So, now we're finally going to find a way to solve the apples and bananas problem.
And along the way,
we're going to find out about a thing called
the inverse of a matrix and a method for finding it.
So, let's go back to the apples and bananas problem.
We said that I walked into a shop and I bought two apples and three bananas,
and that that cost 8 dollars or euros or whatever my currency is.
And I went to the shop on another day and I buy ten apples and one banana,
and I got charged 13 euros for that.
So I wrote that down as a matrix times the vector.
And I could call that matrix, A,
and I call this vector, r,
and I could call that output vector, s.
So, A operates on r and gives me s.
What I want to do is I want to find out what Rr is,
what this vector a b is,
in order to give me the right numbers for the output vector.
Now, let's just think about this matrix, A, for a moment.
Let's see if I can think of another matrix.
I'm going to call this matrix A to the minus one,
that when I multiply it by A,
gives me the identity matrix.
I call this matrix here the inverse of A because
what it does to A is it exactly reverses whatever A does and gives me just the identity.
So, if I take this equation, Ar equals s,
if I multiply it on the left by A to the minus one on both sides, well,
what I find is,
is that this is just going to be the identity which is the matrix that does nothing.
So, when I multiply it by I, just get r,
so then I've got r is equal to A to the minus one times s,
the inverse matrix of a times s. So if I could find the inverse of A,
the matrix where multiplying by A
gives me the inverse, if I could find the inverse of A,
I can solve my problem and I can find what my a and b are,
and I've solved my apples and bananas problem.
Now, hold that thought about inverses for just a moment.
So, we said the inverse of A times A was the identity matrix.
We'll just park that up there for a moment.
Now, I actually don't have to go all
the way to the inverse to solve the apples and bananas problem.
I can do it just by substitutions.
So let's look at a slightly more complicated problem and see what I mean.
So, take a matrix that looks like like this: 1 1 3, 1 2 4, 1 1 2.
And I'm going to say that I have apples and bananas and carrots, say.
Okay they're not a fruit but they're another sort of fruit like object that we eat.
So, I'm saying that I bought one apple, one banana,
and three carrots, and that cost me,
let's say 15 euros.
Another day I went and I bought 1, 2 and 4, and that cost me 21 euros.
I don't know that I bought 1, 1, and 2,
and that cost me 13 euros.
Now, the thing to notice about a system of equations like this is that I can,
if I take this row off of another row,
I haven't changed anything, right?
So, I know that an apple, a banana,
and three carrots cost 15.
So if I take an apple, a banana, and three carrots,
and 15 off of the next row,
I haven't changed anything really about that row,
and the same for the third row.
So, I can make this problem simpler by going and doing that,
that sort of a process of elimination.
So if I call this row one,
row two, and row three,
if I take row one off of row two,
then I would have, take row one off of row two,
I'll have 0 1 1.
If I take row one off of row three,
I'll have 0 0 minus 1.
I'd still have a, b, and c. If I've taken on the right-hand side,
I've taken row one,
I've taken row one off of row two,
so I've taken 15 off of 21, so I'll have six left.
I'm taking row one off of row three,
so I'll have 13 minus 15 is minus two.
Ah, interesting.
So now, I can see that actually minus C is equal to minus two,
so I can multiply that row three by minus one.
Let's change pens.
And I've got a solution now for c,
that c is equal to two, so that's interesting.
Now, I've got what's called a triangular matrix,
that is, everything below the body diagonal is zero.
And that's a very interesting sort of matrix.
I've reduced it to what's called Echelon form.
All the numbers below the leading diagonal is zero.
Now, I can do what's called back substitution.
I can take my answer for c here and I
can put my answer for c back into the first two rows.
So, if I take one c equals two,
I know that one times c here is two here,
so I could take that off there and I could do the same.
I can take three times that of the first row.
So, if I go take three times it off the first row,
I'm going to get 1 1 0, a, b, c,
and I've taken three times the first one off of here,
so I've taken three twos off of here, so I'm going to get nine.
And if I take that third row off of the second row,
I'm going to get 0 1 0.
And I've got two off of six is four,
and I haven't changed the last row at all.
So now, I know that c here is equal to two.
I know that if I multiply that row by that column,
I've got just b is equal to four.
And I've got a plus b is equal to nine,
but I can tell you my answer for b back off this first row.
So, I've got then 1 0 0, 0 1 0, 0 0 1, a, b, c, and I've taken one b,
I've taken four off of the first row,
so that's 5 4 2.
So, my solution here for my apples,
bananas and carrot problem is the apples cost five euros,
bananas cost four euros,
and carrots cost two euros.
And that's solved the problem for me.
Very nice. So I didn't really have to compute the inverse at all.
But I've only found out the answer for this specific output vector.
So we said that this was A times r is equal to s.
I've only found it for this specific s. If I did the inverse in the general case,
I could do it for any s. I could find r for any s.
So, what I've done here is I've done what's first, elimination.
I've taken multiples of rows and I've taken them off of each
other to get to having a triangular form where the bottom corner here is zero.
So I've just got ones on the leading diagonal.
And then, I've done back substitution. Back substitution.
I've gone back up
of putting the numbers for c back into the first two rows, and then for b back into the first row,
and so on, to get my solution to my problem.
And that's very nice. This is actually one of
the most computationally efficient ways to solve this problem,
and it's going to work every time,
and it's very simple to understand how we can do it in relatively few operations.
So it's really, really nice.
And one thing to notice here is that, in doing so,
what I've done is I've transformed A into the identity matrix.
This is just the identity matrix.
There's ones only on leading diagonals and zeros everywhere else.
So, this is going to be a key to a way
to find the inverse which is what we'll look at next.
