3 and 4 .Determinants and Matrices
medium

Find values of $x$, if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

A

$ x=\pm \sqrt{2}$

B

$ x=\pm \sqrt{7}$

C

$ x=\pm \sqrt{6}$

D

$ x=\pm \sqrt{3}$

Solution

$\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

$\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4$

$\Rightarrow 2-20=2 x^{2}-24$

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

$\Rightarrow x^{2}=3$

$\Rightarrow x=\pm \sqrt{3}$

Standard 12
Mathematics

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