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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using properties of determinants, prove the following : $ \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x) $

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Toolbox:
  • If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
  • By this property we can take out any common factor from any one row or any one column of the determinant.
  • Elementary transformations can be done by
  • 1. Interchanging any two rows or columns.
  • 2. Mutiplication of the elements of any row or column by a non-zero number
  • The addition of any row or column , the corresponding elements of any other row or column multiplied by any non zero number.
Step 1:
Let $\Delta=\begin{vmatrix}x&y&z\\x^2&y^2&z^2\\x^3&y^3&z^3\end{vmatrix}$
Taking $xyz$ as the common factor from $C_1$
$\Delta=xyz\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}$
Step 2:
$C_1\rightarrow C_1-C_2,C_2\rightarrow C_2-C_3$
$\Delta=\begin{vmatrix}0&0&1\\x-y&y-z&z\\x^2-y^2&y^2-z^2&z^2\end{vmatrix}$
$\quad=xyz\begin{vmatrix}0&0&1\\x-y&y-z&z\\(x-y)(x+y)&(y-z)(y+z)&z^2\end{vmatrix}$
Taking $(x-y)$ and $(y-z)$ as a common factor from $C_1$ and $C_2$
$\quad=xyz(x-y)(y-z)\begin{vmatrix}0&0&1\\1&1&z\\x+y&y+z&z^2\end{vmatrix}$
Step 3:
On expanding we get,
$\Delta=xyz(x-y)(y-z)[1(y+z-x-y)]$
$\therefore \Delta=xyz(x-y)(y-z)(z-x)$
Hence proved.
answered Dec 2, 2013 by sreemathi.v
 

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