JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q5

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MathematicsSingle correct+4 / -1Question 5 of 75

Let $A=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}$ and $B=[b_{ij}]$ , $1\le i,j\le3$ . If $B=A^{99}-I$ , then the value of $\dfrac{b_{31}-b_{21}}{b_{32}}$ is :

199

149

159

Answer and worked solution

Official answer: C

Nilpotent part. $A=I+N$ with $N=\begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix}$ and $N^2=\begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix}$ , $N^3=0$ .

Power. $A^{99}=I+99N+\binom{99}2N^2$ , so $B=99N+4851N^2$ :

$b_{31}=99(9)+4851(9)=44550$ .
$b_{21}=b_{32}=99(3)=297$ .

Value. $\dfrac{44550-297}{297}=150-1=149$ .

(C)

Topics:Matrix algebra Binomial expansion

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JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 5