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← JEE Main 2022 · 25 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 11 of 90

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 11

Let A=[0−220]A=\begin{bmatrix}0&-2\\2&0\end{bmatrix}. If M and N are two matrices given by M=∑k=110A2kM=\displaystyle\sum_{k=1}^{10}A^{2k} and N=∑k=110A2k−1N=\displaystyle\sum_{k=1}^{10}A^{2k-1} then MN2MN^2 is :

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Answer and worked solution

Official answer: A

Powers. A2=−4IA^2=-4I, so A2k=(−4)kIA^{2k}=(-4)^kI and A2k−1=(−4)k−1AA^{2k-1}=(-4)^{k-1}A.

So M=mIM=mI and N=nAN=nA with scalars m=∑(−4)k≠0m=\sum(-4)^k\ne0, n=∑(−4)k−1n=\sum(-4)^{k-1}; N2=n2A2=−4n2IN^2=n^2A^2=-4n^2I.

Product. MN2=−4mn2IMN^2=-4mn^2I, a scalar matrix: symmetric, and not II (its entry is a large integer).

(A)