Question

The value of $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|$ is:

• A. $1+a+b+c$
• B. $1+a^3+b^3+c^3$
• C. $abc$
• D. $1+a^2+b^2+c^2$

Answer 1

The correct option is D $1+a^2+b^2+c^2$

$\Delta =\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|$
Multiplying $C_1,C_2,C_3$ by $a,b,c$ respectively

$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}a(a^2+1)& a{b}^2& a{c}^2\\ a^{2}b& b(b^2+1)& b{c}^2\\ a^{2}c& b^{2}c& c(c^2+1)\end{array}\right|$
Taking common $a,b,c$ from $R_1,R_2,R_3$ respectively

$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+1& b^2& c^2\\ a^2& b^2+1& c^2\\ a^2& b^2& c^2+1\end{array}\right|$
Using the transformation, $C_1\to C_1+C_2+C_3$
$\Delta =\left|\begin{array}{ccc}1+a^2+b^2+c^2& b^2& c^2\\ 1+a^2+b^2+c^2& b^2+1& c^2\\ 1+a^2+b^2+c^2& b^2& c^2+1\end{array}\right|$

$=(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& b^2& c^2\\ 1& b^2+1& c^2\\ 1& b^2& c^2+1\end{array}\right|$
Applying $R_2\to R_2-R_1\text{and}{R}_3\to R_3-R_1$
$\Delta =(1+a^2+b^2+c^2)\left|\begin{array}{ccc}1& b^2& c^2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

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