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← JEE Main 2022 · 29 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 3 of 90

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 3

Let A=(2−102)A=\begin{pmatrix}2&-1\\0&2\end{pmatrix}. If B=I−5C1(adj⁡A)+5C2(adj⁡A)2−⋯−5C5(adj⁡A)5B=I-{}^5C_1(\operatorname{adj}A)+{}^5C_2(\operatorname{adj}A)^2-\dots-{}^5C_5(\operatorname{adj}A)^5, then the sum of all elements of the matrix B is

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Answer and worked solution

Official answer: C

Binomial. B=(I−adj⁡A)5B=(I-\operatorname{adj}A)^5 with adj⁡A=(2102)\operatorname{adj}A=\begin{pmatrix}2&1\\0&2\end{pmatrix}, so I−adj⁡A=(−1−10−1)=−(I+N)I-\operatorname{adj}A=\begin{pmatrix}-1&-1\\0&-1\end{pmatrix}=-(I+N), N2=0N^2=0.

Power. B=−(I+5N)=(−1−50−1)B=-(I+5N)=\begin{pmatrix}-1&-5\\0&-1\end{pmatrix}; sum −7-7.

(C)