Question is to Diagonalize a 3x3 matrix:
$$ \begin{bmatrix} -4 & -4 & -8 \\ 4 & 6 & 4 \\ 6 & 4 & 10 \\ \end{bmatrix} $$
After personally computing I get the following eigenvalues:
λ= 12 λ = 2 λ = -2
However these do not appear to be the accepted answers according to the solutions. I am fairly new to linear algebra so highly plausible I am doing something wrong. Recently learnt about rational root theorem. Calculations are as follows:
$$ (-4 -6λ) \begin{bmatrix} 6-λ & 4 \\ 4 & 10-λ \\ \end{bmatrix} +4 \begin{bmatrix} 4 & 4 \\ 6 & 10-λ \\ \end{bmatrix} -8 \begin{bmatrix} 4 & 6-λ \\ 6 & 4 \\ \end{bmatrix} $$
= (-4-λ)(λ^2-16λ+44)+4(-4λ+16)-8(6λ-20) = -λ^3+12λ^2-44λ+48 = (-(λ-12)(λ+2))(λ-2)
λ=12 λ=-2 λ=2
However non of these match the respective values on the diagonal for D according to the prescribed solutions.
I will obviously continue with the rest of the calculations once I can find the correct Eigenvalues.
That $-4-6\lambda$ that you have there is a mistake. It should be simply $4-\lambda$. In the end, you will get the polynomial $-\lambda^3+12\lambda^2-44\lambda+48$, whose roots are $2$, $4$, and $6$. So, these are the eigenvalues of your matrix.