[Rainingshame]

ok so the matrix is:
[1 1 1 ]
[a b c ]
[bc ca ab]
It shows you that the determinant is (a-b)(b-c)(c-a) i managed to do this.
then is say:
Hence or otherwise show that the system of equations
x+y+z = p
3x + 3y +5z = q
15x+15y +9z = r

has no solution whatever the values of p q r,.
I realised that the value can be put into the matrix above
so a= 3 and b=3 c = 5
so a=b and a-b=0. this means that the determinant =0.
I've checked with the mark scheme and that was what they were looking for.
My question is: why does that prove that the equation has no solution whatever the values of p q and r?

[BabyMaths]

p=3, q=11, r=39

x=y=z=1.

[ElMoro]

(Original post by Rainingshame)
...

If a matrix has a determinant of 0 that means it's singular and has no inverse so you wouldn't be able to solve those equations no matter what was on the RHS. Hope this makes sense

[Rainingshame]

(Original post by ElMoro)
If a matrix has a determinant of 0 that means it's singular and has no inverse so you wouldn't be able to solve those equations no matter what was on the RHS. Hope this makes sense

Yes it does thank you. Because of C3 and S1 it's been like 2 months since of done this, i'm incredibly rusty.

[notnek]

I think the question should say, "no unique solution".

[Rainingshame]

(Original post by notnek)
I think the question should say, "no unique solution".

sorry it does :P. i was more concerned with getting the numbers correct.

[notnek]

(Original post by Rainingshame)
sorry it does :P. i was more concerned with getting the numbers correct.

Good, otherwise it would've been a mistake since BabyMaths showed you an example solution.

A singular matrix implies either no solutions or infinitely many solutions.