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Q : 7         Find values of x, if

      (i)         \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}

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Given that \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =2-20 = -18      and   \begin{vmatrix}2x &4 \\6 &x \end{vmatrix} = 2x(x)-24 = 2x^2-24

So, we have then,

-18= 2x^2-24      or    3= x^2       or   x= \pm \sqrt{3}

Posted by

Divya Prakash Singh

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