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Home  >>  CBSE XII  >>  Math  >>  Determinants
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The value of the determinant $\begin{vmatrix}x & x+y & x+2y\\x+2y & x & x+y\\x+y & x+2y & x\end{vmatrix}$ is

$\begin{array}{1 1} 9x^2(x+y) \\ 9y^2(x+y) \\ 3y^2 (x+y) \\ 7x^2(x+y)\end{array} $
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1 Answer

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Toolbox:
  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
$\Delta=\begin{vmatrix}x & x+y & x+2y\\x+2y & x & x+y\\x+y & x+2y & x\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}-2y & y & y\\y & -2y & y\\x+y & x+2y & x\end{vmatrix}$
Take y as the common factor from $R_1$ and $R_2$
$\Delta=y^2\begin{vmatrix}-2 & 1 & 1\\1 & -2 & 1\\x+y & x+2y & x\end{vmatrix}$
Now expanding along $R_1$
$\Delta=y^2[-2(-2x-(x+2y))-1(x-(x+y))+1(x+2y+2(x+y))]$
$\quad=y^2[4x+2x+4y-x+x+y+x+2y+2x+2y]$
$\quad=y^2[9x+9y]$
$\quad=9y^2[x+y]$
Hence B is the correct answer.
answered Mar 26, 2013 by sreemathi.v
 

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