Question

Using properties of determinants, prove that:
$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ba& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=a^2+b^2+c^2+1$

Answer 1

$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ba& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=a^2+b^2+c^2+1$
L.H.S $\Delta =\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ba& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|$
Applying $R_1\to \frac{1}{a}{R}_1,R_2\to \frac{1}{b}{R}_2,R_3\to \frac{1}{c}{R}_3$
$\Delta =abc\left|\begin{array}{ccc}a+\frac{1}{a}& b& c\\ a& b+\frac{1}{b}& c\\ a& b& c+\frac{1}{c}\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}{a}^2+1& b^2& c^2\\ a^2& b^2+1& c^2\\ a^2& b^2& c^2+1\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$\Delta =\left|\begin{array}{ccc}1+a^2+b^2+c^2& b^2& c^2\\ 1+a^2+b^2+c^2& b^2+1& c^2\\ 1+a^2+b^2+c^2& b^2& c^2+1\end{array}\right|$
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& b^2& c^2\\ 1& b^2+1& c^2\\ 1& b^2& c^2+1\end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& b^2& c^2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
Expanding along $C_1$, we get
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$
$=a^2+b^2+c^2+1=$ R.H.S.