Question

By using properties of determinants, show that: 

\(\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix}\)=1+a²+b²+c²

Answer 1

△=\(\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix}\)
Taking out common factors a, b, and c from R₁, R₂, and R₃ respectively, we have: 

△=abc\(\begin{vmatrix}a+\frac{1}{a}&b&c\\a&b+\frac{1}{b}&c\\a&b&c+\frac{1}{c}\end{vmatrix}\)
Applying R₂ \(\to\) R₂ − R₁ and R3 \(\to\) R₃ − R₁, we have: 

△=abc\(\begin{vmatrix}a+\frac{1}{a}&b&c\\-\frac{1}{a}&\frac{1}{b}&0\\-\frac{1}{a}&0&\frac{1}{c}\end{vmatrix}\)

Applying C₁ \(\to\) a C₁, C₂ → b C₂, and C₃ → c C₃, we have

△=abc.\(\frac{1}{abc}\)\(\begin{vmatrix}a^2+1&b^2&c^2\\-1&1&0\\-1&0&1\end{vmatrix}\)

=\(\begin{vmatrix}a^2+1&b^2&c^2\\-1&1&0\\-1&0&1\end{vmatrix}\)
Expanding along R₃, we have: 

△=-1\(\begin{vmatrix}b^2&c^2\\1&0\end{vmatrix}+\begin{vmatrix}a^2+1&b^2\\-1&1\end{vmatrix}\)

=-1(-c²)+(a²+1+b²)=1+a²+b²+c²

Hence, the given result is proved.