00:01
So in this specific problem, we can use a really cool trick.
00:04
It turns out the dimension of all the subspace of vectors such that a times x equals zero can be found by first row reducing a, and then secondly, finding the number of linearly dependent columns in this matrix we've obtained by row reducing.
00:24
So let's go ahead and apply that to these two matrices.
00:27
Let's start with problem a, writing down our matrix, and then we just want to row reduce our matrix.
00:34
So we will zero out the negative 1 in the second row by just adding the first and second row together.
00:42
So we'll copy the first row as is, and then we get 0, 4, and we're adding the first row to the second, so we get a 2 and the negative 1.
00:54
But now at this point, we've row reduced as much as we can, and we see that we have 2.
00:59
Linearly dependent columns, the third and the fourth column.
01:05
So that means our dimension must be equal to two for part a.
01:10
Now if we write the matrix of part b and then we just row reduce, we get the following.
01:17
So let's start by switching the first and the third row...