Question

show that determinant is equal to(b-c)(c-a)(a-b)(a+b+c),where R1=[1 1 1],R2=[a b c],R3=[bc ca ab]

Answer 1

Now this you should really be able to do, as you only have to use Sarrus' rule. Look it up if you don't remember.

Answer 2

The factorisation might be difficult though.

Determinant = det [b c] - det [a c] + det [a b]
[ca ab] [bc ab] [bc ca]
= ab^2 - ac^2 - a^2b + bc^2 + a^2c - b^2c

Considering (b-c)(c-a)(a-b) = (bc-ab-c^2+ac)(a-b)
= abc - b^2c - a^2b + ab^2 - ac^2 + bc^2 + a^2c - abc
= ab^2 - ac^2 - a^2b + bc^2 + a^2c - b^2c

So answer is (b-c)(c-a)(a-b).

Answer 3

plz solve it in easy way!
i can not understand it.

Answer 4

So I make an effort to solve your problem, and then you tell me to solve it again in the 'easy' way?

Answer 5

ok.

Answer 6

plz solve my problem in sooo easy way because i solve that type questions at first time!