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

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

 

answered Mar 14, 2013 by sreemathi.v
edited Mar 14, 2013 by sreemathi.v
 

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