Question

$\left[\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2{a}^2& ca& c+a\\ a^2{b}^2& ab& a+b\end{array}\right]$ is equal to

• A. $\frac{1}{abc}(ab+bc+ca)$
• B. $ab+bc+ca$
• C. $0$
• D. $a+b+c$

Answer 1

Answer: (C)

$0$

$\left[\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2{a}^2& ca& c+a\\ a^2{b}^2& ab& a+b\end{array}\right]$
$R_1\to a^2{R}_1,R_2\to b^2{R}_2,R_3\to c^2{R}_3$
$\frac{1}{a^2{b}^2{c}^2}\left[\begin{array}{ccc}{a}^2{b}^2{c}^2& a^{2}bc& a^2(b+c)\\ b^2{c}^2{a}^2& b^{2}ca& b^2(c+a)\\ a^2{b}^2{c}^2& c^{2}ab& c^2(a+b)\end{array}\right]$
$abc\left[\begin{array}{ccc}1& a& a^2(b+c)\\ 1& b& b^2(c+a)\\ 1& c& c^2(a+b)\end{array}\right]$
$R_1\to R_1-R_2{R}_2\to R_2-R_3$
$abc\left[\begin{array}{ccc}0& a-b& a^{(}b+c)-b^2(c+a)\\ 0& b-c& b^2(c+a)-c^2(a+b)\\ 1& c& c^2(a+b)\end{array}\right]$
$abc\left[\begin{array}{ccc}0& a-b& (a-b)(ab+bc+ca)\\ 0& b-c& (b-c)(ab+bc+ca)\\ 1& c& c^2(a+b)\end{array}\right]$
$\left(abc\right)(a-b)(b-c)\left[\begin{array}{ccc}0& 1& (ab+bc+ca)\\ 0& 1& (ab+bc+ca)\\ 1& c& c^2(a+b)\end{array}\right]$
On expanding the determinant, we have
$\left(abc\right)(a-b)(b-c)(ab+bc+ca-ab-bc-ca)=0$