JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

The characteristic equation of M is given by (M - I) = 0: vmatrix 2 - & -1 \\ 1 & - vmatrix = ^2 - 2 + 1 = ( - 1)^2 = 0 The only eigenvalue is = 1 with algebraic multiplicity 2. For the eigenvector, (M - I)X = 0 bmatrix 1 & -1 \\ 1 & -1 bmatrix bmatrix x \\ y bmatrix = bmatrix 0 \\ 0 bmatrix x = y. Since there is only one linearly independent eigenvector, M is not diagonalizable but can be reduced to its Jordan canonical form J = bmatrix 1 & 1 \\ 0 & 1 bmatrix . Thus, there exists an invertible matrix N such that M = N J N^ -1 MN = N bmatrix 1 & 1 \\ 0 & 1 bmatrix . Statement (A) is TRUE. We can write M = I + A, where A = bmatrix 1 & -1 \\ 1 & -1 bmatrix . Notice that A^2 = bmatrix 1 & -1 \\ 1 & -1 bmatrix bmatrix 1 & -1 \\ 1 & -1 bmatrix = bmatrix 0 & 0 \\ 0 & 0 bmatrix . Using the binomial theorem (since I and A commute), M^k = (I + A)^k = I + kA = bmatrix 1+k & -k \\ k & 1-k bmatrix . For k=26, M^ 26 = bmatrix 27 & -26 \\ 26 & -25 bmatrix = bmatrix p & q \\ r & s bmatrix . The determinant of M^ 26 is ps - qr = 27(-25) - (-26)(26) = 1. Since (M^ 26 ) = 1, its inverse has integer entries. Thus, for any integers m and n, the system bmatrix p & q \\ r & s bmatrix bmatrix x \\ y bmatrix = bmatrix m \\ n bmatrix has a unique integer solution. Statement (C) is TRUE. Now, let's find _ k=1 ^ 26 M^k = _ k=1 ^ 26 bmatrix 1+k & -k \\ k & 1-k bmatrix = bmatrix a & b \\ c & d bmatrix . a = _ k=1 ^ 26 (1+k) = 2 + 3 + + 27 = 26 2 (2 + 27) = 13 29 = 377. Statement (B) is FALSE. We also have: b = _ k=1 ^ 26 (-k) = -351 c = _ k=1 ^ 26 k = 351 d = _ k=1 ^ 26 (1-k) = 26 - 351 = -325 For statement (D), the system of equations is bmatrix a+t & b \\ c & d+t bmatrix bmatrix x \\ y bmatrix = bmatrix 1 \\ -1 bmatrix . The determinant of the coefficient matrix is: = (a+t)(d+t) - bc = (377+t)(-325+t) - (-351)(351) = t^2 + 52t - 122525 + 123201 = t^2 + 52t + 676 = (t+26)^2 For any positive real number t > 0, = (t+26)^2 > 0, meaning the determinant is non-zero. Hence, the system has a unique solution. Statement (D) is TRUE. Answer: There exists a 2 2 invertible matrix N with real entries such that MN = N bmatrix 1 & 1 \\ 0 & 1 bmatrix ; For any two given integers m and n, there exist unique integers x and y such that px + qy = m and rx + sy = n; For each positive real number t, the system of linear equations (a + t)x + by = 1 and cx + (d + t)y = -1 has a unique solution