The next topic is Inverse Matrices.
A square matrix A is invertible so square
means the dimensions have to match, right?
So, I've been calling matrices m by n, so
m rows and n columns.
Square just means that m is equal to n so
I could say it's either m by m or n by n.
And I think most of the time, I'll say n
by n.
so A is invertible if there exists a
matrix, so I'm going
to call the inverse A to the minus 1
power, it doesn't mean 1 over A.
But it's sort of in the same sense that if
I had a real number that's not equal to 0.
And I took 1 divided by that real number,
then
when I multiply those two things together,
I would get 1.
So A inverse is going to be a matrix where
I
multiply either before or after A so I can
take A-inverse A,
that has to be equal to the identity
matrix and
A A-inverse has to be equal to the
identity matrix.
So if I can find this matrix A-inverse,
then I will say that A is invertible.
So the inverse, if it exists, is unique.
And one way you can see that is if, so,
so, what I'm trying to do here is I'm
trying to say is that B and C are
two different matrices that satisfy the
properties in the definition.
So the left-hand inverse, this one here,
will be the
matrix B and the right-hand inverse will
be this matrix C.
And then
you can do a clever little trick.
So
if I have B times AC, but
what is AC?
That's just the identity matrix.
So this is B times the identity matrix,
which is just
equal to B.
But since I know AC, the product
AC is also equal to the identity matrix.
Then I can substitute the value AC in here
for I.
And then for my rules of matrix
multiplication,
it doesn't matter where I put the
parenthesis.
So I'm going to, instead of putting the
parenthesis around AC, I'll put
the parentheses around BA.
But I've already assumed that this is
equal to the identity matrix.
So this is telling me that this is I times
C which is the same thing as C.
So if I assume BA is equal to the identity
matrix and AC is equal to the identity
matrix, then you can make an argument like
this to say that B has to be equal to C.
So there can only be one matrix that
satisfies this property
or these two properties if an inverse
exists.
So, if A is invertible, then the unique
solution to this system of equations
represented by Ax equals b, I'll start off
just by writing Ax equals b.
If A is invertible, then I have this
matrix A inverse.
So I'm, I'm going to multiply both sides
of this, pre-multiply by A-inverse.
So I put a A-inverse before the Ax, and an
A-inverse before the b.
And then A-inverse A, by the definition
of the inverse, that's just the identity
matrix.
So identity matrix times x is equal to x.
So I end up with x, which
is the solution, this is the thing I'm
looking for, is equal
to A-inverse times b.
And if there's a vector x
that's not equal to 0 but the, has the
property that
A times x is equal to 0, then A is not
invertible.
And basically, I want to use this result
here just to show what's going to go
wrong so this, this is a kind of common
argument that you make in mathematics.
I want to assume that x is not equal to 0.
And then based on that assumption, I
want to prove that it is equal to 0.
And so I'm, I'm assuming that something is
true, contradicting that assumption.
And so that's going to be called a proof
by contradiction.
And that's going to allow me to say that A
is not invertible.
Because if A is invertible, I can solve
for this vector x
here using this trick that I've just done
in the third bullet point.
[COUGH]
So if there is a vector, a non-zero vector
x so
at least one element of this vector x has
to be non-zero.
Then I could write x as the identity
matrix times x.
And then if A was invertible, it would
have an inverse and then that would
satisfied
the, the property in the definition so I
could replace this identity matrix with
A-inverse A.
And then I can just, because of my rules
for
matrix multiplication, I can put
parentheses wherever I want to.
So I'm going to put some around this Ax
here but I've assumed that there's
this non-zero vector x that has a property
that Ax is equal to 0.
So x, remember, is non-zero.
It has at least one non-zero component.
But I can replace this Ax with the zero
vector.
And so that tells me that x has to be
equal to A-inverse times 0.
But anything times 0 is equal to 0.
So what I've assumed on the
left is that if A is invertible or I've
assumed on the
left that I have an invertible matrix a
and this non-zero
vector x.
And then I've showed that A-inverse times,
well, so essentially what I've been able
to show with
that assumption is that my non-zero x is
equal to 0.
So that's a contradiction so I can't have
a
non-zero vector x having every element
equal to 0.
And so then whatever I assumed has to be
false.
So I can't assume that A is invertible.
Or if I assume that A is invertible,
I can provide that it's not invertible
essentially.
So that's how you show that if there is
this non-zero vector
x, it satisfies Ax equals to 0, then A is
not invertible.
