For which value of k does the matrix
8 k
-9 -3
have one real engenvalue of multiplicity 2? Explain.
The solution to the problem is based on two key questions:
1. How to find eigenvalues of a given matrix?
2. What are the conditions for a quadratic equation to have real roots of multiplicity two?
We will answer the questions one by one.
1. For a given (square) matrix A, the eigenvalues are the solution to the characteristic equation formed by equating the determinant of A' to zero, namely:
|A'| = 0
where A' is obtained by subtracting the variable λ from each of the diagonal elements.
In the given example,
A=
8 k
-9 -3
so A'=
8-λ k
-9 -3-λ
and |A'|=0 =>
(8-λ)(-3-λ)-(-9)k=0
λ² -5λ -24+9k = 0 ...(1)
The solutions (λ) to the above quadratic equation are the eigenvalues.
2. When does a quadratic equation have real roots of multiplicity two?
Using the quadratic formula:
x=(b²±√(b²-4ac))/2a
we see that when the discriminant is zero, i.e. b²-4ac=0, x has a real root of multiplicity two, i.e. two coincident roots.
Applying this to the equation (1) above, we get:
(-5)²-4(1)(-24+9k)=0
Solve for k.
I get 121/36, and λ=5/2 (multiplicity 2)
The quadratic formula should read:
x=(-b±√(b²-4ac))/2a
To find the value of k for which the matrix has one real eigenvalue of multiplicity 2, we need to determine the eigenvalues of the matrix.
An eigenvalue is a scalar λ for which the following equation holds true:
A * v = λ * v
Where A is the matrix, v is the eigenvector, and λ is the eigenvalue.
Let's find the eigenvalues:
1. Start by setting up the equation:
(A - λ * I) * v = 0
Where I is the identity matrix.
2. Substitute the given matrix and solve for the determinant:
| 8 - λ k |
| -9 -3 - λ | = 0
(8 - λ)(-3 - λ) - (-9)(k) = 0
3. Expand and simplify:
(-24 + 3λ + 8λ - λ^2) + 9k = 0
-λ^2 + 11λ + 9k - 24 = 0
4. To have one real eigenvalue of multiplicity 2, the discriminant (b^2 - 4ac) of this quadratic equation should be equal to zero. Since we're looking for a real eigenvalue, the discriminant should not be negative.
5. Set up the discriminant equation:
b^2 - 4ac = 0
a = -1
b = 11
c = 9k - 24
(11)^2 - 4(-1)(9k - 24) = 0
121 + 36 - 36k + 96 = 0
157 - 36k = 0
6. Solve for k:
36k = 157
k = 157/36
Therefore, the value of k for which the matrix has one real eigenvalue of multiplicity 2 is k = 157/36.
To find the value of k that gives the matrix a real eigenvalue of multiplicity 2, we need to determine the conditions under which the matrix's characteristic polynomial has a repeated root.
The characteristic polynomial of a 2x2 matrix is given by:
P(λ) = (λ - a)(λ - d) - bc,
where a, b, c, and d are the elements of the matrix in the following order:
a b
c d
In this case, the matrix is:
8 k
-9 -3
So we have:
a = 8, b = k, c = -9, d = -3.
Substituting these values into the characteristic polynomial equation, we get:
P(λ) = (λ - 8)(λ - (-3)) - (k)(-9).
Simplifying:
P(λ) = (λ - 8)(λ + 3) + 9k.
To have a repeated root, the discriminant of the characteristic polynomial should be zero:
Discriminant = b^2 - 4ac = 0.
In our case, a = 1, b = -5, and c = 9k + 24. Substituting these values into the discriminant equation, we get:
(-5)^2 - 4(1)(9k + 24) = 0.
25 - 36k - 96 = 0.
-36k - 71 = 0.
-36k = 71.
k = 71/-36.
k = -71/36.
Therefore, the value of k that gives the matrix a real eigenvalue of multiplicity 2 is k = -71/36.