GATE CS 2024 Set 1 — Question 12

The product of all eigenvalues of a square matrix is equal to its determinant. Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. We can calculate the determinant directly:$$\det(A) = 1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7)$$$$\det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$$$$\det(A) = -3 - 2(-6) + 3(-3) = -3 + 12 - 9 = 0$$Alternatively, observe that the rows are in an arithmetic progression, which implies linear dependence ($R_1 + R_3 = 2R_2$). Since the rows are linearly dependent, the determinant is zero. Since the product of eigenvalues equals the determinant, the product is 0.