Question

Evaluate the following:

$\left|\begin{array}{ccc}a+x& y& z\\ x& a+y& z\\ x& y& a+z\end{array}\right|$

Answer 1

Let $∆=\left|\begin{array}{ccc}a+x& y& z\\ x& a+y& z\\ x& y& a+z\end{array}\right|$.
$$\begin{gathered} ∆=\left|\begin{array}{ccc}a+x& y& z\\ x& a+y& z\\ x& y& a+z\end{array}\right| \\ =\left|\begin{array}{ccc}a+x+y+z& y& z\\ a+x+y+z& a+y& z\\ a+x+y+z& y& a+z\end{array}\right|\left[\mathrm{Applying}{C}_1\to C_1+C_2+C_3\right] \\ =\left(a+x+y+z\right)\left|\begin{array}{ccc}1& y& z\\ 1& a+y& z\\ 1& y& a+z\end{array}\right|\left[\mathrm{Taking}\left(a+x+y+z\right)\mathrm{common}\mathrm{from}{C}_1\right] \\ =\left(a+x+y+z\right)\left|\begin{array}{ccc}1& y& z\\ 0& a& 0\\ 0& 0& a\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ =\left(a+x+y+z\right)a^2\left[\mathrm{Expanding}\mathrm{along}\mathrm{first}\mathrm{column}\right] \\ \therefore ∆=\left(a+x+y+z\right)a^2 \end{gathered}$$