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Find its characterstics equation by find trace and minors and determinant 

The given matrix is:

\[
A = \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}
\]

### Step 1: Trace of the matrix \(A\)

The trace of a matrix is the sum of its diagonal elements. For matrix \(A\), the diagonal elements are \(2, 2, 2\). Therefore, the trace is:

\[
\text{Trace}(A) = 2 + 2 + 2 = 6
\]

### Step 2: Minors of the matrix

The minors are the determinant of submatrices that result when one row and one column are removed.

- **Minor of the first element \(A_{11} = 2\):**
  Removing the first row and first column, the submatrix is:

  \[
  \begin{pmatrix}
  2 & 1 \\
  0 & 2
  \end{pmatrix}
  \]

  The determinant of this \(2 \times 2\) matrix is:

  \[
  (2)(2) - (1)(0) = 4
  \]

- **Minor of the second element \(A_{22} = 2\):**
  Removing the second row and second column, the submatrix is:

  \[
  \begin{pmatrix}
  2 & 0 \\
  0 & 2
  \end{pmatrix}
  \]

  The determinant of this \(2 \times 2\) matrix is:

  \[
  (2)(2) - (0)(0) = 4
  \]

- **Minor of the third element \(A_{33} = 2\):**
  Removing the third row and third column, the submatrix is:

  \[
  \begin{pmatrix}
  2 & 1 \\
  0 & 2
  \end{pmatrix}
  \]

  The determinant is the same as for \(A_{11}\):

  \[
  4
  \]

### Step 3: Determinant of the matrix \(A\)

We can find the determinant of the entire \(3 \times 3\) matrix by expanding along the first row. This gives:

\[
\text{Det}(A) = (2) \cdot \text{det}\left(\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\right) - (1) \cdot \text{det}\left(\begin{pmatrix} 0 & 1 \\ 0 & 2 \end{pmatrix}\right) + (0) \cdot \text{det}\left(\begin{pmatrix} 0 & 1 \\ 0 & 2 \end{pmatrix}\right)
\]

We already calculated the determinant of the submatrix as \(4\). So,

\[
\text{Det}(A) = (2)(4) - (1)(0) + 0 = 8
\]

### Step 4: Characteristic equation

The characteristic equation is given by:

\[
\det(A - \lambda I) = 0
\]

Where \(I\) is the identity matrix and \(\lambda\) is the eigenvalue. The matrix \(A - \lambda I\) is:

\[
A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 & 0 \\ 0 & 2 - \lambda & 1 \\ 0 & 0 & 2 - \lambda \end{pmatrix}
\]

Now, calculate the determinant of this matrix:

\[
\det(A - \lambda I) = (2 - \lambda) \left[ (2 - \lambda)(2 - \lambda) - (1)(0) \right] - (1)(0) + (0)
\]

\[
= (2 - \lambda)((2 - \lambda)^2) = (2 - \lambda)^3
\]

Thus, the characteristic equation is:

\[
(2 - \lambda)^3 = 0
\]

### Final Answer

The characteristic equation of the matrix is:

\[
(2 - \lambda)^3 = 0
\]

Or,

\[
\lambda = 2 \quad \text{(with multiplicity 3)}
\]

Would you like more details on any part of the solution?

Here are 5 related questions:
1. How do you find eigenvalues of a matrix?
2. What is the geometric interpretation of the characteristic equation?
3. Can a matrix have repeated eigenvalues?
4. What are the properties of a matrix with a repeated eigenvalue?
5. How can the characteristic polynomial help in solving systems of equations?

**Tip:** The trace of a matrix is always equal to the sum of its eigenvalues.

Find its characteristic equation by finding trace, minors, and determinant.

Discover how to find the characteristic equation of a 3x3 matrix through step-by-step calculations involving trace, minors, and determinant expansion. This problem highlights matrix algebra concepts, making it ideal for Grades 11-12 or early college students.

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