Question about the properties of the rank of a matrix

[skyturnred]

Homework Statement

Matrix A is a 4 row by 5 column matrix. Matrix B is a column vector in R[itex]^{4}[/itex]. We are supposed to decide whether the following are (a) no solution, (b) one solution, or (c) infinitely many solutions, or whether (d) the data do not give enough information to tell, or (e) the data are impossible.

For the following six cases:

rank(A) rank([A | b])
(i) 4 5
(ii) 4 4
(iii) 4 3 .
(iv)3 4
(v) 3 5
(vi)3 3

***numbers on the right are the rank for augmented matrix, numbers on left are for matrix A alone

Homework Equations

The Attempt at a Solution

I THINK I know the answer to all of them.. but I am not entirely certain. I have always found rank just a little confusing. Can someone please tell me where I went wrong in the following and why?

(i) The data is impossible: rank exceeds number of rows.
(ii) There are 5 unkowns (because matrix A has 5 columns) and the rank is 4. So that means there is ONE free variable, so there are infinitely many solutions.
(iii) Not possible: augmenting Matrix A by a column vector B cannot possible DECREASE the rank. Can only be equal to or ONE more than rank(A).
(iv) No solutions. Since augmenting by a column vector increased the rank by one, that means that the augmented matrix is inconsistent.
(v)Not possible: rank exceeds number of rows
(vi)Infinitely many solutions: 5 unknowns, rank is 3, so 2 free variables.

Thanks!

 

[mighty2000]

Your answers for (i), (ii), (iv), and (vi) are correct. However, for (iii) and (v), your reasoning is incorrect.

(iii) It is possible for the augmented matrix to have a rank lower than the original matrix. This means that there is a linear dependence between the columns of the original matrix, and adding the column vector B does not change this. Therefore, there are infinitely many solutions.

(v) The rank of the augmented matrix can be equal to the number of rows, as long as the rank of the original matrix is less than the number of rows. In this case, there is a unique solution. However, if the rank of the original matrix and the augmented matrix are both equal to the number of rows, then there are infinitely many solutions. So the correct answer for this case is (c) infinitely many solutions.