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Question 687067: I have no clue where to start with this one... I'm assuming it can't be done.
Use the Gauss-Jordan method to solve the system of equations. If the system has infinitely many solutions, let the last variable be the arbitrary variable.
x - y + 2z + w = 4
y + z = 1
z - w = 2
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! x - y + 2z + w = 4
y + z = 1
z - w = 2
is the same as
1x - 1y + 2z + 1w = 4
0x + 1y + 1z + 0w = 1
0x + 0y + 1z - 1w = 2
which converts to the matrix
Now convert to rref form
| 1 | 0 | 3 | 1 | 5 | R1 + (1)*R2 | | 0 | 1 | 1 | 0 | 1 | | 0 | 0 | 1 | -1 | 2 |
| 1 | 0 | 0 | 4 | -1 | R1 + (-3)*R3 | | 0 | 1 | 1 | 0 | 1 | | 0 | 0 | 1 | -1 | 2 |
| 1 | 0 | 0 | 4 | -1 | | 0 | 1 | 0 | 1 | -1 | R2 + (-1)*R3 | | 0 | 0 | 1 | -1 | 2 |
The matrix is now in rref form.
The last line is 0 0 1 -1 2
So 0x + 0y + 1z - 1w = 2
Solve for z to get z = 2 + w
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The second line is 0 1 0 1 -1
So 0x + 1y + 0z + 1w = -1
Solve for y to get y = -1 - w
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The first line is 1 0 0 4 -1
So 1x + 0y + 0z + 4w = -1
which means x = -1 - 4w after we solve for x
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Answer:
In the end, we get
x = -1 - 4w
y = -1 - w
z = 2 + w
w = free variable
So the answer as an ordered tuple is (-1 - 4w, -1 - w, 2 + w, w)
Optionally, you can let w = t, where t is any real number to get (-1 - 4t, -1 - t, 2 + t, t). Either way, you get the same answer pretty much.
Note: this tells us that there are an infinite number of solutions and the system is consistent and dependent. The fact that there are more variables than equations means that you'll have a consistent and dependent system.
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