\ 8&x \end{vmatrix} = \begin{vmatrix} 4&5 \\ 8&3 \end{vmatrix},then\; x\; is\; equal\; to$ $(A)\; 4\qquad(B)\; \pm 6\qquad (C)\; -6\qquad (D)\;0$

Toolbox:
• A determinant of order $2\times 2$ can be evaluated as $\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$
• $\mid A\mid=a_{11}a_{22}-a_{21}a_{12}$
If $\begin{vmatrix}2x & 5\\8 & x\end{vmatrix}=\begin{vmatrix}4 & 5\\8 & 3\end{vmatrix}$
then x is
We know the value of determinant of order $2\times 2$ is $(a_{11}a_{22}-a_{21}a_{12})$
Hence LHS=$2x^2-(8\times 5)$
$\qquad\qquad=2x^2-40$
$RHS=4\times3-8\times 5$
$\qquad=12-40$
Equating LHS and RHS
$\Rightarrow\:2x^2=40=12-40$
$\Rightarrow\;x^2=6$
$x=\pm \sqrt 6$
Hence the correct answer is B.
edited Jan 31, 2014