Linear Algebra : Rank of a matrix

[lax1113]

Homework Statement

Given the following conditions, determine if there are no solutions, a unique solution, or infinite solutions. (Matrix A|B = augmented matrix).

Just in case anyone viewing needs a little refresher... Rank = number of non zero rows in the matrix.

1)

# of equations : 3
# of unknowns : 4
Rank of Matrix A : -
Rank of Matrix A|B: 2

2)

# of equations : 4
# of unknowns : 4
Rank of Matrix A : 4
Rank of Matrix A|B: -


2. Homework Equations and attempt

So I know that for a system to be consistent, the rank of A has to be equal to the rank of A|B. At this point, if this is true, we can then go on to say that if the rank of A is less than the # of unknowns, then A has infinite solutions (Unknowns - Rank(A) = free parameters). The only thing that I am not sure about is how to determine the Rank of A given the other 3, or the rank of A|B. Is this question asking for multiple answers?

For example, for number one, would I say that in the case that Rank(A) is less than Rank A|B, the system is inconsistent, while if the rank of A is equal to that of A|B then the system is consistent with infinite solutions because A<unknowns? I feel like with the number of given equations I should be able to figure out the missing part, since with the way that I just looked at it I am basically completely ignoring the information given in the equations.

I am really stuck on how to find out the other rank (if it is even possible!) and I have an exam on this tomorrow, so any hint would be greatly appreciated. I couldn't find problems like this online or in my book at .

Thanks

 

[lanedance]

Just in case anyone viewing needs a little refresher... Rank = number of non zero rows in the matrix.

not quite, from wiki
"The rank of a matrix A is the maximal number of linearly independent rows or columns of A"
http://en.wikipedia.org/wiki/Rank_(linear_algebra )

 

[lax1113]

LaneDance,
Thank you for your reply. Would it be accurate to say than that in the first problem the rank of matrix A could be interpreted as being 3? Since there will be 3 rows so the maximum would be 3? And I don't know how to interpret when the rank of A is greater than the augmented matrix, I was under the impression that it was either equal to or less than A|B.

Also, for the rank of A|B as shown in problem 2, 4eqs,4unknowns, A has a rank of 4, I feel like A|B also would have to be 4.To add on to my first statement, I noticed that it said it was the lesser of mxn, so we would look at the rows to define the max rank.
And max rank = Rank A?

 

[lanedance]

LaneDance,
Thank you for your reply. Would it be accurate to say than that in the first problem the rank of matrix A could be interpreted as being 3? Since there will be 3 rows so the maximum would be 3? And I don't know how to interpret when the rank of A is greater than the augmented matrix, I was under the impression that it was either equal to or less than A|B.

The max it could be is 3, but it should always be Rank (A) <= Rank(A|B)

Also, for the rank of A|B as shown in problem 2, 4eqs,4unknowns, A has a rank of 4, I feel like A|B also would have to be 4.

I would agree, it has to be 4, as A has 4 rows it can't be any larger

To add on to my first statement, I noticed that it said it was the lesser of mxn, so we would look at the rows to define the max rank.
And max rank = Rank A?

After a quick think, here's some points I hope help
# equations = rows of A = m
# unknowns = columns of A = n

So you have Rank(A) <= min(m,n).
and Rank(A|B) <= min(m,n+1)

Note when you augment A to A|B you effectively add a column vector, so Rank(A|B) >= Rank(A),

 

[lax1113]

Lane,
Thank you so much for your reply. Sure enough it was on my exam today (which is killed!)
Was very uncertain about small parts of the rank, but you cleared it up.

Thanks!

Ben