JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The characteristic equation of a 2 2 matrix A = bmatrix a & b \\ c & d bmatrix is given by: A^2 - Tr (A)A + (A)I = 0 We are given that the sum of the diagonal elements is 4, so Tr (A) = a + d = 4. Substituting this into the characteristic equation, we get: A^2 - 4A + (ad - bc)I = 0 We are also given that the matrix satisfies: A^2 - 4A + 3I = 0 Comparing the two equations, we must have: ad - bc = 3 We need to find the number of quadruples (a, b, c, d) with elements from the set \ 0, 1, 2, 3, 4\ such that a + d = 4 and ad - bc = 3. Let us analyze the possible values for a and d: Case 1: a = 0, d = 4 ad - bc = 3 0 - bc = 3 bc = -3 Since b, c \ 0, 1, 2, 3, 4\ , bc 0. Thus, there are no solutions. Case 2: a = 1, d = 3 ad - bc = 3 3 - bc = 3 bc = 0 For bc = 0, either b = 0 or c = 0. If b = 0, c can take any of the 5 values from the set. If c = 0, b can take any of the 5 values from the set. The pair (0, 0) is counted twice, so the number of pairs (b, c) is 5 + 5 - 1 = 9. Case 3: a = 2, d = 2 ad - bc = 3 4 - bc = 3 bc = 1 The only possible pair from the given set is b = 1, c = 1. This gives 1 solution. Case 4: a = 3, d = 1 ad - bc = 3 3 - bc = 3 bc = 0 Similar to Case 2, there are 9 possible pairs for (b, c). Case 5: a = 4, d = 0 ad - bc = 3 0 - bc = 3 bc = -3 Again, there are no solutions. Adding the valid matrices from all cases, the total number of matrices is 9 + 1 + 9 = 19. Answer: 19