JEE Main 2025 Apr 4 Shift 2, Mathematics Q11: Algebra Of Matrices

JEE Main2025Apr 4, Shift 2Mathematics

Let the matrix $A = 100101010$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is :-

Solution

The recurrence $A^{n} - A^{n - 2} = A^{2} - I$ telescopes. Applying it repeatedly for $n = 50, 48, 46, \dots, 4$ :

$A^{50} = A^{48} + (A^{2} - I) = A^{46} + 2 (A^{2} - I) = \dots = A^{2} + 24 (A^{2} - I) = 25 A^{2} - 24 I$ .

Compute $A^{2}$ : direct multiplication gives $A^{2} = 100110101$ .

Therefore $A^{50} = 25100110101 - 24 I = 10025102501$ .

Sum of entries $= 1 + 25 + 1 + 25 + 1 = 53$ .

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