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(linear algebra) is this rref? or did I solve it wrong?

  • Thread starter kira137
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  • #1
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Homework Statement


I was given
| 1 2 -1 2 |
| 2 4 1 -2 | = A
| 3 6 0 -6 |

and reduce it to rref.

Also I had to find all four fundamental subspaces of A.

2. The attempt at a solution

so I did

R2 - 2R1
R3 - 3R1
R3 - R2

and got

| 1 2 -1 2 |
| 0 0 3 -6 |
| 0 0 0 -6 |

... Arent rref suppose to have all zero row at the bottom? I don't think this is rref because of that -6 in row 3...

and I'm not very sure about how to go about four fundamental subspaces. Am I suppose to solve for C(A), N(A), C(A^T), N(A^T)? and am I suppose to solve for dim for each of them or special solutions?

thank you for replies in adavance
 

Answers and Replies

  • #2
vela
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No, you don't always end up with rows of zeros at the bottom. By the way, you're not done reducing the matrix. The first non-zero entry in a row should be equal to 1.

Your textbook should tell you what the four fundamental subspaces are and how to find them. Have you looked in there for the answer to your question?
 
  • #3
cronxeh
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Since last row has -6 you can divide it by -6, i.e R3 = -1/6 * R3

Row 2 is clearly divisible by 3, so R2 = 1/3 * R2, you get [0 0 1 -2]. By adding 2*R3 + R2 = R2 you eliminate the -2. Long story short you can check your results by plugging it back into the original equation. Your system of linear equations is inconsistent and has NO solution, because you have a row of zeroes and 1 at the end, i.e

0*x1 + 0*x2 + 0*x3 = 1

If you had a row of all zeroes at the end, your system has infinitely many solutions, i.e
0*x1 + 0*x2 + 0*x3 = 0

**Make sure you take a minute to re-read what I just wrote, tattoo it on your hand somewhere and don't ever forget it**

This is the result of reduced row echelon formation performed on your matrix:
| 1 2 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
 
  • #4
vela
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Your system of linear equations is inconsistent and has NO solution, because you have a row of zeroes and 1 at the end
This is only true if A is the augmented matrix for a system of linear equations. It's not true if A is a mapping from R4 to R3.
 
  • #5
cronxeh
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This is only true if A is the augmented matrix for a system of linear equations. It's not true if A is a mapping from R4 to R3.
I'm sorry. I wasn't aware there exists a reason why one would want to rref a transformation matrix?
 
  • #6
vela
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I'm sorry. I wasn't aware there exists a reason why one would want to rref a transformation matrix?
One reason would be to find the four fundamental subspaces. :) You can also reduce it to determine the rank and nullity of the mapping.
 

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