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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
edited Mar 14, 2013