Question

Solve a system of 4 equations:
w+x+y+z=1
2w+x-y-z=4
w-x-2y+2z=2
3w+2x+y+z=7

The answer is supposed to be 7. How do I get there?

Answer 1

Oops. not 7

Answer 2

just 7?...interesting

Answer 3

haha

Answer 4

w is supposed to equal 7

Answer 5

canceling out variables.

Answer 6

I need to find w z y & z

Answer 7

i mean: w x y and z

Answer 8

7
-8
4.25
-2.25

Answer 9

Whoa. how did you get to the answers that quickly?

Answer 10

use excel and Minverse, (array result requires you hold command when press enter), then MMult with the solution vector.

Your TI-83 can also do matrices. Don't bother with elimination, it takes too long, there's better things to do.

Answer 11

I need the steps. I'm working on a project.

Answer 12

do you know how to build a matrix from a set of equations?

Answer 13

@Tonks No I don't.

Answer 14

hello?

Answer 15

Ok, take the co-efficient of each term, if there isn't one, then it is 1:
2w+x-y-z=4
becomes
2, 1, -1, -1
in a coefficient matrix and the 4 goes in your solution vector (a matrix that is only one column wide is called a vector)

w+x+y+z=1
2w+x-y-z=4
w-x-2y+2z=2
3w+2x+y+z=7
becomes:
1, 1, 1, 1
2, 1, -1, -1
1, -1, -2, 2
3, 2, 1, 1
for your coefficient matrix
your solution vector is:
1
4
2
7

Answer 16

So far so good?

Answer 17

Exact solutions from Mathematica:\[\left\{w = 7,x = -8,y = \frac{17}{4},z = -\frac{9}{4}\right\} \]

Answer 18

That might be a little complicated for the level I am on. I was thinking more in the direction of adding them together to eliminate variables.

Answer 19

oof, the problem with using elimination is that there are so many opportunities for errors, while there are only opportunities for entry errors with the matrix method.

Answer 20

That's what my teacher wants us to do...

Answer 21

The following is a matrix solution using Mathematica. LinearSolve finds an x which solves the matrix equation m.x==b. \[\text{LinearSolve}\left[\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 1 & -1 & -1 \\ 1 & -1 & -2 & 2 \\ 3 & 2 & 1 & 1 \\ \end{array} \right),\left( \begin{array}{c} 1 \\ 4 \\ 2 \\ 7 \\ \end{array} \right)\right]\to \left\{7,-8,\frac{17}{4},-\frac{9}{4}\right\} \]

Answer 22

Well I have to go. Thanks for your help!