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← JEE Main 2022 · 25 Jul · Shift 2
MathematicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 23

Let A=[1aa01b001]A=\begin{bmatrix}1&a&a\\0&1&b\\0&0&1\end{bmatrix}, a,b∈Ra,b\in\mathbb R. If for some n∈Nn\in\mathbb N, An=[14821600196001]A^n=\begin{bmatrix}1&48&2160\\0&1&96\\0&0&1\end{bmatrix} then n+a+bn+a+b is equal to ______.

Answer and worked solution

Official answer: 24

Powers. AnA^n has nana, nbnb and na+n(n−1)2abna+\frac{n(n-1)}2ab above the diagonal: na=48na=48, nb=96nb=96, 48+24⋅96n−1n=216048+24\cdot96\frac{n-1}n=2160, so n=12n=12, a=4a=4, b=8b=8.

Sum 24.

Answer: 24