Solution: Note that both the matrices are upper-triangular matrices.
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Since the matrices $A$ and $B$ are upper-triangular matrices, the eigenvalues are the diagonal elements. Thus, the characteristic polynomial of $A$ and $B$ will be
\[
p_A(t) = (t-2)^2 (t-1) \quad \text{and} \quad p_B(t) = (t-1)(t-2)^2.
\]
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To find the minimal polynomial of \( A \), we need to find the polynomial of the least degree whose roots are all the eigenvalues and also the matrix $A$ satisfies the polynomial. The possible minimal polynomials are
\[
(t-2)(t-1) \quad \text{ or } \quad (t-2)^2(t-1).
\]
We will check if the matrix $A$ satisfies the polynomial $(t-2)(t-1)$.
\begin{align*}
(A - 2I)(A- I) & =
\begin{bmatrix}
0 & 0 & 5 \\
0 & 0 & -4 \\
0 & 0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 5 \\
0 & 1 & -4 \\
0 & 0 & 0 \\
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}.
\end{align*}
Thus the minimal polynomial for the matrix $A$ will be
\[
m_A(t) = (t-2)(t-1).
\]
Similarly, for the matrix $B$, the possibilities are
\[
(t-1)(t-2) \quad \text{ or } \quad (t-1)(t-2)^2.
\]
We will check if the matrix $B$ satisfies the polynomial $(t-1)(t-2)$.
\begin{align*}
(B - I)(B- 2I) & =
\begin{bmatrix}
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
-1 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix} \neq
\begin{bmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}.
\end{align*}
Thus, the minimal polynomial will be
\[
(t-1)(t-2)^2.
\]
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We know that similar matrices have the same minimal polynomials (see problem for 09-02-2025). Since the minimal polynomials of \( A \) and \( B \) are different, \( A \) and \( B \) are not similar.