https://file.aiquickdraw.com/mathbot/file/1729701991799ij3qngrq.jpg

The matrix displayed in the quiz is:

\[
A = \begin{pmatrix} 
1 & 2 & 3 \\
0 & 4 & 5 \\
0 & 0 & 6 
\end{pmatrix}
\]

To find the eigenvalues of this matrix, we look at the diagonal elements because the matrix is upper triangular (the elements below the diagonal are all zero). For an upper triangular matrix, the eigenvalues are simply the entries on the diagonal.

Therefore, the eigenvalues of matrix \( A \) are:
\[
\lambda_1 = 1, \quad \lambda_2 = 4, \quad \lambda_3 = 6
\]

Thus, the eigenvalues of \( A \) are \( [1, 4, 6] \).

### Is the matrix diagonalizable?

Yes, the matrix is diagonalizable. Since the eigenvalues are distinct (no repeated eigenvalues), the matrix has a full set of linearly independent eigenvectors, which guarantees that the matrix is diagonalizable.

Let me know if you would like further details or clarifications.

Here are five related questions:
1. What are the conditions for a matrix to be diagonalizable?
2. How can you compute the eigenvectors of matrix \( A \)?
3. What is the significance of eigenvalues and eigenvectors in linear algebra?
4. How do upper triangular matrices simplify the computation of eigenvalues?
5. What does the determinant of matrix \( A \) tell us about its properties?

**Tip:** Always check if the matrix is triangular (upper or lower), as this allows you to directly identify the eigenvalues from the diagonal elements.

Find the eigenvalues of the matrix A = [1 2 3; 0 4 5; 0 0 6] and determine if it is diagonalizable.

Learn how to find the eigenvalues of a 3x3 upper triangular matrix and determine if it is diagonalizable. The problem involves linear algebra concepts such as eigenvalues, eigenvectors, and matrix diagonalization. Suitable for university-level students.

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