JEE Main 2026 Apr 6 Shift 2, Mathematics Q5: Algebra Of Matrices

JEE Main2026Apr 6, Shift 2Mathematics

Let $A = 100310931$ and $B = [b_{ij}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b _{31} - b _{21}}{b _{32}}$ is:

Solution

Write $A = I + P$ where $P = 000300930$ .

$P^{2} = 000000900$ and $P^{3} = 0$ (nilpotent).

By the binomial theorem (since $I$ and $P$ commute):

$A^{99} = (I + P)^{99} = I + 99 P + (2 99) P^{2}$ (all higher terms vanish).

$B = A^{99} - I = 99 P + (2 99) P^{2}$ . Read off:

$b_{21} = 99 \cdot 3 = 297$ ; $b_{32} = 99 \cdot 3 = 297$ ; $b_{31} = 99 \cdot 9 + (2 99) \cdot 9 = 891 + 9 \cdot 4851 = 891 + 43659 = 44550$ .

$\frac{b _{31} - b _{21}}{b _{32}} = \frac{44550 - 297}{297} = \frac{44253}{297} = 149$ .

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