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