# rank and nullity

• Jan 26th 2011, 07:32 PM
Taurus3
rank and nullity
Compute the rank and nullity of the matrix A and state what this implies about the existence and/or uniqueness of solutions x of Ax=b.

A = [1 1 3]
1 2 3
3 2 1

Sol: Reduced row echelon form = [1 0 -1]
0 1 2
0 0 0

Therefore, Rank(A)=2, and Nullity (A) = 1. But I don't get the second part though. Does it mean there's only one solution?
• Jan 26th 2011, 07:48 PM
harish21
Firstly, your rank is not correct..

The row reduced echleon form of the given matrix should be a $I_3$

try doing it again..
• Jan 26th 2011, 07:51 PM
Taurus3
my bad. The first row should be 1 1 1. Not 1 1 3.
• Jan 26th 2011, 07:59 PM
dwsmith
Quote:

Originally Posted by Taurus3
Compute the rank and nullity of the matrix A and state what this implies about the existence and/or uniqueness of solutions x of Ax=b.

A = [1 1 3]
1 2 3
3 2 1

Sol: Reduced row echelon form = [1 0 -1]
0 1 2
0 0 0

Therefore, Rank(A)=2, and Nullity (A) = 1. But I don't get the second part though. Does it mean there's only one solution?

The zero vector is also a solution and a scalar times your nullspace vector are solutions.
• Jan 26th 2011, 08:05 PM
Taurus3
Can I also give the same reason if
A= [0 1 1 0, 2 -1 -1 2, 1 2 2 4] where the reduced row echelon form is [1 0 0 0, 0 1 1 0, 0 0 0 1] as the rank would be 4 and nullity would be 0.
• Jan 26th 2011, 08:08 PM
dwsmith
Quote:

Originally Posted by Taurus3
Can I also give the same reason if
A= [0 1 1 0] where the reduced row echelon form is [1 0 0 0] as the rank would be 4 and nullity would be 0.
2 -1 -1 2 0 1 1 0
1 2 2 4 0 0 0 1

You need to learn latex. You have a row vector and then some random numbers mashed together.
• Jan 26th 2011, 08:15 PM
Taurus3
I've edited my previous post. Is it better now?