Question

Prove the following using properties of determinants:
$\left|\begin{array}{ccc}{1+a}^2& ab& ac\\ ab& 1+b^2& bc\\ ca& cb& 1+c^2\end{array}\right|=1+a^2+b^2+c^2$

Answer 1

Given that,

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

Now,

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

Multiplying $c_1,c_2,c_3$ by $a,b,c$

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

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

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

$\begin{array}{c}{R}_2\to R_2-R_1\\ R_3\to R_3-R_1\end{array}$

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

Hence,Proved