Question

If a, b, c and x are positive integers, then $\left|\begin{array}{ccc}{a}^2+x^2& ab& ac\\ ab& b^2+x^2& bc\\ ac& bc& c^2+x^2\end{array}\right|$ is divisible by

• A. $x^3+1$
• B. $x^2$
• C. $x^4$
• D. $a^2+b^2+c^2+x^2$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x^2$
C $x^4$
D $a^2+b^2+c^2+x^2$

$\left|\begin{array}{ccc}{a}^2+x^2& ab& ac\\ ab& b^2+x^2& bc\\ ac& bc& c^2+x^2\end{array}\right|$
Taking a,b,c common from $C_1,C_2,C_3$
$=abc\left|\begin{array}{ccc}\frac{a^2+x^2}{a}& a& a\\ b& \frac{b^2+x^2}{b}& b\\ c& c& \frac{c^2+x^2}{c}\end{array}\right|$
Multiplying $R_1,R_2,R_3$ by a,b,c we get
$=\left|\begin{array}{ccc}{a}^2+x^2& a^2& a^2\\ b^2& b^2+x^2& b^2\\ c^2& c^2& c^2+x^2\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}{x}^2& 0& a^2\\ -x^2& x^2& b^2\\ 0& -x^2& c^2+x^2\end{array}\right|$
$=x^4\left|\begin{array}{ccc}1& 0& a^2\\ -1& 1& b^2\\ 0& -1& c^2+x^2\end{array}\right|$
$R_2\to R_2+R_1$
$=x^4\left|\begin{array}{ccc}1& 0& a^2\\ 0& 1& a^2+b^2\\ 0& -1& c^2+x^2\end{array}\right|$
$=x^4[x^2+b^2+c^2+a^2]$
Hence, given determinant is divisible by $x^2,x^4,(x^2+b^2+c^2+a^2)$