Question

Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .

Answer 1

Solving L.H.S., we get : \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\] \[C_1 \to C_1 + C_2 + C_3\]

=  \[\begin{vmatrix}a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z\end{vmatrix}\]

Taking common a+x+y+z from C₁, we get:
(​a+x+y+z)​

\[\begin{vmatrix}1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z\end{vmatrix}\]

\[R_1 \to R_1 - R_2 , R_2 \to R_2 - R_3\]

=(​a+x+y+z)​​

\[\begin{vmatrix}0 & - a & 0 \\ 0 & a & - a \\ 1 & y & a + z\end{vmatrix}\]

Expanding along C₁, we get:
(​a+x+y+z)​​

\[\left\{ 1 \times 1 \times \left( a^2 - 0 \right) \right\}\]

= \[a^2 \left( a + x + y + z \right)\]

L.H.S=R.H.S
Hence proved.