We will now embark on what is
probably my least favorite exercise or computation in
mathematics-- and I think you'll see why-- where we will
invert a 3 by 3 matrix. And in my mind, the only thing
less pleasant than inverting a 3 by 3 matrix is inverting
a 4 by 4 matrix. It very quickly becomes obvious
to you that it's probably better for a
computer to do this. But you need to learn
how to do it. And it's a good exercise
for me to do. And if I keep doing it my whole
life, it'll prevent my brain from degrading. But as you'll see, this is
almost an exercise in not making careless mistakes. So let's start with a
3 by 3 matrix and try to take the inverse. So let's say I have matrix a. I think I'm going to need a lot
of space here, so I'll try to do this small, without
being confusing. Matrix a. Let's say it's 1, 0-- and I'm
specifically choosing this matrix because the numbers are
reasonably non-hairy-- 0, 2, 1, 1, 1, 1. So the first thing that I do
when I take an inverse of a 3 by 3 matrix, I create what I
call-- or not what I call, what everyone calls--
a matrix of minors. So let me write that down. Matrix of minors. So what's a minor? Let me draw that out. So it's going to be another
3 by 3 matrix. And what it is, so this
element, this top left element, is essentially going
to be the determinant. If I were to take my original
matrix, and I were to cross out this position's
row and column. So for example, this 1, 1
position, row 1, column 1. I cross out row 1
and column 1. What numbers do I have left? I have this 2, 1, 1, 1. I have this right here. So it's the determinant
of 2, 1, 1, 1. And actually maybe I'll
write that down. So it's the determinant
of 2, 1, 1, 1. So I'm gonna run out
of space I'm sure. It's going to be 2, 1, 1, 1. The determinant. The absolute value sign says
it's the determinant. Remember, all I did is, I said,
OK, in position 1,1, let me cross out the column and
the row 1,1, and take the determinant of what's left. Or the minor of this matrix. And then I will take the
determinant-- so when I go to this position, I'm in
row 1, column 2. I'm essentially going to
take the determinant. If I were to cross out row
1 and column 2, what do I have left over? I have 0, 1, 1, 1. It might be confusing, but just
remember-- I wish I had something I could
cover this with. Unfortunately my fingers can't
show up on this video. But if you cross this row and
this column out, you're just left with this 0, this 1,
this 1, and this 1. And you take the determinant
of that minor. And we'll keep going. I'm probably going to run out of
space here, but I will try my best. And so when you go to
this position-- row 1, column 3-- what do you do? Well you cross out
row 1, column 3. And then the determinant, or the
minor that you have to do, is 0, 2, 1, 1. So the determinant of that
two by two matrix. And then you keep doing that,
so forth and so on. And I'm going to run
out of space. But what I'm going to
do is I'm just going to calculate it. I think you understand
how to do it. Well you don't understand how to
do it, but I think when we calculate it it'll make
a little more sense. Let me actually just
calculate it out. Because if I were to write these
2 by 2 matrices, I would run out of space. But anyway, let's go back
to this position 1, 1. Cross out the first row, first
column, I want the determinant of this thing right here. So what's the determinant
of this 2 by 2? That's not too hard. It's 2 times 1, minus
1 times 1. So what's 2 times 1,
minus 1 times 1? Well it's just 1. Then when we go to row 1,
column 2, I want the determinant of 0, 1, 1, 1. So it's 0 times 1,
minus 1 times 1. So 0 times 1 is 0, minus
1 times 1 is minus 1. And that's just this determinant
right here. I'm just kind of reshowing you
how I visualize when I cross out the rows and columns. So it's 0 times 1,
minus 1 times 1. And in this position of course,
you cross out this row, this column, and 0 times
1 minus 1 times 2. So that's minus 2. Let's keep going. All right. So now, when we're in row 2,
column 1, we cross out row 2, cross out column 1. And we're left with this 0,
this 1, this 1 and this 1. So it's 0 times 1, which is 0. Minus 1 times 1. So we're at minus 1. Then when we get row 2, column
