 Solution

Subtract the identity matrix and putDirect matrix multiplication givesFor 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 $N\ne 0$ but $N^2=0$. If both are nonzero, then $N^2\ne 0$ while $N^3=0$. Consequently the complete answer isNo lower power can annihilate the matrix in any of these cases, as the displayed nonvanishing matrix powers show.