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# If $\begin{vmatrix}2x & 5\\8 & x\end{vmatrix}=\begin{vmatrix}6 & -2\\7 & 3\end{vmatrix}$,then value of x is $\begin{array}{1 1}(A)\quad3 & (B)\quad\pm 3\\(C)\quad \pm 6 & (D)\quad 6\end{array}$

Toolbox:
• If $A=\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$
• Then $|A|=a_{11}\times a_{22}-a_{12}\times a_{21}$
Given:$\begin{vmatrix}2x & 5\\8 & x\end{vmatrix}=\begin{vmatrix}6 & -2\\7 & 3\end{vmatrix}$
On expanding we get,
$2x^2-40=18+14$
$\Rightarrow 2x^2-40=32$
$2x^2=40+32=72$
$x^2=\frac{72}{2}=36$
x$=\pm 6$
Hence the correct answer is C