Proving Rank(A)=2 for a 3x3 Matrix with Non-Zero Elements

[dietcookie]

Homework Statement

Assume that a, b, and c do not equal zero.

Let matrix A=

0 a b
-a 0 c
-b c 0

Prove that Rank(A)=2

Homework Equations

Definition: The rank of a matrix A is the number of linearly independent rows or columns of A

The Attempt at a Solution

I've attempted to get it into reduced row echelon form with the following row ops (as row ops preserve the rank)

r2<-->r1
(-b)r1-->r1
(c)r3+r1-->r3
(1/2)r3-r2-->r3
(1/c)r3-r2-->r3
(1/2)r3-->r3

I end up with this matrix

ab 0 -bc
0 a b
0 0 b

If I could get r3 all zeros, then I would have two non-zero rows in rref form, which would mean the rank(a)=2, but I'm stuck here. Am I approaching this right? Thanks.

I keep staring at this and now I'm thinking perhaps I wrote down the problem wrong. As we know there are the equivalent conditions for square matrices, one of them being, det(a) does not equal zero and Rank(A)=n. Well the determinant of the example does not equal zero, det(A)=-2abc so doesn't that mean that the rank of this matrix should equal three?? (as it is 3x3?)

 

[tiny-tim]

hi dietcookie!

I keep staring at this and now I'm thinking perhaps I wrote down the problem wrong. … Well the determinant of the example does not equal zero, det(A)=-2abc so doesn't that mean that the rank of this matrix should equal three?? (as it is 3x3?)

yup, i think that bottom c should be -c

 

[tiny-tim]

in fact, if you multiply the matrix by the vector (c,b,-a), you can see that that's an eigenvector with eigenvalue 0 …

this matrix is the standard matrix for eg a magnetic field … it simply gives the cross-product of any vector with (c,b,-a)