00:01
We are asked to find the null space of a.
00:03
The definition of the null space is given at the top here with the null space of an n -by -n matrix is given by the set of vectors x in rn, such that the matrix a times the vector x is equal to the zero vector.
00:19
The problem gives the matrix a as a 3 -by -3 matrix shown here.
00:25
We can see from there that the value of n is 3, which means we're in r3.
00:30
And so we can write our x vector as a 3 by 1 matrix with x1 x2 and x3 so now we can start to solve this linear equation by using row reduction to get a in row echelon form so i've rewritten a in augmented form down here and we can start our row reduction by seeing that we can get we can get a zero here if we multiply minus 4 times the first row and add that to the second row.
01:25
So we're going to add minus 4 times row 1 and add it to 2.
01:40
So that gives our first row will stay the same.
01:44
1 minus 2 1 and 0.
01:51
And so we'll have a minus 4 times 1 added to 4 which gives 0.
01:57
We have a minus 4 times a minus 2 plus a minus 7 which gives 1.
02:04
We have minus 4 times 1 plus a minus 2 which is minus 6.
02:10
And our third row stays the same, minus 1, 3, 4.
02:15
And then we have a minus 4 times 0, which is 0, and the 0 of the 3rd row.
02:27
So now we can get a 0 here if we add 1 times row 1 to row 3.
02:39
So we'll add 1 times row 1 to row 3.
02:52
So that will give our first row will stay the same, and our second row will stay the same.
03:04
And then our third row will be 1 plus and minus 1, which is 0.
03:11
Then we have a minus 2, minus 2 plus 3, which is 1.
03:23
And then we have 1, let's see 1 plus 4, which is 5.
03:32
And then a 0 plus 0, which is 0.
03:44
Now we can get, let's see, we can multiply the second row by minus 1 and add that to 3, and that will give us a 0 here.
04:00
So we'll multiply, or we will add minus 1 times row 2 and add that to 3.
04:18
So our first row stays the same.
04:26
Our second row will stay the same...