Question

$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=1+a^2+b^2+c^2$

Answer 1

$LHS=\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|$
Taking out common factors a, b and c from $R_1,R_2$, and $R_3$ respectively, we get

$=abc\left|\begin{array}{ccc}a+\frac{1}{a}& b& c\\ a& b+\frac{1}{b}& c\\ a& b& c+\frac{1}{c}\end{array}\right|=abc\left|\begin{array}{ccc}a+\frac{1}{a}& b& c\\ -\frac{1}{a}& \frac{1}{b}& 0\\ -\frac{1}{a}& 0& \frac{1}{c}\end{array}\right|(using{R}_2\to R_2-R_{1}and{R}_3\to R_3-R_1)$

Multiply and divide $C_1$ by a, $C_2$ by b and $C_3$ by c and take common $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$ out from $C_1,C_2$ and $C_3$ respectively, we get

$=abc\times \frac{1}{abc}=\left|\begin{array}{ccc}{a}^2+1& b^2& c^2\\ -1& 1& 0\\ -1& 0& 1\end{array}\right|=\left|\begin{array}{ccc}{a}^2+1& b^2& c^2\\ -1& 1& 0\\ -1& 0& 1\end{array}\right|$

Expanding along $R_3$, we get
$=-1\times (-c^2)+1\left[1\right(a^2+1)+1(b^2\left)\right]=(a^2+b^2+c^2+1)=1+a^2+b^2+c^2=RHS$