Understanding the Rank of a Matrix: Explained Simply

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    Matrix rank
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Homework Help Overview

The discussion revolves around understanding the concept of the rank of a matrix within the context of linear algebra. Participants are exploring the definition of rank, particularly in relation to leading ones and non-zero rows in a given matrix.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to clarify their understanding of rank by asking about leading ones in a specific matrix. They question whether the last row contributes to the rank. Other participants discuss the implications of having identical rows on the rank.

Discussion Status

The discussion includes some guidance on the definition of rank and its calculation based on non-zero rows. Participants are engaging with each other's interpretations and clarifying concepts without reaching a definitive conclusion.

Contextual Notes

There is a focus on the definition of leading ones and linear independence, with some participants questioning their understanding of these concepts in relation to the rank of a matrix.

Taryn
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Hey I am just wondering about this question... I have reduced it as much as I can and the second part of the question is asking about the rank of the matrix... which means the leading number of ones right?

SO if I had this matrix 2 5 0
0 2 1
0 0 0

Wat would be the leading ones and would the last row be classed as leading ones? If you can just giv a brief explanation that would be great! I understand wat I meant to be finding just am a lil unsure of the concept!

Would the rank be 2?
 
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the matrix was meant to be

2 5 0
0 2 1
0 0 0

sorry if it doesn't look right in the above post!
 
If [tex]A = \begin{bmatrix}<br /> 2 & 5 & 0 \\<br /> 0 & 2 & 1 \\<br /> 0 & 0 & 0 \\<br /> \end{bmatrix}[/tex]

then the rank is 2. The rank is just the number of non-zero rows.
 
ahhh that's easy! Thanks... also just wonderin if you had 2 rows that were identical then the rank would just be 1 right?
 
yes, because they are not linearly independent
 

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