The product of all eigenvalues of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ is

The product of all eigenvalues of a square matrix $A$ equals the determinant of $A$, i.e., $\prod_{i} \lambda_i = \det(A)$.

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$.

Applying row operations:

• $R_2 \leftarrow R_2 - R_1 = \begin{bmatrix} 3 & 3 & 3 \end{bmatrix}$

• $R_3 \leftarrow R_3 - R_2 = \begin{bmatrix} 3 & 3 & 3 \end{bmatrix}$

Since the second and third rows of the transformed matrix are identical, the rows of matrix $A$ are linearly dependent. Therefore:

Thus, the product of all eigenvalues of the matrix is 0.