JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $A = [3 - 4 1 - 1]$ and $B$ be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is ______ .

Your answer

Solution

Write $A = I + M$ where $M = [2 - 4 1 - 2]$ .

Compute $M^{2} = [2 - 4 1 - 2] [2 - 4 1 - 2] = [0000]$ .

So $M$ is nilpotent of index 2: $M^{k} = 0$ for all $k \geq 2$ .

By the binomial theorem (valid since $I$ commutes with $M$ ): $A^{100} = (I + M)^{100} = I + 100 M$ .

Comparing with $A^{100} = 100 B + I$ gives $B = M$ . Since $M^{100} = 0$ (zero matrix), the sum of all elements of $B^{100}$ is $0$ .

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.