Question

The determinant
$\Delta =\left|\begin{array}{ccc}{a}^2+x^2& ab& ac\\ ab& b^2+x^2& bc\\ ac& bc& c^2+x^2\end{array}\right|$ is divisible by

• A. $(a^2+b^2+c^2+x^2)$
• B. $x^4$
• C. $x^3$
• D. All of these

Answer 1

The correct option is D All of these

$\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+x^2& ab& ac\\ ab& b^2+x^2& bc\\ ac& bc& c^2+x^2\end{array}\right|=\frac{1}{abc}\left|\begin{array}{ccc}a\left(a^2{+x}^2\right)& a^{2}b& a^{2}c\\ b{ab}^2& \left(b^2{+x}^2\right)& b^{2}c\\ a{c}^2& b{c}^2& \left(c^2{+x}^2\right)\end{array}\right|\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+x^2& a^2& a^2\\ b^2& b^2+x^2& b^2\\ c^2& c^2& c^2+x^2\end{array}\right|=\left|\begin{array}{ccc}{a}^2+b^2+c^2+x^2& a^2+b^2+c^2+x^2& a^2+b^2+c^2+x^2\\ b^2& b^2+x^2& b^2\\ c^2& c^2& c^2+x^2\end{array}\right|R_1\to R_1{+R}_2+R_3{a}^2+b^2+c^2+x^2\left|\begin{array}{ccc}1& 1& 1\\ b^2& b^2+x^2& b^2\\ c^2& c^2& c^2+x^2\end{array}\right|c_2\to c_2-c_1,c_3\to c_3-c_1(a^2{+a}^2{+a}^2+x^2)\left|\begin{array}{ccc}1& 0& 0\\ b^2& x^2& 0\\ c^2& 0& x^2\end{array}\right|$

expanding from $R_1$

$△=\left(1\right)(x^4{-0\left)\right(a}^2+b^2+c^2+x^2)$

$△=x^4{(a}^2+b^2+c^2+x^2)$

Hence $△$ is divisible by${(a}^2+b^2+c^2+x^2),x^4,x^3$