00:01
Okay, so with our given system here, while we can keep the first equation, and we can eliminate our isub 1 term from the remaining equation.
00:10
So what we're going to do is do negative 3 times the first equation, and then add that to our second equation to produce a new second equation.
00:19
So what we get here is, well, our first equation stays as i sub 1, not plus, but i sub 1 minus i .72.
00:32
Plus i sub 3 is equal to 0.
00:37
Okay, the second equation then becomes 5 i.
00:41
Sub 2 minus 3 i.
00:45
3 is equal to 7, and then we have 2 i.
00:49
2 plus 4 i sub 3 is equal to 8.
00:56
Okay, now to eliminate our i sub 2 by working on the second column.
01:02
So let's eliminate i sub 2 from the third equation.
01:06
So we're going to do negative 2 times the second, and then add that to 5 times the third.
01:14
So what we get is, while we have i .1 minus i.
01:21
2 plus i.
01:25
Sub 3 is equal to 0...