We're given matrix A,
which is a three by three
matrix and asked to find
the eigenvalues of A, which
means you want to find
the values of lambda that
satisfy the equation,
the determinant of A minus
lambda I equals zero,
or the equivalent equation,
the determinant of lambda
I minus A equals zero.
So for a quick review,
if A is an N by N matrix,
suppose vector X is a
nonzero vector in RN,
and lambda is a number or scalar such that
AX equals lambda X.
This means if AX is a scalar
multiple of the vector X,
where the vector X is
called the eigenvector of A,
and lambda is called the eigenvalue of A.
Let's go ahead and set this
up using this equation here.
So we'll have the determinant
of vector A which is given,
minus lambda times I.
Lambda times the three
by three identity matrix
would give us this matrix here,
and this determinant must equal zero.
Let's go ahead and find the difference
of these two matrices,
and write the determinant
using vertical bars.
So the first row would
be two minus lambda,
negative five minus zero
which is negative five,
five minus zero is five.
Second row we have zero,
three minus lambda,
and negative one minus
zero, which is negative one.
Third row we have zero minus zero,
negative one minus zero,
and three minus lambda.
Now let's use expansion by
minors, or the cofactor method,
to evaluate this three
by three determinant.
So if we use the first row,
we start with the first
element in row one,
which is two minus lambda.
Now we're going to multiply this by
the two by two determinant,
formed by eliminating
the row and column of two minus lambda.
So we eliminate row one, column one,
so the first row is three
minus lambda and negative one.
Second row is negative one
and three minus lambda.
And then, we'll have, minus,
the next element in row
two is negative five.
To form the next two by two
determinant, called the minor,
we eliminate the row and
column of negative five,
so eliminate row one, column two.
So the elements in the
two by two determinant
are zero, negative one,
zero, three minus lambda,
and we have plus.
The last element in row
one is positive five,
times the two by two determinant
formed by eliminating the
row and column of five.
So you eliminate row one, column three,
so we have zero, three minus lambda,
zero, negative one,
and this must equal zero.
Let's continue on the next slide.
Now the value of each
two by two determinant
is equal to this product
minus this product.
So here we have the
quantity two minus lambda,
times, here we have the
quantity three minus lambda
times the quantity three minus lambda,
minus negative one times negative one,
and this becomes plus five, times,
here we have zero minus zero,
and we have plus five
times zero minus zero,
equals zero.
So here we have the
quantity two minus lambda
times, let's go ahead and
find this product here.
We have nine, minus three
lambda, minus three lambda is
minus six lambda, plus lambda squared.
This ends up being minus,
simplifies to minus one.
Of course this is zero and this zero.
So this product here is equal to zero.
Let's go ahead and combine
the like terms here.
We have two minus lambda,
times the quantity,
you'll have lambda squared
minus six lambda, plus eight
equals zero.
Looks like this'll factor again.
We have the quantity two minus lambda,
and we'll have two more binomial factors.
The factors of lambda squared
are lambda and lambda.
The factors of positive eight
that add to negative six
are negative two and negative four.
So this product equals zero,
and lambda equals two,
or lambda equals two,
or lambda equals four.
In our homework system,
we're asked to answer
lambda sub one, lambda sub
two and lambda sub three,
and we're asked to enter
the greatest values
of lambda first, so we'll
enter four, two and two.
I hope you found this helpful.
