00:01
Alright, we're going to use our inverses to solve the following, so writing out our matrices.
00:05
Our a matrix would be 1 -1 -1, 3 -2 -negative 1, 3 -1, our x -materics would be our variables, so x, y, z.
00:18
Our b matrix would be our answers, so 2, 7 -thirds, 10 -thirds, and our a inverse, our...
00:31
Would be from our previous one.
00:34
So let's say we've got 1 -1 -1 that we're looking for.
00:38
So we've got all over 7.
00:41
Negative 5 -13.
00:45
Negative 5 -1 -3.
00:51
And then 9 -1 -4.
00:57
9 -1 -84.
01:04
And then 3 -2 -1.
01:09
3 -9 -2 -1.
01:11
Alright, so i'm going to write these a little differently.
01:18
So b, i'm actually going to take a one -third out of it.
01:24
So 2 divided by 1 -third is 6, and the other ones are 7 and 10.
01:30
And for my a inverse matrix, i'm going to take a 1 -7th out.
01:36
So negative 5, 1 -3, 9, 1, negative 4, 3 -2 -1.
01:44
So when i actually go to multiply these, i'm going to have a one -third times a one -seventh out front.
01:51
That's going to have to get multiplied to my matrix answer in the end.
01:55
So 5 -13, 9 -1 -4, 3 -2 -1, 2, 1 times 6, 7, 10.
02:07
So 1 3.
02:08
3rd times 1 -7th, i'm going to multiply my final answer by 1 over 21.
02:12
But let's see what we get.
02:15
So we get negative 5 times 6 plus 1 times 7 plus 3 times 10...