00:01
Now in order to find the basis for the set of solutions to the system of equations, we want to find a basis for the null space of this matrix a, which is a coefficient matrix.
00:09
So in order to do so, we're going to perform elimination first.
00:13
So here we have 1 -1 -1.
00:16
So for row 2, we're going to subtract 3 of row 1.
00:20
To get 0 here, a negative 1, and a negative 5.
00:27
Now for row 3, we're going to subtract 4 of row 1.
00:29
So we're going to get a negative 1 here and a negative 5.
00:36
Finally for row 4, we're going to subtract 6 times row 1.
00:39
So we're going to have a negative 1 and then a negative 5.
00:43
So we see these last three rows are all the same afterwards.
00:50
So for the next step, 1 -1, 1, we're going to divide the second row by negative 1 so that we can get a pivot of 1...