The answer is "yes".

A matrix $A$ is invertible if and only if none of its eigenvalues are zero. This is because the determinant of a matrix is the product of its eigenvalues, and a matrix is invertible if its determinant is non-zero.

In this case, the eigenvalues of $A$ are 3, 2, and 2. Since none of these eigenvalues is zero, the determinant of $A$ is non-zero. The determinant can be calculated as:

$\text{det}(A) = 3 \times 2 \times 2 = 12$

Since the determinant is non-zero, the matrix $A$ is invertible.

To summarize:

• Eigenvalues of $A$ are 3, 2, and 2.
• The determinant of $A$ is non-zero.
• Therefore, $A$ is invertible.

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Do you want any more details, or have further questions?

Here are five related questions:

1. How is the determinant of a matrix related to its eigenvalues?
2. What does it mean geometrically if a matrix has an eigenvalue of 0?
3. How can you determine if a matrix is singular based on its eigenvalues?
4. How do the eigenvalues affect the condition number of a matrix?
5. How can you find the eigenvalues of a matrix from its characteristic equation?

Tip: The product of eigenvalues equals the determinant of a matrix, and their sum equals the trace of the matrix.