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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using properties of determinants,solve for x,\[\begin{vmatrix}a+x & a-x& a-x\\a-x & a+x & a-x\\a-x & a-x & a+x\end{vmatrix}=0\]

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Toolbox:
  • Elementary transformation can be done by
  • (i)$R_i \leftrightarrow R_j$ or $C_i\leftrightarrow C_j.$
  • (ii)$R_i\leftrightarrow kR_i$ or $C_i\leftrightarrow kC_i$
  • (iii)$R_i\rightarrow R_i+kR_j$ or $C_i\rightarrow C_i+kC_j$
  • The value of the determinant can be obtained by expanding it along any of its row(or column) and multiplying by its cofactors.
Let $\Delta=\begin{vmatrix}a+x & a-x & a-x\\a-x & a+x &a-x\\a-x & a-x & a+x\end{vmatrix}$
 
Apply $C_1\rightarrow C_1+C_2+C_3$,we get
 
$\Delta=\begin{vmatrix}3a-x & a-x & a-x\\3a-x & a+x &a-x\\3a-x & a-x & a+x\end{vmatrix}$
 
Now take (3a-x) as the common factor from $C_1$
 
$\Delta=(3a-x)\begin{vmatrix}1 & a-x & a-x\\1 & a+x &a-x\\1 & a-x & a+x\end{vmatrix}$
 
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
 
$\Delta=(3a-x)\begin{vmatrix}0 & -2x & 0\\0 & 2x &-2x\\1 & a-x & a+x\end{vmatrix}$
 
Now expanding along $R_1$,we get
 
$\Delta=(3a-x)\begin{vmatrix}2x & 0\\2x &-2x\end{vmatrix}=(3a-x)(4x^2-0)=4x^2(3a-x)$
 
Now it is given $\Delta=0.$
 
$4x^2(3a-x)=0\Rightarrow x=0$ or x=3a.
 
Hence the value of x are 0 or 3a.

 

answered Mar 13, 2013 by sreemathi.v
 

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