00:01
If we want to show that this matrix here is invertible if in aal a bd minus bc is non -zero and the inverse is equal to what we have here the first thing we can do is set this up to this augmented matrix here and then go ahead and reduce the left side as far as we can into a row echelon for so the first thing we would want to do is clear out the c.
00:40
And doing that is going to give a over 1 row 1 minus, and then we would have want to have c right here.
01:00
Now, in there they also give us this little hint of assuming that this is not equal to, or a is not equal to zero because of right here.
01:13
So let's just go ahead and do that.
01:14
So a not equal to zero.
01:18
Then we can do the next step for if it is zero later.
01:26
And then we're going to add this to row two.
01:28
So the top row is not going to change.
01:32
So it will be ab10.
01:35
Now here, this is going to become zero.
01:39
This is going to become d minus cb over a.
01:48
And then this here is going to become minus c over a.
01:52
And then this is just going to stay as one.
01:56
All right.
01:57
Now, in order for this to be a matrix that is invertible, one, we need to have where this is going to have full rank, and this is going to have full rank if this is not equal to zero.
02:16
So that implies that d minus cb over a is not equal to zero.
02:23
Well, again, remember we're assuming a is not zero.
02:25
So we can multiply this whole thing by a, giving us ad minus cb, is not equal to zero.
02:35
And this is exactly what we have right there.
02:41
So this is how we can kind of show this.
02:46
So maybe i should say something right here.
02:48
So if invertible, full rank.
03:03
Now, first to get that next part, we can just finish going through and canceling things out here.
03:14
So let's come down here, and i'm just going to multiply row 2 by a.
03:20
So a, row 2 is just going to make things look a little bit prettier, i think.
03:25
So a, b, 1, 0, and this becomes ad minus cb, and then that's negative c, and then this is going to be a.
03:38
Next, we need to go ahead and clear this b.
03:43
And we can do that by first, dividing by what we have done there.
03:54
So it would be 1 over ad minus cb.
03:59
But then we need to multiply by negative b to get that to cancel out.
04:05
And so that's going to be row 2 plus row 1.
04:14
Okay.
04:15
So the bottom row should stay the same.
04:21
So 0, ad minus c, b, minus c, a.
04:27
And then the top row, so a stays the same.
04:32
This is going to become zero.
04:35
Now here is going to become, so multiply negative c by this.
04:42
So that's just going to be bc plus 1.
04:46
So it would be 1 plus 1 plus bc over ad minus cb.
05:02
And then here, let me get rid this for right now.
05:07
So i'm not trying to squeeze it in.
05:12
And then multiplying this by negative b over ad minus cb.
05:18
That's just going to be negative a.
05:23
B over a .d minus c.
05:32
Next, we would go ahead and divide this whole top row by a.
05:37
Divide this whole bottom row by a .d.
05:44
So first, 1 over a, row 1, and then, also 1 over ad minus c b row 2.
06:04
So the left side is just going to become 1 .001 to the identity matrix.
06:11
So 1 .0.
06:16
And then on this left side we divide everything by a.
06:19
So this is going to become 1 over a plus b .c over a, a, a, a, b, c over a, a, d minus c b.
06:37
This is going to just become negative b over ad minus cb.
06:47
Now this is going to become 0 .1...