We were given matrix A is diagonalizable
and we were asked to find matrix A to the power of 33.
We're told to enter the elements as expressions
and not use scientific notation.
We'll begin by diagonalizing matrix A
by writing matrix A in the form of matrix P times matrix D
times the inverse of matrix P.
Once we have matrix A in this form,
we can determine matrix A to the power of 33
by determining matrix P times matrix D
raised the power of 33 times the inverse of matrix P.
To start to diagonalize matrix A,
we need to find the eigenvalues of matrix A.
We do this by solving the equation,
the determinant of the difference of matrix A
and the product of lambda
and the identity matrix equals zero,
which I've already set up below.
Notice the result of matrix A minus lambda times
the two by two identity matrix
has entries four minus lambda,
one, negative two, one minus lambda.
Next, we find the determinant,
set it equal to zero and solve.
The determinant is equal to the product of
four minus lambda and one minus lambda
minus the product of one and negative two.
Multiplying and simplifying,
we have the characteristic equation
lambda squared minus five lambda plus six equals zero,
which is factorable giving us lambda sub one equals two,
lambda sub two equals three.
We use these eigenvalues to form the diagonal matrix D
where we place the eigenvalues along the main diagonal.
and the other entries are zero.
And now we find corresponding eigenvectors
that will form the columns in matrix P.
Let's begin with lambda sub one equals two
to find an eigenvector.
We solve the equation
the difference of matrix A and the product of lambda
in the identity matrix times vector X
equals the zero vector,
which I've already set up below.
Next, let's write the corresponding matrix equation,
where we have the coefficient matrix times the vector X
is the vector X1X2 equals zero vector,
which is the vector zero, zero.
Let's check this coefficient matrix.
The first entry in row one is four minus two times one,
which is two.
The second entry remains one.
In the second row, the first entry remains negative two.
And the second entry in row two is one minus two times one,
which is negative one.
Next, we write the augmented matrix
and write the augmented matrix and reduce to echelon form.
Notice how there is no pivot in column two,
which indicates X2 is a free variable.
The first row indicates X1 plus 1/2 X2 equals zero.
Solving for X1
we have X1 equals negative 1/2 X2, X2 equals X2.
To parametrize these solutions let's let X2 equal T.
All the eigenvectors are in the form of the vector
negative 1/2 T, T,
or we could say T times the vector negative 1/2 one,
which indicates the eigenvectors
are all the scalar multiples of the vector
negative 1/2 one, except T can't equal zero,
because the zero vector can't be an eigenvector.
So we can use any scalar multiple as a vector
corresponding to lambda equals two.
To avoid fractions, let's let T equal two
and use the vector negative one, two
for the eigenvector corresponding to lambda equals two.
Which means we can let the first column of matrix P
be negative one, two.
Because we can use any scalar multiple of the vector
negative 1/2, one
as the vector corresponding to lambda equals two,
matrix P is not unique.
And now let's find an eigenvector corresponding to
lambda sub two equals three.
We go through the same process.
Set up an augmented matrix.
Write the augmented matrix in reduced row echelon form,
and we get X1 equals negative X2.
Again, X2 is a free variable.
X2 equals X2.
Parameterizing the solution,
let's let X2 equal S.
All the eigenvectors are in the form of the vector
negative S, S or S times the vector negative one, one
indicating the eigenvectors corresponding to
lambda sub two equals three
are all the scalar multiples of the vector negative one, one
except the zero vector.
Let's just go ahead and let S equal one
and use the eigenvector negative one, one
for the eigenvector corresponding to
lambda sub two equals three,
which means the second column of matrix P
is negative one, one.
So last step to finish the digitalization
is to find the inverse of matrix P.
Using the formula for the inverse of
a two by two invertible matrix,
we have P inverse equals one, one,
negative two, negative one.
You may want to pause the video if you need to review this.
So now that we have P inverse,
we know the diagonalization of matrix A
where matrix A is equal to matrix P times matrix D
times the inverse of matrix P.
From here A to the power of 33 is equal to matrix P
times matrix D raised to the 33rd power
times the inverse of matrix P.
Because matrix D is a diagonal matrix,
to determine D to the power of 33,
we raise the entries along the main diagonal
to the 33rd power.
And now to find A to the power of 33,
we need to find the product of these three matrices.
And again, we don't want to convert to scientific notation.
We're told to leave the elements or entries as expressions.
Let's multiply this out on the next slide.
So multiplying the first two matrices,
in row one column one of the product
we have negative one times two to the 33rd
plus negative one times zero.
In row one column two,
we have negative one times zero
plus negative one times three to the 33rd.
In row two column one we had
two times two the power of 33 plus one times zero.
In row two column two
we have two times zero plus one times three to the 33rd.
And now let's simplify these expressions
as much as possible.
Of course we can drop the products that are zero.
So in row one column one,
we have negative one times two the power of 33.
In row one column two,
we have negative one times three to the 33rd.
In row two column one,
we have two times to the 33rd,
which is two to the 34th.
In row two, column two
we just have three to the power of 33.
And now we need to find the last product.
In row one column one, we have negative one
times two to the 33rd times one
plus negative one times three to 33rd times negative two.
In row one column two we have
negative one times two to the 33rd times one
plus negative one times three to 33rd
times negative one.
In row two column one,
we have to the 34th times one
plus three to the 33rd times negative two.
And in row two, column two,
we have two to the 34th times one
plus three to the 33rd times negative one.
And again let's simplify it as much as we can.
In row one column one,
we have negative one times to the 33rd
plus two times three to the 33rd.
In row one, column two
we have negative one times two to the 33rd
plus three to the 33rd.
In row two column one,
we have two to the 34th
plus negative two times three to the 33rd.
And finally in row two column two, we have two to the 34th
plus negative one times three to the 33rd.
And now we have found matrix A raised to the power of 33
using the digitalization of matrix A.
I hope you found this helpful.