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

[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)

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?

 

[rock.freak667]

\left(<br /> \begin{array}{Ccc}<br /> 9 &amp; -1 &amp; 11\\<br /> -6 &amp; 5 &amp; -16 \\<br /> 3 &amp; 2 &amp; k\<br /> \end{array}<br /> \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.

 

[snoggerT]

\left(<br /> \begin{array}{Ccc}<br /> 9 &amp; -1 &amp; 11\\<br /> -6 &amp; 5 &amp; -16 \\<br /> 3 &amp; 2 &amp; k\<br /> \end{array}<br /> \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.

- 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?

 

[rock.freak667]

Try these operations
3R_3-R_1,9R_2+6R1that should give you something better.

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

 

[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!

 

[snoggerT]

yes, that made it much easier. thanks.

 

[sanjeevece]

just do determine of matrix zero and find value of k

 

[rock.freak667]

just do determine of matrix zero and find value of k

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

 

[Karthikmk8791]

Heyya Snogger

 

[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...