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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the value of $x$, if $ \begin{vmatrix} 2&4 \\ 5&1 \end{vmatrix} = \begin{vmatrix} 2x&4 \\ 6&x \end{vmatrix}$

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  • 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=$2\times 1-5\times 4$
RHS=$2x\times x-6\times 4$
Now equating the both,
$\Rightarrow 2x^2=24-18$
$\Rightarrow x^2=\frac{6}{2}=3$
$x=\pm\sqrt 3$.


answered Feb 20, 2013 by sreemathi.v
edited Feb 20, 2013 by sreemathi.v

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