Question

$\text{Express the det}A=\left|\begin{array}{ccc}a& bc& abc\\ b& ca& abc\\ c& ab& abc\end{array}\right|\text{as product of factors}$

• A. $abc(a+b)(b+c)(c+a)$
• B. $abc(a-b)(b-c)(c-a)$
• C. $ab(a-b)(b-c)$
• D. $abc(a+b)(b-c)(c-a)$

Answer 1

The correct option is B

$abc(a-b)(b-c)(c-a)$

$\text{By factor theorem, if we but b = a}\left|\begin{array}{c}A\end{array}\right|=0\text{Therefore (a-b) is or factor of A similarly c = a, b = a also gives}\left|\begin{array}{c}A\end{array}\right|=0.\text{Implies (a - b) (b - c) (c - a) are factors, But this will give highest power of 3}\text{But the det(A) has a highest power of 6}\text{So if you see carefully if we put abc = 0 the}\left|\begin{array}{c}A\end{array}\right|=0\to \text{abc is also a factor. Therefore abc, (a - b), (b -c ),(c - a) are the factors of}\left|\begin{array}{c}A\end{array}\right|$