In the 2 x 2 case, there's actually a nice
formula for computing the inverse.
So, a 2 x 2 matrix will be invertible if
and only if ad minus bc.
So I've, I'm just labeling the elements.
So I have a, b, c, d as the, the so ab is
the first row, cd is the second row.
And it needs to satisfy this property, so
the ad minus bc has to be non-zero.
If that's not equal to 0, then the inverse
of A
is going to equal to 1 divided by ad minus
bc.
So this is why this value has to be
non-zero because
I'm going to divide by it times d minus b
minus ca.
So that's going to give me a matrix that
when I multiply it either
in front of or after A, is going to give
me the identity matrix.
And the number ad minus bc is called the
determinant of A.
So essentially, if this number is not
equal
to 0, then A is going to be invertible.
So a matrix is invertible if its
determinant is not equal to 0.
And then just, as a special case, I want
to point
out the easiest type of matrix to compute
the inverse of.
So the 2 x 2, I have this nice
formula here, not nice necessarily, but I
have a formula.
A diagonal matrix just means that any
off-diagonal element.
So any, any element where I is not equal
to
J, so above the diagonal or below the
diagonal, all of
those have to be 0.
If I have a matrix like that,
then the inverse is just 1 over the
diagonal elements.
And so this also supposes that, so
A-inverse will
exist if all of these diagonal elements
are non-zero.
If if one of these, so suppose the first
one was equal
to 0, then there's nothing I could put in
this cell here so
that, that matrix wouldn't have an
inverse.
So if, if the diagonal elements of a
diagonal matrix are all non-zero, then I
can very easily calculate the inverses
just
by taking the reciprocal of each diagonal
element.
>>
[INAUDIBLE]
>> All of the rest of them stays 0.
>>
[INAUDIBLE]
>> Yeah.
And you can just convince yourself of this
when you do the multiplication.
I'll have first first row times the first
column here so I'd have d1 times 1 over
d1 and then just a whole bunch of zeros
here times a whole bunch of zeros here.
And if the I is not equal to J, then the
non-zero element here will
just crash into a 0 on this side and the
non-zero element here will hit
a 0 element on this side.
So everything ends up being 0 off the
diagonal so I get the identity matrix
back.
So then we have some rules for inverses.
So these are probably the most important
things
you're going to want to remember about the
inverse.
generally, if you ever need to compute the
inverse of a matrix, you're going to do
that
using a computer so even this formula
here,
it's not something I have committed to
memory.
If I need it, I'll look it up or it's
usually,
if you remember just the determinant, you
can figure out what the,
the arrangement of the rest of the letters
has to be pretty quick, quickly.
[COUGH]
What is useful though is thinking about
matrices sort of more abstractly.
So instead of thinking about the actual
elements,
I'm trying to do mathematics with the
matrices themselves.
And so if I end up with a product AB and I
need to take the inverse of
that product, that ends up being B-inverse
times A-inverse.
And the way to convince yourself that's
true is to just plug that into the
definition of the inverse.
So remember, my, my definition of the
inverse said that if I
have the inverse times the matrix, that
has to give me the identity.
So in this case, it would be AB inverse
times the matrix product AB.
And now this, I can write without the
parentheses and that would be the same
thing.
The AB inversed, the order
swapped, so it becomes B-inverse A-inverse
and then I'll
put some parentheses around what I get on
the inside.
A-inverse A is just equal to the identity
matrix
so it's going to just sort of collapse in
upon itself.
So this I, I can multiply that I times B
that's identity matrix times anything just
gives me the anything back.
So I times B gives me B back.
I end up with
B-inverse B which is by the definition of
an inverse, the identity matrix.
And then I can do the same thing for the,
for the right-hand inverse.
Again, because it switches the order, all
that's going to happen is I
end up eliminating the B first so the B
multiplied by B-inverse becomes identity.
And I end up with AIA or AIA-inverse which
becomes
AA-inverse, which becomes I.
And you can do exactly the same thing for
a longer products as well.
So, if I have ABC-inverse, I just reverse
the order
of the product so that becomes C-inverse
B-inverse A inverse.
So if you wanted to prove that that's
true, you're just putting ABC
over here and then its going to collapse
three times rather than just two times
but what you're left with is the identity
matrix.
So now, let's think about how would I
actually find A-inverse.
So I need to find a matrix so that when
I multiply A by A-inverse, I get the
identity matrix.
