If 0 is an eigenvalue, this means that there is an eigenvector v = [x,y] so that
Av = 0
Let’s multiply A by the column vector v.
Av =
[-9x-6y]
[-2x+ky]
Since we need Av = 0, we need
-9x-6y = 0
and
-2x+ky = 0
The first equation tells us that x = -2y/3. Substituting -2y/3 for x in the second gives us
-2x+ky = -2(-2y/3)+ky = 4y/3 + ky = 0
So k = -4/3.
A second method is to find the characteristic equation for the matrix A.
The characteristic equation is det(A – Iλ) = 0
A – Iλ is the matrix
[-9-λ -6]
[ -2 k-λ]
det(A-Iλ) = (-9-λ)(k-λ) – (-2)(-6) = -9k + 9λ – kλ + λ^2 – 12
So we set this equal to 0, plug in 0 for λ (since we want λ to be an eigenvalue) and solve for k.
-9k + 9*0 – k*0 + 0^2 – 12 = 0
-9k – 12 = 0
-9k = 12
k = -12/9 = -4/3