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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve for $x$ if $\begin{vmatrix} 1 & 4 & 4 \\ 1 & -2 & 1 \\ 1 & 2x & x^2 \end{vmatrix} =0 $

$\begin{array}{1 1} (A) x=2,-1 \\(B) x=-2,1 \\ (C) x=-2,-1 \\ none\;of \;the \;above \end{array}$

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  • The deteminant value can be found by expanding along any of its rows or columns
Let $\Delta=\begin{vmatrix} 1 & 4 & 4 \\ 1 & -2 & 1 \\ 1 & 2x & x^2 \end{vmatrix} $
Apply $ C_2 \to C_2 - C_1$ and $C_3 \to C_3-C_1$
$\Delta=\begin{vmatrix} 1 & 4 & 4 \\ 0 & -6 & -3 \\ 0 & 2x-4 & x^2-4 \end{vmatrix} $
Expanding along $C_1$ we get
$\Delta =1 \bigg[-6(x^2-4)+3(2x-4)\bigg]$
But given$|\Delta|=0$
Therefore $0=-6(x^2-x-2)$
on factorizing we get
$(x-2)(x+1)=0\;=>x=2\;or-1$
answered Apr 8, 2013 by meena.p
 

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