Discussion
The product of all eigenvalues of the matrix \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) is
Answer: B
The product of the eigenvalues is 0.
The product of the eigenvalues of a square matrix equals its determinant. Here the rows are in arithmetic progression: row 2 − row 1 = (3, 3, 3) = row 3 − row 2, so row 1 − 2(row 2) + row 3 = 0 and the rows are linearly dependent:
\[ \begin{aligned} \det &= 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \\ &= -3 + 12 - 9 = 0 \end{aligned} \]
So one eigenvalue is 0 and the product is 0. The values −1, 1 and 2 would require a non-singular matrix.
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