2, we cross those two out, and we take the matrix of the
minor that's left. So that's 1 times 1,
minus 1 times 1. So that's 0. Almost. We're halfway done. OK, so then we're in
row 2, column 3. So we cross out row
2, column 3. And what we have left is 1
times 1, minus 1 time 0. So that is just 1. Last row. OK, so we're in row
3, column 1. So we cross out row
3, column 1. You're left with
0 times 1 is 0. Minus 2 times 1. So that's minus 2. Then we're in row 3, column 2. So we cross out row
3, column 2. And you have 1 times
1, minus 0 times 1. So that's just 1. Last one. Row 3, column 3. So we cross out row 3, we
cross out column 3. And you're just left with 1
times 2, minus 0 times 0. So that is 2. And if I haven't made any
careless mistakes, that is our matrix of minors. Now we now have to convert
this to what we call the matrix of cofactors. And actually this step is
fairly straightforward. So to convert from a matrix
of minors to a matrix of cofactors, you just have to
remember this pattern. This pattern applies to
any 3 by 3 matrix. Plus, minus, plus, minus,
plus, minus, plus, minus, plus. And so you can kind of just
imagine this as kind of a checkerboard of pluses
and minuses. And you apply that to this. So what do I mean by that? Well that means, when you start,
it's a checkerboard, and you start with a plus
at the top left. And then you just keep
alternating plus, minus. So if you applied this
to this, you get the matrix of cofactors. Let me write that down. You can imagine this is a bit of
a marathon of computation. OK, so the matrix of cofactors
is essentially applying this pattern to the matrix
of minors. So what do you do? You say this plus
1 times 1 is 1. But now we have a minus. So that's minus times minus
1 is positive 1. Then you have plus times
minus 2 is minus 2. Then you have a minus here. Minus times minus
1 is positive 1. Plus times 0 is still 0. Minus times 1 is minus 1. Plus times minus 2 is minus 2. Minus applied to 1 is minus 1. And then plus applied
to 2 is just 2. And we have our matrix
of cofactors. And we are more than
halfway done with inverting this matrix. And I just want to
take a note here. What we're doing is kind of
just a magic formula. It might seem a little bit
like voodoo for you. But I just want you to keep in
mind that in future videos, I will show you where
this comes from. Although it will be quite
hairy to prove it for a 3 by 3. But I'll definitely show
it to you for a 2 by 2. And actually, I'll show you
other algorithms that might make a little bit more intuitive
sense for doing it for a 3 by 3. But I just wanted to show you
how to do it this way, so that at least when you see it on your
Algebra 2 exam-- because I think they actually teach this
in Algebra 2-- you could at least, if the teacher asks
you, solve for the matrix of minors or the cofactors or solve
for the determinant of the inverse, you can do it. And then we'll worry about
getting the intuition, which is not how I normally like
to teach things. But this is an exception. But anyway, back
to the problem. This is the matrix
of cofactors. Now from this, we take the
adjoint of matrix a-- or I learned from Wikipedia, the
correct term is the adjugate of matrix a. But this is determined the
notation is the adjugate of a. And all this is is
the transpose of the matrix of cofactors. And I know I'm throwing out a
lot of weird terminology here. But the transpose, all that
means is that you switch the rows and the columns. So this one right here is
in row 1, column 1. But you know, so the rows and
columns are the same, so that just stays the same. So actually anything on the
diagonal stays the same. Because this is row
2, column 2. This is row 3, column 3. So the diagonals
stay the same. And then you switch places. You kind of flip across
the diagonal. And what do I mean by that? Well this 1 was in
row 1, column 2. So it'll then be moved
to row 2, column 1. So this 1 right here
will go here. So you can kind of say that it
flipped over the diagonal. And similarly, this right here
is in row 1, column 3. It's going to be switched
to row 3, column 1. So it's going to go here. And you can kind of see that it
just flipped over that end. So this minus 2 isn't
this one. It's this one over here. And actually, we see that this
