We're asked to find the greatest value
of lambda that makes matrix A singular.
A square matrix that is not invertible,
meaning it doesn't have an inverse,
is a singular matrix.
And a square matrix is singular
if the determinant of
the matrix equals zero.
So to answer this question,
we'll find the determinant
of this two by two matrix,
set it equal to zero,
and solve for lambda.
So we'll use vertical bars
to represent the two by two determinant.
And again, this must equal zero
if the matrix is singular.
And now the value of the
two by two determinant
is equal to this product
minus this product.
So we'll have the quantity
negative nine minus lambda
times the quantity
negative two minus lambda
minus two times negative six.
And this must equal zero.
We'll go ahead and multiply this out.
We'll have four products.
One, two, three, four.
Negative nine times
negative two is positive 18.
Negative nine times negative
lambda is plus nine lambda.
Negative lambda times negative
two is plus two lambda.
And then we have negative
times negative lambda.
That's plus lambda squared.
And then we have plus 12 equals zero.
Combining like terms,
we have lambda squared
plus 11 lambda plus 30 equals zero.
So we have a quadratic equation,
which is factorable.
The left side factors
into two binomial factors.
The factors of lambda squared
are lambda and lambda.
The factors of 30 that add to 11
are positive six and positive five.
So this product equals zero
when lambda is equal to negative six
or when lambda is equal to negative five.
So both of these values
would make the determinant
equal to zero, but this question only asks
for the greatest value of lambda,
and because negative is to
the right of negative six,
negative five is the
greatest value of lambda
that makes the given matrix singular.
So our answer is lambda
equals negative five.
This is the type of equation
that we'll end up solving
when we're trying to find the eigenvalues
of a two by two matrix.
I hope you found this helpful.
