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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $x,y,z$ are all different from zero and $\begin{vmatrix}1+x & 1 & 1\\1 & 1+y & 1\\1 & 1 & 1+z\end{vmatrix}$ = 0, then value of $x^{-1}+y^{-1}+z^{-1}$ is

$\begin{array}{1 1} xyz \\ x^{-1}y^{-1}z^{-1} 1 \\ -x-y-z \\ -1\end{array} $

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1 Answer

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Toolbox:
  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=(-1)^{1+1}a_{11}\begin{vmatrix}a_{22} &a_{23}\\a_{32} &a_{33}\end{vmatrix}+(-1)^{1+2}.a_{12}\begin{vmatrix}a_{21} &a_{23}\\a_{31} &a_{32}\end{vmatrix}+(-1)^{1+3}a_{13}\begin{vmatrix}a_{21} &a_{22}\\a_{31} &a_{32}\end{vmatrix}$
  • If each row or column of a determinant is multiplied by k,then its value is multiplied by k,then the value of the determinant is k|A|.
Let $\Delta=\begin{vmatrix}1+x& 1& 1\\1 & 1+y &1\\1 & 1&1+z\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}x& -y& 0\\ & y &-z\\1 & 1&1+z\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=x[y(1+z)+z]+y[(0+z)]+0$
$\quad=xy+xyz+xz+yz$
Given $|\Delta|=0$
xy+xz+yz+xyz=0
$\Rightarrow xy+yz+xz=-xyz$
Divide by xyz on both sides
$\frac{1}{z}+\frac{1}{y}+\frac{1}{x}$=-1
$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$
Hence D is the correct answer.

 

answered Mar 26, 2013 by sreemathi.v
 

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