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.