For the given matrix, find the values of $b$ for which the matrix might not be diagonalizable. Out of those values, find the ones which produce a diagonalizable matrix and diagonalize the matrix.
$$\begin{pmatrix} 5 & 1 & 2 \\ 0 & b & 4\\ 0 & 1 & b \end{pmatrix}$$
Note: I've translated this problem and to clarify: the problem asks to find for which values of $b$ does the matrix have repeated eigenvalues but independent eigenvectors.
My take is this: Find the determinant of the matrix:
$$\begin{vmatrix} 5 - λ & 1 & 2 \\ 0 & b - λ & 4\\ 0 & 1 & b - λ \end{vmatrix}$$
Solve the characteristic polynomial equation: $(5-λ)(b-λ-2)(b-λ+2) = 0$.
Well this is where I kinda get lost. Well one eigenvalue is $λ = 5$, but how am I supposed to find the value of $b$ ? plugging the $λ = 5$ we get: $$ 0 \cdot(b - 7)(b-3) = 0$$ So can $b$ be any value? Since it's already being multiplied by $0$? Or does $b$ have to equal $7$ and $3$? Why?
The way I understand this is that we're looking for repeated values of lambda (Since $3$ different eigenvalues will always produce a diagonalizable matrix). One possibility is $λ = 5$, in which case $b$ can't produce any new values of $λ$ since we got the value of $b$ from $λ$ in the first place. So we have only one eigenvalue $λ = 5$, but I'm unsure about the value of $b$ in this case.
And the other possibility is when $λ \neq 5$, in which case $(λ-5)(b-λ-2)(b-λ+2) = 0$ can produce two different eigenvalues for some value of $b$? But how do I even find the value of $b$? I'm very lost and I understand that I might not be making any sense.
I'm not sure i'm getting this right. Please help me understand this problem.