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

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SUMMARY

The discussion centers on the process of reducing a matrix A to its reduced row echelon form (RREF) and finding its four fundamental subspaces. The user initially misinterprets the RREF condition, believing that a row of zeros must always appear at the bottom. However, the correct RREF for the given matrix is confirmed to be | 1 2 0 0 |, | 0 0 1 0 |, | 0 0 0 1 |. The conversation emphasizes that a row of zeros at the bottom indicates infinitely many solutions, while a row of zeros with a non-zero entry signifies no solution. Furthermore, the user is guided on how to compute the four fundamental subspaces: column space C(A), null space N(A), column space of the transpose C(A^T), and null space of the transpose N(A^T).

PREREQUISITES
  • Understanding of matrix operations, specifically row reduction techniques.
  • Familiarity with the concept of reduced row echelon form (RREF).
  • Knowledge of the four fundamental subspaces in linear algebra.
  • Ability to interpret the implications of row echelon forms in relation to solutions of linear systems.
NEXT STEPS
  • Study the process of finding the reduced row echelon form (RREF) using Gaussian elimination.
  • Learn how to compute the four fundamental subspaces: C(A), N(A), C(A^T), N(A^T).
  • Explore the concepts of rank and nullity in relation to linear transformations.
  • Review examples of inconsistent systems of linear equations and their implications.
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to clarify concepts related to matrix operations and linear systems.

kira137
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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
 
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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?
 
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 |
 
cronxeh said:
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.
 
vela said:
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?
 
cronxeh said:
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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