# Rank of matrix problem

1. Apr 1, 2008

### snoggerT

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)
3. The attempt at a solution

- I tried to solve this using row reduction and then solve for k, but I don't think thats right at all. Can someone please explain this problem to me and the technique for solving it?

Last edited: Apr 1, 2008
2. Apr 1, 2008

### rock.freak667

$$\left( \begin{array}{Ccc} 9 & -1 & 11\\ -6 & 5 & -16 \\ 3 & 2 & k\ \end{array} \right)$$

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.

3. Apr 1, 2008

### snoggerT

- 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?

4. Apr 1, 2008

### rock.freak667

Try these operations
$$3R_3-R_1,9R_2+6R1$$

that should give you something better.

Edit: Do you know the easiest method to get to RE form when given a matrix?

Last edited: Apr 1, 2008
5. Apr 1, 2008

### PingPong

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!

6. Apr 1, 2008

### snoggerT

yes, that made it much easier. thanks.

7. Jul 24, 2010

### sanjeevece

just do determine of matrix zero and find value of k

8. Jul 24, 2010

### rock.freak667

You did not really have to bump a 2 year old thread.

9. Jan 30, 2012

### Karthikmk8791

Heyya Snogger

Last edited: Jan 30, 2012
10. Jan 30, 2012

### Karthikmk8791

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...