https://file.aiquickdraw.com/mathbot/file/1727105916461nebir1lh.jpg

To determine the minimal polynomial of the matrix given in the image, we need to perform the following steps:

The matrix is:
\[
A = \begin{pmatrix} 
0 & 0 & 3 \\ 
1 & 0 & 2 \\ 
0 & 1 & 1 
\end{pmatrix}
\]

### Step 1: Find the characteristic polynomial
The characteristic polynomial of a matrix \(A\) is given by:
\[
p_A(\lambda) = \det(A - \lambda I)
\]
where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix of the same size as \(A\).

First, compute \(A - \lambda I\):
\[
A - \lambda I = 
\begin{pmatrix} 
0 & 0 & 3 \\ 
1 & 0 & 2 \\ 
0 & 1 & 1 
\end{pmatrix}
-
\begin{pmatrix} 
\lambda & 0 & 0 \\ 
0 & \lambda & 0 \\ 
0 & 0 & \lambda 
\end{pmatrix}
=
\begin{pmatrix} 
-\lambda & 0 & 3 \\ 
1 & -\lambda & 2 \\ 
0 & 1 & 1 - \lambda 
\end{pmatrix}
\]

Now calculate the determinant of this matrix:
\[
\det(A - \lambda I) = \det\begin{pmatrix} 
-\lambda & 0 & 3 \\ 
1 & -\lambda & 2 \\ 
0 & 1 & 1 - \lambda 
\end{pmatrix}
\]

Using cofactor expansion along the first row:
\[
\det(A - \lambda I) = (-\lambda) \det\begin{pmatrix} -\lambda & 2 \\ 1 & 1 - \lambda \end{pmatrix}
- 3 \det\begin{pmatrix} 1 & -\lambda \\ 0 & 1 \end{pmatrix}
\]

The first determinant is:
\[
\det\begin{pmatrix} -\lambda & 2 \\ 1 & 1 - \lambda \end{pmatrix} = (-\lambda)(1 - \lambda) - (2)(1) = -\lambda + \lambda^2 - 2 = \lambda^2 - \lambda - 2
\]

The second determinant is:
\[
\det\begin{pmatrix} 1 & -\lambda \\ 0 & 1 \end{pmatrix} = 1
\]

Thus, the characteristic polynomial becomes:
\[
\det(A - \lambda I) = -\lambda (\lambda^2 - \lambda - 2) - 3(1) = -\lambda^3 + \lambda^2 + 2\lambda - 3
\]

So the characteristic polynomial is:
\[
p_A(\lambda) = -\lambda^3 + \lambda^2 + 2\lambda - 3
\]

### Step 2: Find the minimal polynomial
The minimal polynomial must divide the characteristic polynomial, and it is the polynomial of the smallest degree such that \(p(A) = 0\), where \(A\) is substituted into the polynomial.

We can check factors of the characteristic polynomial to find the minimal polynomial. However, for a matrix of size \(3 \times 3\), the minimal polynomial can be of degree 1, 2, or 3. Based on the structure of the matrix, the minimal polynomial is generally of degree 3, and in this case, it will match the characteristic polynomial:
\[
m_A(\lambda) = -\lambda^3 + \lambda^2 + 2\lambda - 3
\]

Therefore, the minimal polynomial associated with the given matrix is:
\[
m_A(\lambda) = \lambda^3 - \lambda^2 - 2\lambda + 3
\]

Would you like further details on this process or have any questions?

Here are 5 related questions:
1. How is the characteristic polynomial related to the eigenvalues of a matrix?
2. What is the difference between the characteristic polynomial and the minimal polynomial?
3. How can we compute eigenvalues from the characteristic polynomial?
4. Why does the minimal polynomial have to divide the characteristic polynomial?
5. How do we verify if a given polynomial is the minimal polynomial?

**Tip:** The minimal polynomial is always unique and gives important information about the diagonalizability of a matrix.

Find the minimal polynomial associated with the matrix \(\begin{pmatrix} 0 & 0 & 3 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{pmatrix}\).

Learn how to compute the minimal polynomial of a 3x3 matrix. This solution covers key linear algebra concepts such as characteristic polynomials and minimal polynomials, complete with step-by-step guidance. Ideal for undergraduate students.

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