Question

If $a,b,c$ are real then the value of determinant $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|=1$ if

• A. $a+b+c=0$
• B. $a+b+c=1$
• C. $a+b+c=-1$
• D. $a=b=c=0$

Answer 1

Answer: (D)

Multiply $R_1$ by a, $R_2$ by b & $R_3$ by $c\&$ divide the determinant by abc. Now take $a,b$ & c common from $c_1,c_2\&c_3$. Now use $C_1\to C_1+C_2+C_3$ to get ]
$\left(a^2+b^2+c^2+1\right)\left|\begin{array}{ccc}1& 1& 1\\ b^2& b^2+1& b^2\\ c^2& c^2& c^2+1\end{array}\right|=1$.
Now use $c_1\to c_1-c_2\&c_2\to c_2-c_3$
we get $1+a^2+b^2+c^2=1\Rightarrow a=b=c=0\Rightarrow$ (D)