00:01
All right, so for this problem, the first thing that we want to do is try creating a matrix with each one of our given vectors as one of the columns.
00:11
So we have 1, 4, negative 2, 1, 1, 5, negative 3, 1, 1, 7, negative 5, 1, 1, 5, negative 2, 1.
00:21
Now, if we try doing just a single row operation, if we do, for instance, row one minus row four, we can see, or actually i'll do that the other way around, row four minus row one, we can see that we end up with a row at the bottom constructed entirely of zeros.
00:45
So i'll copy and paste over all of the other values.
00:49
But we can see that we have a zero row.
00:53
Now, because we have a zero row, and actually, i mean, it doesn't really matter which order we do that operation in.
01:01
For instance, if we did row one minus row four, we got a zero row at the top, we can see immediately that the determinant of the matrix would have to be equal to zero.
01:15
We would have, when we do our co -factor expansion, we'd have a zero coefficient on each one of the sub -determinants.
01:22
So the determinant of the matrix is zero, therefore the columns are not linearly independent.
01:41
So in terms of trying to exhibit one as a linear combination of the others, let's, for instance, just take that first vector.
01:50
So let's say we want four to be equal to 5x1 plus 7x2 plus 7x2 plus, 5x3, we want, or actually, pardon me, for doing this, actually, so i'll take a slightly different approach here...
