Matrices Help

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Matrices Help

Postby Edward » Thu Sep 01, 2011 4:50 pm

I am having quite a bit of trouble solving my Discrete Math homework. I have two systems of equations to solve using augmented matrices. One has been frustrating me ever since I was handed the assignment, and the other I have no clue how to even begin solving.

Here's the first problem:

2x + y - 2z = 7
x - 2y - 5z = -1
4x + y + z = -1

And now the second:

x + 2y = 5
3x + 4z = 2
2y + 3w = -2
3z - 2w = 1

I'm going to need a response very soon, because the problems are due tommorow, and we're going to go over them in class, and by then I won't need any help from you guys. Thanks.
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Postby ABlipinTime » Thu Sep 01, 2011 5:54 pm

I'm assuming you know about matrix multiplication.

Set up your matrices first:
In the first one, your matrices are (note: you'll have to write these out to see them. ";" designates next row in the matrix)
[2,1,-2 ; 1,-2,-5 ; 4,1,7] * [x ; y ; z] = [7 ; -1 ; -1]

This can be simplified to a notation: A*B = C, where A, B, and C just represent the matrices I just showed you.
Since we want to find x, y, and z, we multiply both sides of the equation by A-1 (inverse matrix of A). This gives us: B = (A-1)*C

Wolfram alpha can help you at this point:
http://mathworld.wolfram.com/MatrixInverse.html
The boxes within that big box are the determinants, which you will find how to solve here:
http://en.wikipedia.org/wiki/Determinant#3-by-3_matrices

At that point, apply matrix multiplication to the 3x3 matrix A-1 and the 1x3 matrix C. Then incorporate the constant (which distributes to all the elements) and you'll have in each column of the result the value corresponding to the variable in the matrix on the other side of the equation.

I'm not sure how to solve the second one there, but I can tell you that each equation has four variables, two of which are always set to zero - so there slots in the matrix should be zero).
Written out (notice I've made "w" last):
[1,2,0,0 ; 3,0,4,0 ; 0,2,0,3 ; 0,0,3,-2] * [x ; y ; z] = [5;2;-2;1]

Hope that helps.
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Postby blkmage » Thu Sep 01, 2011 6:04 pm

Set up your systems of equations as augmented matrices:

Code: Select all
2  1 -2  7
1 -2 -5 -1
4  1  1 -1


and

Code: Select all
1 2 0  0  5
3 0 4  0  2
0 2 0  3 -2
0 0 3 -2  1


and do row reduction on them to solve.
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Postby Edward » Thu Sep 01, 2011 6:13 pm

ABlipinTime (post: 1501421) wrote:Wolfram alpha can help you at this point:
http://mathworld.wolfram.com/MatrixInverse.html
The boxes within that big box are the determinants, which you will find how to solve here:
http://en.wikipedia.org/wiki/Determinant#3-by-3_matrices

At that point, apply matrix multiplication to the 3x3 matrix A-1 and the 1x3 matrix C. Then incorporate the constant (which distributes to all the elements) and you'll have in each column of the result the value corresponding to the variable in the matrix on the other side of the equation.

Hope that helps.


Yikes. That stuff with the determinants is scary. I was trying to solve it with the Gaussian method, and couldn't figure anything out, so I don't think that's going to help. I don't have the time to do it the regular way. It doesn't really matter anyway, as the questions most likely won't be graded. But I think I could use the link you gave me in the future, so I guess this thing wasn't completly useless. :p

edit: And thanks for telling me how to solve the second problem. [s]That one looks like I could do it.[/s] Nope! I can't.
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Postby ABlipinTime » Sat Sep 03, 2011 8:16 pm

Sorry I don't know much about solving matrices.
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