If I can find this, I've already argued
that it's going to be
unique and it would also be the matrix
that can go on the front.
So, I don't need to compute both of them,
I just need to compute one of them.
If there's one matrix that have this
property then it automatically
has the property that if I pre-multiply by
A-inverse, that's also
going to turn A into the identity matrix.
So, let's make some vectors here.
So, I've used e before to mean a column
vector of all ones.
I'm going to add a subscript now to mean
it's a vector of all zeros
except it's got a 1 in whatever
position, this little script, subscript
corresponds to.
So e sub 1 would be 1 0 0.
and if it's in general in N dimensions,
then I have e1 up through en.
And it just means that it's all 0 except
for
the, the ith element and that's going to
be equal
to 1.
So then, I have e2 is 0 1 0 and e3 is 0 0
1.
And then, if I bind all three of these
vectors
together, e1, e2, e3, that's just going to
be the identity matrix.
So, you can almost kind of see it here.
I just have ones going down the diagonal.
And now I can imagine that I have an
inverse, so A-inverse, an inverse of this
matrix A.
And the columns of my matrix A-inverse,
I'm going to call x1, x2, and x3.
So x1 is a column vector with three
entries, x2 is a column
vector with three entries, and x3 is a
column vector with three entries.
So I can write out my A A-inverse is equal
to A times these three column vectors
bound together into a matrix.
And I know that has to equal e1, e2, e3,
which is the identity matrix.
And so what I end up having to do is solve
three systems of equations.
I have to solve Ax1 equals e1, Ax2 equals
e2, and Ax3 equals e3.
So, if I was actually computing the
inverse, remember, before I said I could
compute the solution to Ax equals B by
pre-multiplying by A-inverse.
So if I already knew what the inverse was,
I could just
multiply and find that, well, A-inverse B
is going to be the solution.
But if I just want to actually solve that
system, I would just be solving one of
these.
If I want to find the A-inverse and use
A-inverse to
solve that system, I end up solving three
of these.
So it ends up being three times as much
work to find a inverse to
solve that problem as it would be
just to directly solve that problem using
elimination.
So computing A-inverse three times
as much work as solving Ax equals b.
So the, this is going to be sort of one of
the
morals of the story here is quite often if
you need to
compute an inverse, if you see an inverse
in a matrix formula,
there's a better, faster way to solve the
problem without using an inverse.
so in this case, I, I was looking at a 3 x
3 case.
And I said that you'd have to solve this
three times.
So it turns out that if you look at a 4 x
4 case or a 5
x 5 case, the, the three times ends up
being an upper bound on the amount of
work you have to do because you've done
enough elimination on A in solving these
first three,
that you can kind of recycle that you
know,
if you're are intelligent about the
algorithm you use.
And so the algorithm is called the
Gauss-Jordan
method, which I'm not going to go through
this here but it's essentially just
doing this elimination problem and you can
do this in n cubed elimination steps.
But if you compare that to solving Ax
equals b, this requires n
cubed divided by 3 or roughly n cubed
divided by three elimination steps.
So you're going to be doing less work.
sorry, not n cubed.
This is not quite right.
Not elimination steps, but mathematical
operations.
but the point I want you take away from
this
is, it's 1 3rd as much work to solve Ax
equals b using elimination as it would be
to find
an inverse and then use that to solve Ax
equals b.
And I remember I said if I, if I was
doing elimination and I ran into a row
that was all
zeroes, I couldn't find a pivot in that
row so
that particular system does not have a
full set of pivots.
Oops.
And now, I have this idea of
invertibility.
It says some matrices have inverses and
some matrices don't.
So if a matrix has an inverse that I can't
find a non-zero vector
that has the property that if I say a
times that vector, I get zero.
And so it turns out these two things are
related,
so let's supposed A is a square matrix and
suppose
there are n pivots.
Then I could solve each one of these
equations and that would
give me these xis that would be the
columns of my inverse.
So I can put the, these column vectors xi
together
to make an matrix that's the inverse of my
matrix A.
And so elimination then gives a complete
test for A-inverse to exist.
So there has to be n pivot.
So, this isn't a complete proof.
This is just the result.
So in a, a Linear Algebra textbook, they
will tell you exactly how to,
how to do this because it's sort of
something you have to do both directions.
But essentially if a matrix A is
invertible,
then you can find a set of n pivots.
So you say, it has a full set of pivots.
And if a matrix is non-singular, that
means every row has
a pivot then that matrix is also going to
be invertible.