Question 687067
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


<table cellpadding = 15><tr><td>1</td><td>-1</td><td>2</td><td>1</td><td>4</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>2</td></tr></table>


Now convert to rref form


<table cellpadding = 15><tr><td>1</td><td>-1</td><td>2</td><td>1</td><td>4</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>2</td></tr></table>



<table cellpadding = 15><tr><td>1</td><td>0</td><td>3</td><td>1</td><td>5</td><td>R1 + (1)*R2</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>2</td></tr></table>



<table cellpadding = 15><tr><td>1</td><td>0</td><td>0</td><td>4</td><td>-1</td><td>R1 + (-3)*R3</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>2</td></tr></table>



<table cellpadding = 15><tr><td>1</td><td>0</td><td>0</td><td>4</td><td>-1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>1</td><td>-1</td><td>R2 + (-1)*R3</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>2</td></tr></table>


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.