Question

Let $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|=k+a^2+b^2+c^2$
then $4k$ is

Answer 1

Let $\Delta =\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|$
First multiplying $R_1,R_2,R_3$ by $a,b,c$ respectively and taking
$a,b,c$ common from $C_1,C_2,C_3$ respectively, we get
$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+1& a^2& a^2\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|$
Now Applying $R_1\to R_1+R_2+R_3$ then
$\Delta =\left|\begin{array}{ccc}1+a^2+b^2+c^2& 1+a^2+b^2+c^2& 1+a^2+b^2+c^2\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|$
Taking $1+a^2+b^2+c^2$ from $R_1$, then
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& 1& 1\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|$
Applying $C_2\to C_2-C_1$, then
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& 0& 1\\ b^2& 1& b^2\\ c^2& 0& c^2+1\end{array}\right|$
Expanding along $C_2$, then
$\Delta =(1+a^2+b^2+c^2).1.\left|\begin{array}{cc}1& 1\\ c^2& c^2+1\end{array}\right|$
$=(1+a^2+b^2+c^2).1.1$
Thus, $\Delta =1+a^2+b^2+c^2$

$\Rightarrow k=1$
Hence, $4k=4$.