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Matrix Row Reduction Question

  • Thread starter Yosty22
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  • #1
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Homework Statement



Write and row reduce the augmented matrix to find out whether the given set of equations has exactly one solution, no solutions, or infinitely many solutions.

-x+y-z=4
x-y+2z=3
2x-2y+4z=6

Homework Equations





The Attempt at a Solution



I saw right away that Row 3 and Row 2 are the same equation, off by a factor of 2. Because of this, I was able to make the matrix with a zero row for Row 2, which shows that row 2 and row 3 are linearly dependent. However, my question arises here. I know that a zero row and linear dependence of these two equations means that there is either 0 solutions or infinitely many solutions. Since they are linearly dependent, there is not one unique solution. However, How can I tell whether there is 0 solutions or infinitely many?
 

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  • #2
Ray Vickson
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Homework Statement



Write and row reduce the augmented matrix to find out whether the given set of equations has exactly one solution, no solutions, or infinitely many solutions.

-x+y-z=4
x-y+2z=3
2x-2y+4z=6

Homework Equations





The Attempt at a Solution



I saw right away that Row 3 and Row 2 are the same equation, off by a factor of 2. Because of this, I was able to make the matrix with a zero row for Row 2, which shows that row 2 and row 3 are linearly dependent. However, my question arises here. I know that a zero row and linear dependence of these two equations means that there is either 0 solutions or infinitely many solutions. Since they are linearly dependent, there is not one unique solution. However, How can I tell whether there is 0 solutions or infinitely many?
Have you carried through the row-reduction procedure, as the question asked you to do? Doing that, or some equivalent thing, is necessary if you want a final answer.
 
  • #3
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Yes, I continued a bit. Here is what I have done and what I think so far: I have simplified the matrix down to:

Row 1: -1 1 -1 4
Row 2: 0 0 0 0
Row 3: 0 0 2 14

Once I do this, it tells me in the bottom equation that z=7. However, I'm not too sure how to tell anything about how many solutions there are, besides the fact that there is NOT just one.
 
  • #4
Ray Vickson
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Yes, I continued a bit. Here is what I have done and what I think so far: I have simplified the matrix down to:

Row 1: -1 1 -1 4
Row 2: 0 0 0 0
Row 3: 0 0 2 14

Once I do this, it tells me in the bottom equation that z=7. However, I'm not too sure how to tell anything about how many solutions there are, besides the fact that there is NOT just one.
OK, so now what is y? What is x?
 
  • #5
185
4
So for row 3, x = 0, y = 0, z = 7, but for row 1 you have -x+y-z=4, so i'm a little stuck as what to do next.
 
  • #6
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No. For R3
0x+0y+2z=14
 
  • #7
HallsofIvy
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So for row 3, x = 0, y = 0, z = 7, but for row 1 you have -x+y-z=4, so i'm a little stuck as what to do next.
Row 3 reduced to 2z= 14 which tells you that z= 7 but says nothing about what x and y are! Knowing that z= 7, -x+ y- z= 4 becomes -x+ y= 7+4= 11 or y= x+ 11.
 

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