00:01
So we're given that t of x is equal to this vector, which depends on x1 and x2.
00:09
And we want to find an x1 and an x2, such that t of x is equal to the vector negative 1 for 9.
00:18
So first, let's write this in the form of a matrix a times x.
00:27
So we want some matrix times x1, x2 to equal this vector here.
00:37
So we need to multiply x1 by 1 and x2 by minus 2 in the first component.
00:43
For the second, we have negative x1, so minus 1, plus 3x2.
00:50
And for the last component, we have 3x1 and minus 2x2.
00:57
So this is our vector, or our matrix a that we need to deal with.
01:02
And we want ax to equal negative 1 -4 -9.
01:07
So we need to row reduce the augmented matrix of a, and we need to put the vector we want on the right hand side.
01:21
So minus 1, 4, 9.
01:25
So to row reduce, first we want these two values to be 0.
01:33
So i'm going to take the first row and add it to the second row.
01:39
So i get a 0 here...
