Question

Prove that

\[\begin{vmatrix}a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1\end{vmatrix} = 1 + a^2 + b^2 + c^2\]

Answer 1

\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 + 1 & ab & ac\\ab & b^2 + 1 & bc\\ca & cb & c^2 + 1 \end{vmatrix}\] 
\[ = \left( abc \right) \begin{vmatrix} a + \frac{1}{a} & b & c\\a & b + \frac{1}{b} & c\\a & b & c + \frac{1}{c} \end{vmatrix} \left[\text{ Taking out a, b and c common from }R_1 , R_2\text{ and }R_3 \right]\] 
\[ = \left( abc \right) \begin{vmatrix} a + \frac{1}{a} & b & c\\ - \frac{1}{a} & \frac{1}{b} & 0 \\ - \frac{1}{a} & 0 & \frac{1}{c} \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\] 
\[ = \left( abc \right) \left( \frac{1}{abc} \right)\begin{vmatrix} a^2 + 1 & b^2 & c^2 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{vmatrix} \left[\text{ Applying }C_1 \to a C_1 , C_2 \to b C_2\text{ and }C_3 \to c C_3 \right]\] 
\[ = \begin{vmatrix} a^2 + 1 & b^2 & c^2 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{vmatrix}\] 
\[ = \left( - 1 \right) \begin{vmatrix} b^2 & c^2 \\ 1 & 0 \end{vmatrix} + \left( 1 \right) \begin{vmatrix} a^2 + 1 & b^2 \\ - 1 & 1 \end{vmatrix} \left[\text{ Expanding along }R_3 \right]\] 
\[ = \left( - 1 \right) \left( - c^2 \right) + \left( a^2 + 1 + b^2 \right)\] 
\[ = \left( a^2 + 1 + b^2 + c^2 \right)\] 
\[ = \left( a^2 + b^2 + c^2 + 1 \right)\] 
\[ = RHS\]