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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve for x : $ \begin{vmatrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \end{vmatrix} = 0$

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Toolbox:
  • The value of the determinant can be found by expanding along any rows or columns.
  • Elementary transformation in a determinant can be made by
  • (a) Interchanging two rows or columns.
  • (b) By adding or subtracting two or more rows or columns
Step 1:
$\Delta=\begin{vmatrix}3x-8 & 3 & 3\\3 & 3x-8 & 3\\3 & 3 & 3x-8\end{vmatrix}$
Apply $R_1\rightarrow R_1+R_2+R_3$
$\Delta=\begin{vmatrix}3x-2 & 3x-2 & 3x-2\\3 & 3x-8 & 3\\3 & 3 & 3x-8\end{vmatrix}$
Take (3x-2) as the common factor from $R_1$
$\Delta=3x-2\begin{vmatrix}1 & 1 & 1\\3 & 3x-8 & 3\\3 & 3 & 3x-8\end{vmatrix}$
Apply $C_2\rightarrow C_2-C_3$ and $C_3\rightarrow C_3-C_1$
$\Delta=3x-2\begin{vmatrix}1 & 0 & 0\\3 & 3x-1 & 0\\3 & -3x+11 & 3x-11\end{vmatrix}$
Step 2:
Expanding along $R_1$ we get,
$\Delta=(3x-2)[1(3x-1)^2-0]$
But it is given $|\Delta|$=0.
$0=(3x-2)(3x-11)^2$
Either 3x-2=0 0r 3x-11=0.
If 3x-2=0 $\Rightarrow x=\frac{2}{3}$
If 3x-11=0 $\Rightarrow x=\frac{11}{3}$
x=$\frac{-2}{3},\frac{11}{3}$
answered Apr 8, 2013 by sreemathi.v
 

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