matrix is symmetric. When you flip it, you actually
get the same thing. So maybe it was a bad example. But I want you to understand
that the transpose is where-- if something like this number,
if it's in a row 1, column 2, then it moves to row
2, column 1. So you're switching the
rows and columns. But anyway, we could
keep doing that. But essentially you're just
flipping over the diagonal. So let's see. So then this number will be
flipped to this position, so it goes there. This is in row 2, column 1. So it will go to column 2,
row 1, which is that. And then if we go here, that's
going to be flipped down here, flipped across the diagonal. So that's minus 1. This is going to be flipped
all the way up there. So that's minus 2. And then this will
be flipped there. This is minus 1. We are almost done. So this is the adjoint
of matrix a. So to get the inverse of a-- and
let me actually erase some of this, because we're going to
run out of space otherwise. And as you can see, I'll be very
impressed if I have not introduced a careless
mistake yet. So let me erase all of this. I'm building an appetite just
doing this problem. It's so taxing on me. So the inverse of matrix a
is equal to 1 over the determinant of a times
the adjugate, or adjoint, of matrix a. So we solved for this part. So now let's solve for
the determinant. So the determinant of a-- and I
kept the matrix of cofactors here for a reason-- the
determinant of a is-- if you go across-- you can actually go
across any row-- but just for simplicity, just remember
it this way. You go across the top row, and
you multiply each term times its corresponding cofactor,
and you add them. So in this case, it'll be 1
times its corresponding cofactor, which is 1. Plus 0 times its corresponding
cofactor, which is 1. Plus 1 times its corresponding
cofactor plus minus 2. So this is 1 plus 0 minus 2. It equals minus 1. And thank God it was a
relatively straightforward determinant. And if you didn't have this
matrix of cofactors, the other way you could think about it--
and this is good because it gives you an intuition of how
we even got to the matrix of cofactors-- you could view this
as the same thing as 1 times the determinant
of its minor. So if you cross out the row
and the column, it's this determinant. So it's 2, 1, 1, 1. And remember there
was that pattern. You have plus, and then
you go minus. So minus 0 times the determinant
of its minor. So you cross out that
row, that column. So 0, 1, 1, 1. And then we switch again. We go back to plus. Plus 1 times the determinant
of its minor. So you cross out that
row, that column. You get 0, 2, 1, 1. And you could compute
this out. And this is this cofactor. This, with a minor sign,
this is just a minor. And then when you apply
the minus sign, it becomes this cofactor. And then this is that minor. And since it's a plus sign
there, that's that cofactor. But anyway, I just wanted to
explain that, and hopefully it hasn't confused you. But we're ready now to solve
the inverse of a. We know that the determinant
of a is equal to minus 1. We know that the adjugate of
a is this number here. So we now can solve
for the inverse. And let's do that. Let me erase all
of this stuff. Cause actually, after I solve
for the inverse, I want to prove to you that it is
the inverse-- maybe. If I have enough time. Because I just realized I'm
running pretty long. That might be a good
exercise for you. OK. So the inverse of a is equal
to 1 over the determinant. We figured out the determinant
is negative 1 times the adjugate of a. 1, 1, minus 2. 1, 0, minus 1. Minus 2, minus 1, 2. So this is just minus
1, right? So we just apply minus
1 times everything. So we get-- if I haven't made
any careless mistakes-- minus 1, minus 1, plus 2, minus
1, 0, 1, 2, 1, minus 2. I think that I have-- let's see,
I just did a minus times everything. That looks right. And so that is a inverse. And it only took
us 17 minutes. And I will leave you there,
because it will probably take me another 5 or 10 minutes
to [? seed ?] and prove. But that might be a good
exercise for you. To multiply this matrix times
this matrix, and make sure that you get the identity
matrix. I will see you in
the next video.
