0:00
Hello.
00:02
So from our given system of equations, we can have our matrix here.
00:07
A is just the coefficients on our variables.
00:11
X is, well, the column vector of our variables.
00:14
We have three variables here.
00:15
And b is our column vector of the constants or the solutions.
00:20
So we then take the system of equations can then be rewritten as the matrix equation where we have a times x.
00:30
Is equal to b.
00:34
Okay, we then take the matrix of our matrix a and multiply on both sides, and we get that x is equal to, well, the column vector x, y, which is equal to a inverse, a inverse times b.
00:54
Okay, well, a inverse we already found in a prior problem.
00:57
So we just concatenate with the 3 by 3 identity, and then use, you know, elementary row operations.
01:04
And we found that a inverse was equal to, well, 3, negative 3, 1, negative 2, 2, negative 1, negative 4, 5, negative 2...