Question

prove that determinant is equal to (a-1)^3(a+3),where R1=[a 1 1 1],R2=[1 a 1 1],R3=[1 1 a 1],R4=[1 1 1 a].

Answer 1

plz solve my question!

Answer 2

Laplace's formula :D

Answer 3

can you solve it by column or row operation?

Answer 4

I'm doing it as follows:

first I found determinant of

(b 1 1)
(1 a 1)
(1 1 a)

and got (a-1)(ab+b-2).

Then your determinant is equal to

a det (a 1 1) - 1 det (1 1 1) + 1 det (1 a 1) - 1 det( 1 a 1)
(1 a 1) (1 a 1) (1 1 1) (1 1 a)
(1 1 a) (1 1 a) (1 1 a) (1 1 1)

First determinant is (a-1)(a^2+a-2)
Second determinant is (a-1)(a+1-2) = (a-1)^2
Third determinant is -(a-1)^2 (swapping two rows changes sign of determinant)
Fourth determinant is (a-1)^2 (same reason)

Hence answer is a(a-1)(a^2+a-2) - (a-1)^2 - (a-1)^2 - (a-1)^2
= a(a-1)(a^2+a-2) - 3(a-1)^2
= (a-1)[a(a^2+a-2) - 3(a-1)]
= (a-1)[a^3 + a^2 - 2a - 3a + 3]
= (a-1)[a^3 + a^2 - 5a + 3]
= (a-1)(a+3)(a^2-2a+1) using long division
= (a-1)(a+3)(a-1)^2
= (a-1)^3(a+3) as required.