= Solution
{c}
Subtract the <identity matrix> and put
$$
N=A_{\lambda,\mu}-I=
\begin{pmatrix}
\lambda&\lambda&0\\
\mu-\lambda&\mu-\lambda&\mu\\
-\mu&-\mu&-\mu
\end{pmatrix}.
$$
Direct <matrix multiplication> gives
$$
N^2=\lambda\mu
\begin{pmatrix}1&1&1\\-1&-1&-1\\0&0&0\end{pmatrix},\qquad N^3=0.
$$
For the last equality, each column of $N^2$ is a multiple of $(1,-1,0)^T$, and $N(1,-1,0)^T=0$. Thus $N$ is a <nilpotent matrix>. The <minimal polynomial> of $A_{\lambda,\mu}$ divides $(t-1)^3$ and equals $(t-1)^s$, where $s$ is the <nilpotency index> of $N$.
The first row of $N$ forces $\lambda=0$ if $N=0$, and its third row then forces $\mu=0$. Conversely $\lambda=\mu=0$ gives $N=0$. If exactly one parameter is nonzero, then $N\ne0$ but $N^2=0$. If both are nonzero, then $N^2\ne0$ while $N^3=0$. Consequently the complete answer is
$$
\boxed{
m_{A_{\lambda,\mu}}(t)=
\begin{cases}
t-1,&\lambda=\mu=0,\\
(t-1)^2,&\lambda\mu=0\text{ and }(\lambda,\mu)\ne(0,0),\\
(t-1)^3,&\lambda\mu\ne0.
\end{cases}}
$$
No lower power can annihilate the <matrix> in any of these cases, as the displayed nonvanishing <matrix powers> show.
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