Question

Using the properties of determinants, show that:

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

Answer 1

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

Taking $a,b,c$ common from $R_1,R_2,R_3$ respectively

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

Multiplying $C_1,C_2,C_3$ by $a,b,c$ respectively

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

$C_1\to C_1+C_2+C_3$

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

Taking $(1+a^2+b^2+c^2)$ common from $C_1$

$=(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& b^2& c^2\\ 1& b^2+1& c^2\\ 1& b^2& c^2+1\end{array}\right|$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

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

Expanding along first column,

$=(1+a^2+b^2+c^2)\left|\begin{array}{cc}-1& 0\\ b^2& -1\end{array}\right|$

$=1+a^2+b^2+c^2$