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.