00:01
So the first thing we're going to want to do to solve this problem is put our systems of equations in a matrix form.
00:07
And this is much easier to do if we combine like terms.
00:11
As you can see in the second row, we have y in there twice and no z.
00:16
So this equation, or the systems of equations, is really equivalent to writing it like so.
00:32
So then we first and third row are left unchanged, but in the second row, negative 12 y plus 7y gives us a negative 5 y and i've included a zero for z that'll make it a little bit easier to put it into matrices so on the left -hand side we create an a matrix which has all of our coefficients so it's first from the left -hand side so the first row would be 2561 negative 12 the second row 18 negative 5 and 0 and then we're you take the third row the same way, and that's going to be times our x matrix, which contains our variables.
01:13
It has x, y, and z.
01:16
And this is equal to our b matrix, which is just a single column matrix, 10, negative 9, and 12 from the right hand side.
01:26
Now to solve for x, we multiply both sides by a inverse...