Rank of Matrix Problem: Finding k for Rank=2 | Explanation & Solution

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Homework Help Overview

The discussion revolves around determining the value of k for which a given 3x3 matrix has a rank of 2. The matrix is presented with specific entries, and participants are exploring methods to find k while adhering to the constraints of linear algebra.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using row reduction techniques to analyze the matrix and determine conditions for rank. Some suggest finding k such that the third row is a linear combination of the first two rows. Others express uncertainty about how to proceed after initial row operations.

Discussion Status

There are various approaches being explored, including row reduction and linear combinations. Some participants have offered specific row operations, while others are questioning the effectiveness of their methods. A few have indicated that they found certain suggestions helpful, but there is no clear consensus on the best approach yet.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a mention of a potential misunderstanding regarding the need to bump an older thread, indicating a focus on maintaining the relevance of the discussion.

snoggerT
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find the value for k for which the matrix

A=
| 9 -1 11 |
|-6 5 -16 |
| 3 2 k |

has rank= 2


* the spacing on the matrix doesn't seem to want to stay formatted, but it's a 3X3 with row 1= (9, -1, 11), row 2= (-6, 5, -16) and row 3=(3,2, k)

The Attempt at a Solution



- I tried to solve this using row reduction and then solve for k, but I don't think that's right at all. Can someone please explain this problem to me and the technique for solving it?
 
Last edited:
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[tex]\left(<br /> \begin{array}{Ccc}<br /> 9 & -1 & 11\\<br /> -6 & 5 & -16 \\<br /> 3 & 2 & k\<br /> \end{array}<br /> \right)[/tex]

What you need to is row reduction to Row echelon form, the number of non-zero rows is the rank,2 in this case. So there should be 2 non-zero rows.
 
rock.freak667 said:
[tex]\left(<br /> \begin{array}{Ccc}<br /> 9 & -1 & 11\\<br /> -6 & 5 & -16 \\<br /> 3 & 2 & k\<br /> \end{array}<br /> \right)[/tex]

What you need to is row reduction to Row echelon form, the number of non-zero rows is the rank,2 in this case. So there should be 2 non-zero rows.

- That is what I tried to do, but then I don't know how to find k. I got this:

row 1= (3, 2, k)
row 2= (0, 9, -16+2k)
row 3= (0, -7, 11-3k)

I don't know how to solve for k at this point, would it just be guess work, or is there a technique to use?
 
Try these operations
[tex]3R_3-R_1,9R_2+6R1[/tex]that should give you something better.

Edit: Do you know the easiest method to get to RE form when given a matrix?
 
Last edited:
There are a few ways of approaching this, but this is the first that came to my mind.

For it to be rank 2, you should try finding k such that the third row is a linear combination of the first two. So try finding c1 and c2 such that:

c1 R1 + c2 R2 = R3.

The first two components will give you a 2x2 system, and you can then solve for c1 and c2. Then, the third component will give you k.

Hopefully that makes sense and is helpful!
 
yes, that made it much easier. thanks.
 
just do determine of matrix zero and find value of k
 
sanjeevece said:
just do determine of matrix zero and find value of k

You did not really have to bump a 2 year old thread.
 
Heyya Snogger
 
Last edited:
  • #10
A = 9 -1 11
-6 5 -16
3 2 k

= taking 3 common from C1

3 -1 11
-2 5 -16
1 2 k

= R2->R2+2(R3)

3 -1 11
0 9 k-16
1 2 k

=R3->3(R3)-R1

3 -1 11
0 9 k-16
0 7 3k-11

=R2->R2/9

3 -1 11
0 1 k-16/9
0 0 3k-11

=R3->R3+7R1

3 -1 11
0 1 k-16
0 0 3k-66

Since,the rank of the matrix is 2

3k-66=0
3k=66
k=22

Sorry if I m wrong just gave a try...
 

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