Matrices, algebra, please answer my question?

Question

which of the following matrices have linearly independent columns:
Matrix A:
row 1: 1 1 0 0
row 2: 1 1 0 0
row 3: 0 0 1 0
row 4: 0 0 1 1

Matrix B:
row 1: 1 0 0
row 2: 1 0 0
row 3: 0 1 0
row 4: 0 0 1

Matrix D:
row 1: 1 0
row 2: 1 0
row 3: 2 5
row 4: 0 0

Matrix N:
row 1: 1
row 2: 2
row 3: 3
row 4: 4

Matrix F:
row 1: 1 5 6
row 2: 1 5 6
row 3: 0 7 2
row 4: 0 0 9

Matrix H:
row 1: 1 0 2 2 6
row 2: 1 0 2 2 6
row 3: 7 9 3 9 -1
row 4: 0 0 0 3 -7

matrix G:
row 1: 1 0 5 2
row 2: 1 0 5 2
row 3: 2 5 7 9
row 4: 0 0 0 3

If you have a bs answer i will report u

Answer 1

B, D, N, F...These are the only ones.

Answer 2

♥ legend: d for dependent, n for independent;
Matrix A:
boo !! d d n n
row 1: 1 1 0 0
row 2: 1 1 0 0
row 3: 0 0 1 0
row 4: 0 0 1 1

Matrix B:
boo !! n n n
row 1: 1 0 0
row 2: 1 0 0
row 3: 0 1 0
row 4: 0 0 1

Matrix D:
boo !! n n
row 1: 1 0
row 2: 1 0
row 3: 2 5
row 4: 0 0

Matrix N:
Nothing to depend on!
row 1: 1
row 2: 2
row 3: 3
row 4: 4

Matrix F:
boo !! n n n
row 1: 1 5 6
row 2: 1 5 6
row 3: 0 7 2
row 4: 0 0 9

Matrix H:
boo !! n n n n n
row 1: 1 0 2 2 6
row 2: 1 0 2 2 6
row 3: 7 9 3 9 -1
row 4: 0 0 0 3 -7

matrix G:
boo !! n n n n
row 1: 1 0 5 2
row 2: 1 0 5 2
row 3: 2 5 7 9
row 4: 0 0 0 3
♥ and what is a BS answer? Am I reported?

Answer 3

For vectors to be linearly independent means in essence that there's no way to write one as linear combination of the others. Or the vectors span a space with the same dimension as the number of vectors. Or the determinant is non-zero, if the dimensions of the matrix are right.

Some things to look for are duplicates and multiples and sets of unit vectors.

In A the first two colums are the same, so they aren't l.i.

In B the columns are l.i. because you can't form the first from the other two, or vice versa (but you should do something to prove that it's so). Same with D: the second column is not a multiple of the first.

N has only one column vector, so that set of vectors is linearly independent trivially.

In F, you cannot get a 9 in the fourth place of the last column from the zeroes in the other two. So it's l.i.

H has 5 columns, 4 rows. Any time the number of vectors (five of them) exceeds their dimension (4) you automatically lose linear independence.

G has 2 rows that match, which means it's determinant is zero, so it won't have l.i. columns. In fact, you can easily write a combination of the other 3 columns (I'll let you work out what it is) that gives you the second column. So they aren't l.i.