Question

Using properties of determinants, prove the following:

\[\begin{vmatrix}x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1\end{vmatrix} = 1 + x^2 + y^2 + z^2\] .

Answer 1

Let \[\bigtriangleup = \begin{vmatrix}x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1\end{vmatrix}\]

Multiplying R₁, R₂ and R₃ by x, y and z, respectively, we get:

\[\bigtriangleup = \frac{1}{xyz}\begin{vmatrix}x\left( x^2 + 1 \right) & x^2 y & x^2 z \\ x y^2 & y\left( y^2 + 1 \right) & y^2 z \\ x z^2 & y z^2 & z\left( z^2 + 1 \right)\end{vmatrix}\]

Taking x, y and z common from the columns C₁, C₂ and C₃, respectively, we get:

\[\bigtriangleup = \frac{xyz}{xyz}\begin{vmatrix}\left( x^2 + 1 \right) & x^2 & x^2 \\ y^2 & \left( y^2 + 1 \right) & y^2 \\ z^2 & z^2 & \left( z^2 + 1 \right)\end{vmatrix}\]

Applying R₁  \[\to\] + R₂ + R₃, we get: 

\[\bigtriangleup = \begin{vmatrix}\left( 1 + x^2 + y^2 + z^2 \right) & \left( 1 + x^2 + y^2 + z^2 \right) & \left( 1 + x^2 + y^2 + z^2 \right) \\ y^2 & \left( y^2 + 1 \right) & y^2 \\ z^2 & z^2 & \left( z^2 + 1 \right)\end{vmatrix}\]

\[\Rightarrow \bigtriangleup = \left( 1 + x^2 + y^2 + z^2 \right)\begin{vmatrix}1 & 1 & 1 \\ y^2 & \left( y^2 + 1 \right) & y^2 \\ z^2 & z^2 & \left( z^2 + 1 \right)\end{vmatrix}\]

Applying

\[C_2 \to C_2 - C_1\text { and } C_3 \to C_3 - C_1\] we get:

\[\bigtriangleup = \left( 1 + x^2 + y^2 + z^2 \right)\begin{vmatrix}1 & 0 & 0 \\ y^2 & 1 & 0 \\ z^2 & 0 & 1\end{vmatrix} = \left( 1 + x^2 + y^2 + z^2 \right) \times 1 = \left( 1 + x^2 + y^2 + z^2 \right)\]

Hence proved.