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

## 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

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vela
Staff Emeritus
Homework Helper
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?

cronxeh
Gold Member
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 |

vela
Staff Emeritus
Homework Helper
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.

cronxeh
Gold Member
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?

vela
Staff Emeritus