Let A be the matrix 5 -2, 6 -2
and x be the vector 2 3.
And we'll try to compute following: A times x
A squared times x, and A to the 1000 times x.
If you look at part a and part b of this question
they are quite straightforward; you can
just carry out the multiplication.
But if you look at part c, do we really
want to compute
A to the 1000, assuming that 
we're doing this by hand?
So let's see what we can do here.
Maybe we can get lucky.  So let's first
compute A times x.
So if we write down A and x,
this is what we have.  And so,
the first entry will be 5 times 2 plus
-2 times 3, and the second entry will be 
6 times 2
plus -2 times 3,
and that will be
so 10 minus 6 is 4 and 12
minus 6 is 6.
Ok so that's A times x, that's part a.
Now what about part b?
Let's work out what A squared is.
A squared is 5 -2, 6 -2
times 5 -2, 6 -2
and that will give us this matrix here.
And so A squared times x
is 8 and 12.
Now, it doesn't look like that we can write
down a formula for A to the 1000 right away.
But if you look at 4 6,
that's simply 2 times x. 
And if you look at
8 12, that's 4 times x.
The fact that A times x is 2 times x
means if you multiply both sides again by A,
you get A squared equal
2 times A x, which again gives you 2
times 2 times x.  Right, because A
times x is 2 times x, and that's
2 squared x.  Now, you can repeat this
and get that if k
is a positive integer, then this is simply
2 to the k times x.
So, we can write down A to the 1000 times x
to be 2 to the 1000
times 2 3.
And that will be a good enough answer 
for this question.
So in this case we got lucky because
if you try to compute A to the 1000
times 1 1
we don't get anything this nice because
A times 1 1 is simply 3 4
and 3 4 is not a scalar multiple of 1 1, so
this trick here will not apply to the vector 1 1.
But is there something that we can do here?
It turns out that yes there is.
And I'm gonna do this by writing A differently.
We're going to let P be the matrix 2 1, 3 2,
D be the matrix 2 0, 0 1.
If you compute the inverse of P,
you get this.
And A can be written as P times D times
P inverse.
This is something that you can check.
Now, how does that help us?
So let's look at A squared
So A squared is P times D times P inverse.
times P times D times P inverse again.
Right? So this is A and this is A.
P inverse times P, that's the identity.
So this, after you simplify,
will be P, D  squared, because this
has become the identity, so I've D times I times D, that
gives me D squared times P inverse.
And what about A cubed?
A cubed is A times A squared.
And so this is P times D
times P inverse times...A squared is now
this thing over here...
P times D squared times P inverse,
Again this becomes the identity.
So what we'll get is P times D cubed times
P inverse.  And as you can see,
if we continue in this fashion,
we can deduce that A to the k is P times
D to the k times P inverse.
D is a diagonal matrix.
When you take power of a diagonal
matrix, you simply take powers of the
diagonal entries.
So for example, if I want to look at what
D to the fifth power is,
I just take the diagonal entries to the
fifth power.
And that will give us 32 0, 0 1.
And so if you want to know what
A to the 1000 is, well it's simply P times
D to the 1000 times P to the -1.
And this will be P times
two to 1000 0, 0 1, P inverse.
So with this, you can easily compute A to 
1000 times whatever vector
that you have.
But the question is, how did I come up
with P and D?
What I've done here is just I gave you
these two matrices.
I haven't shown you how to obtain P and D.
Going to answering this question
will take us to the topic of eigenvalues
and eigenvectors.
