Question

For which value of k does the matrix

8 k
-9 -3

have one real engenvalue of multiplicity 2? Explain.

Answer 1

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)

Answer 2

The quadratic formula should read:
x=(<b>-b</b>±√(b²-4ac))/2a