Question

Using properties of determinants, prove 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

LHS :

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

$=\frac{1}{abc}\left|\begin{array}{ccc}a(a^2+1)& a^{2}b& a^{2}c\\ a{b}^2& b(b^2+1)& b^{2}c\\ c^{2}a& c^{2}b& c(c^2+1)\end{array}\right|$ $(R_1\to a{R}_1,R_2\to b{R}_2,R_3\to c{R}_3)$

$=\left|\begin{array}{ccc}{a}^2+1& a^2& a^2\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|$ $(C_1\to \frac{1}{a}{C}_1,C_2\to \frac{1}{b}{C}_2,C_3\to \frac{1}{3}{C}_3)$

$=\left|\begin{array}{ccc}1+a^2+b^2+c^2& 1+a^2+b^2+c^2& 1+a^2+b^2+c^2\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|$ $(R_1\to R_1+R_2+R_3)$

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

$=(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& 0& 0\\ b^2& 1& 0\\ c^2& 0& 1\end{array}\right|$ $(C_2\to C_2-C_1,C_3\to C_3-C_1)$

Now, on expanding along $R_1$, we get,

$=(1+a^2+b^2+c^2)1[1-0]=1+a^2+b^2+c^2$ = RHS