Thread: Matrix problem 2

Im having another matrix problem...



I cant find the solution to this one, every time I try it seems to go wrong. The answer i get is x=15 y=11 and z=30, and that cant be correct in any way

How do I find x,y, and z? And is there really a solution to this one? Thanks!

Gauss, Gauss-Jordan.

show your progress.

Originally Posted by Krizalid

Gauss, Gauss-Jordan.

show your progress.

Easier to just do it with a matrix I think, Gaussian equations are an unnecessary waste of time IMO.

(-5 2 1)( X) = (-5)
(1 -4 1)( Y) (1)
(1 1 -1)(Z) (2)

Sorry for the messy brackets there, not sure how to setup a matrix on the forums here.

Anyways those are your original values input into a matrix, now what you have to do is find the inverse of(Using a calculator):

(-5 2 1)
(1 -4 1)
(1 1 -1)

Which will give you the identity matrix:

(1 0 0)
(0 1 0)
(0 0 1)

Therefore on the left side you will be left with:


(1 0 0)(X)
(0 1 0)(Y)
(0 0 1)(Z)

Just remove the identity matrix from your final answer.

Now what you have to do is multiply the inverse for(Obviously your 3x3 matrix has to be in front of the other):


(-5 2 1)
(1 -4 1)
(1 1 -1)

By:

(-5)
(1)
(2)

Which will answer your problem.