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

## 1 Answer

Toolbox:
• 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,

$\Delta=(x+y+z)[1(2y+x)(2z+x)-(x-z)(x-y)]-(x+y)[0]+(-x+z)(0)$

$\Delta=(x+y+z)[(2y+x)(2z+x)-(x-z)(x-y)]$

On simplifying we get,

$\Delta=(x+y+z)[4yz+2xy+2xz+x^2-x^2+xy+xz-yz]$

$\quad=(x+y+z)[3yz+3xy+3xz]$

$\Delta=3(x+y+z)(xy+yz+xz)$

answered Mar 14, 2013

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer