00:01
We're going to solve the following system using inverse matrices.
00:04
So we have the formulas written on the right there.
00:08
So our matrix a would be b, 3, b2, our matrix x would be x and y.
00:20
Our matrix b would be our answers.
00:24
So 2b plus 3, 2b plus 2.
00:28
And our inverse we found in the previous sections, so i have them all written on the x page.
00:34
We're looking for b3, b2.
00:38
So right there we've got b3, b2, so our inverse is negative 2 over b 3 over b, 1 and negative 1.
00:47
So negative 2 over b, 3 over b, 1 and negative 1.
00:54
We'll just double check that, and that's what we've got.
00:59
So if i take my inverse matrix, so negative 2 over b, 1, 3 over b, negative 1.
01:07
And multiply that by my answers.
01:11
2b plus 3, 2b plus 2.
01:16
I should get down to what my x and y are worth.
01:20
So i have negative 2 over b times 2b plus 3, plus 3 over b times 2b plus 2, and then 1 times 2b plus 2, plus a negative 1 times 2b plus 2...