We're given matrix A,
which is a four by four matrix
and we're asked to find
the eigenvalues of A.
The eigenvalues of A
are the values of lambda
that satisfy the equation the determinant
of the different of matrix A
and lambda times i equals zero,
which can also be expressed
as the determinant
of the difference of lambda
i in matrix A equals zero
as shown here.
Let's go ahead and use
this form of the equation.
I've already set some of this up.
We have the determinant of this difference
where this matrix here is lambda times i,
where matrix i is the four
by four identity matrix
and then here's matrix
A, the given matrix,
so we want to find the
determinant of these two matrices,
which gives us this determinant here.
So we want to find the values of lambda
that make this determinant equal to zero.
To evaluate the determinant,
we'll use the co-factor method.
It will be best to select
the row with the most zeros,
but in this case, because all
the rows contain two zeros,
let's go ahead and use row one.
Starting with row one, the first element
is lambda plus three,
so we have the quantity
of lambda plus three times negative one,
raised to the power of the
sum of the row and column
of this element,
which is in row one, column one.
One plus one is two.
Times the determinant
form my eliminating
row one and column one,
which will give us a three
by three determinant,
with the first row would
be lambda minus four, zero, zero.
Second row is zero lambda plus five, four,
the third row is zero,
two, lambda minus two.
Then we have plus the
next element in row one,
which is negative two,
so plus negative two times negative one,
raised to the power of three,
because this element
is row one, column two,
one plus two is three
times the determinant
formed by eliminating
the row and column of negative two,
so we eliminate row one column two.
Remaining elements make
up the determinant,
so we have three, zero, zero,
zero, lambda plus five, four,
and zero, two lambda minus two.
And we can actually stop here
because notice how the remaining elements
in row one are zero,
so the rest of the
expansion would be zero.
So we want to know when
all this is equal to zero.
So now we need to evaluate the
three by three determinants,
and again we'll use the co-factor method.
So negative one to the
second is positive one,
so we have the quantity
lambda plus three times,
let's go and expand using row one.
So we have the quantity
lambda minus four times
negative one to the second
because this element is
in row one, column one,
times the two by two determinant,
which is called the minor,
formed by eliminating row one column one,
so we have lambda plus five, four, two,
and lambda minus two.
And again, notice how
the remaining elements
in row one are zero,
so we can go ahead and stop here,
the rest of the expansion
would be equal to zero.
Here we have plus negative
two, times negative one,
that's positive two,
so we have plus two times,
let's use the co-factor
method using row one again,
so we'd have three, times
negative one to the second
because this three is
in row one, column one,
and one plus one is two,
times the two by two determinant,
formed by eliminating row one column one.
So we have lambda plus five,
four, two, lambda minus two.
And once again the remaining
elements in row one are zero,
so we can go ahead and stop the expansion.
So all this on the left must equal zero.
So for the next step,
we'll evaluate the two by two determinants
by finding this product
minus this product.
Let's do this on the next slide.
So we have the quantity
of lambda plus three,
let's go ahead and use a bracket here,
and we have negative one to the second,
that's positive one, so
we have lambda minus four,
times the value of the determinant
which is the quantity lambda plus five,
times the quantity lambda minus two,
then minus four times
two which is minus eight,
so close parentheses here then a bracket,
then plus two,
let's use a bracket here,
then we have positive
three times the value
of the determinant, which
again is the quantity
lambda plus five, times
the quantity lambda minus
two, minus four times two,
minus eight, and then
all this equals zero.
For our next step, let's go ahead
and simplify here and here.
We have the quantity of
lambda plus three, times
we have lambda minus four, times,
here we're going to have lambda squared,
and then minus two
lambda plus five lambda,
that's plus three lambda,
and then we have negative 10, minus eight
which is minus 18,
plus two
times the quantity three.
Now this is the same
expression as this expression.
So this is going to simplify
to the quantity lambda squared
plus three lambda, minus 18.
Now we're actually going to factor this.
Notice how this product
and this product contain
a common factor of lambda
squared plus three lambda,
minus 18.
So we're going to go
ahead and factor that out,
so we'd have lambda squared
plus three lambda, minus 18,
now let's be careful with what's left.
If we factor this factor
out of this product,
we would have the
quantity lambda plus three
times the quantity lambda minus four,
and then we'd have plus,
if we factor this trinomial
out of this product,
notice how we'll have
two times three left.
So now for our next step,
let's go ahead and simplify
inside these parentheses here,
so we have lambda squared,
plus three lambda,
minus 18, times, here
we'll have lambda squared,
and then minus four
lambda plus three lambda,
that's minus one lambda or minus lambda,
and then three times
negative four is negative 12,
but then we have plus six,
so minus six, equals zero.
Both of these trinomials will factor,
into two binomial factors.
The factors of lambda squared
are lambda and lambda,
the factors of negative
18 add a positive three,
would be positive six and negative three.
And now for this trinomial,
the factors of lambda squared
are lambda, and lambda,
the factors of negative six
that add to negative one,
are negative three and positive two.
So this product equals zero
when lambda equals negative six,
when lambda equals positive three,
we should get another lambda
equals positive three,
and here we get lambda
equals negative two.
So we actually have three eigenvalues
even though the eigenvalue
of positive three
has multiplicity two.
Let's get these in order
from least to greatest.
So we'll say lambda sub
one equals negative six,
lambda sub two equals negative two,
and lambda sub three
equals positive three.
Let's go ahead and make a note here,
this eigenvalue has multiplicity...
two.
I hope you found this helpful.
