## Powers of a Matrix

in this video we're gonna discuss powers

of a matrix

the problem is we're given this

two-by-two matrix a and we'd like to

find the matrix a raised to the power

2015 let's start with a squared this

means multiply a times a using usual

matrix multiplication so in other words

we'll multiply a by itself and thinking

back to matrix multiplication normally

we circle the rows of the first matrix

columns of the second Matrix and we're

forming a bunch of dot products so the

product is gonna look like negative 1

negative 1 1 0 alright now that we've

got that let's figure out a cubed so a

cubed would be a squared times a we

figured out a squared in the previous

step so I'll copy that down and if we

multiply that by a doing our usual

matrix multiplication process now we get

negative 1 0 0 negative 1 so let's try

if a to the 4th will be a cubed times a

at some point you can stop circling the

rows and columns once this this matrix

multiplication starts to seem a little

bit more automatic so this product here

is going to give us 0 1 negative 1

negative 1 now to get 8 v will take a to

the 4th times a this time let's try it

without circling the rows and columns

we'll have 1 1 negative 1 and 0 and what

we're altom utley looking for is a

matrix or a certain power of a that

equals the identity matrix so we may

need to do a few more calculations

before we end up with a power that's

equal to the identity so let's try a to

the 6 into the 6 is in the 5th times 8

so we'll recopy in the 5th take our

original matrix a and we'll do this

multiplication we're gonna have 1 0 0

and 1

aha so finally we have a power which is

equal to the identity matrix I so what

we should do now is recap a few

properties of this identity matrix and

we'll see how those properties will help

us to solve the problem so we're looking

at this identity matrix here which has

ones along the main diagonal zeros on

the off diagonal doesn't necessarily

have to be a two-by-two matrix could be

a 3 by 3 4 by 4 matrix etc so let's look

at some properties that that this

identity matrix has first property

I to the N is equal to I for any

positive integer power it's sort of

obvious when the power is one so let's

say for n equal to 3 etc so again this

would be true for the 2 by 2 identity

matrix but also for 3 by 3 4 by 4 etc so

just to convince ourselves of this let's

say for example if we take the 2 by 2

identity matrix and multiply it by

itself we can check that yes we do get

back the original identity matrix all

right second property that I has if we

take our identity matrix times any

matrix B provided that the sizes are

compatible here we're gonna end up with

that same matrix B so let's say for any

for any choice of B and let's just do a

little example to convince ourselves of

this so we take the 2 by 2 identity

matrix let's say B is the matrix that

looks like e F G H for example if we do

this multiplication we end up with E F G

H so in other words we end up with this

original matrix B that we put in here

all right so let's see how these

properties here tie in to the problem

that we're trying to solve so here's the

idea we've discovered that a to the 6 is

the same thing as the identity matrix

and we know that the identity matrix has

some nice properties so the trick will

be to take a to the power 2015 and to

write it as a to the 6 times a to the 6

and we'll want to multiply by a to the 6

a whole bunch of times

feel a little bit better about this we

didn't have these dots here just doing

our usual matrix multiplication so a to

the six times a to the six whole bunch

of times and then at the end here there

will be some some product or some last

factor so I've just recopy to our basic

idea from the previous screen and let's

see how it can write out this product so

first of all I'm going to think about

what happens if I divide 2015 by six

using a calculator you're going to get

something like three thirty five point

eight three repeating might be a little

bit nicer to write 2015 over six is

equal to three thirty-five plus 0.8

three repeating and if we then go

through and multiply both sides by six

then we've got twenty fifteen is equal

to six times three 35 plus will have

this remainder here if we go six times

this expression here we're getting five

so let's take this expression that we've

just figured out and let's say that a to

the 2015 can be re-written as a to the

six times three thirty-five plus five

and now we're going to think about some

exponent rules so first of all when you

have a sum in the exponent that's the

same thing as having a product so we

could rewrite this as a to the six times

three thirty five times a to the fifth

okay now looking at the first factor

when we've got a product in the exponent

we can rewrite that as a to the six

raised to the power three thirty-five so

whenever we've got a power raised to

another power those powers multiply

that's our second exponent rule here and

we'd still need to multiply by a to the

fifth now the reason why we're doing

this is remember we said that a to the

six is equal to the identity matrix so

we can now say this is I raised to the

power three thirty five times a to the

fifth using our first property about the

identity matrix when you raise it to any

power you're getting back the identity

matrix so this is really I times in the

fifth I'm thinking back to the second

property second property of the identity

matrix from our previous screen when you

take I times any matrix you just get

back that original matrix so this gives

us a to the fifth technically if we're

being very careful about the notation

here I want to be careful about this dot

certainly not a dot product so it might

be more correct if we actually put

brackets on here the the notation then

gets a little bit more cluttered but but

we're being more rigor

with the notation this way alright so

what we've determined finally after all

this calculation is that a to the power

2015 is equal to a to the fifth you

could leave your answer like this or

more specifically you could go back to

the previous screen and and just copy

out explicitly what a to the fifth was

it was the matrix 1 1 negative 1 0