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Using the properties of determinants, evaluate $\begin{vmatrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{vmatrix}$

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  • If each element of a row(or column)of a determinant is multiplied by a constant 'k' then |A|=k|A|.
  • If each element of a row(or a column)of A is expressed as the sum of two or more terms,then the determinants of A can be expressed as the sum of determinants of two or more matrices.
Let $\Delta=\begin{vmatrix}3x & -x+y &-x+z\\x-y& 3y & z-y\\x-z & y-z&3z\end{vmatrix}$
Apply $C_1\rightarrow C_1-C_2+C_3$,we get
$\Delta=\begin{vmatrix}x+y+z & -x+y &-x+z\\x+y+z& 3y & z-y\\x+y+z & y-z&3z\end{vmatrix}$
Let us take x+y+z as the common factor from $C_1$,then
$\Delta=(x+y+z)\begin{vmatrix}1 & -x+y &-x+z\\1& 3y & z-y\\1 & y-z&3z\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_1$,$R_3\rightarrow R_3-R_1$
$\Delta=(x+y+z)\begin{vmatrix}1 & -x+y &-x+z\\0& 2y+x & x-y\\0 & x-z&2z+x\end{vmatrix}$
On expanding along $R_1$ we get,
On simplifying we get,


answered Mar 14, 2013 by sreemathi.v

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