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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Matrices

Let A=[ array ll 1 & 2 0 & 1 array ] and B=I+adj(A)+(adj A)^2+...

Let A=\left[1201\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] andB=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}.Then,thesumofalltheelementsofthematrix. Then, the sum of all the elements of the matrixB$ is:

Options

  1. A-124
  2. B22
  3. C-88
  4. D-110

Correct answer

C. -88

Step-by-step solution

B=[1001]+[1−201]+[1−401]+…+[1−2001]B=[11−110011]⇒ sum of elements of B=−88\begin{aligned} & \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -4 \\ 0 & 1\end{array}\right]+\ldots+\left[\begin{array}{cc}1 & -20 \\ 0 & 1\end{array}\right] \\ & \mathrm{B}=\left[\begin{array}{cc}11 & -110 \\ 0 & 11\end{array}\right] \Rightarrow \text{ sum of elements of } \mathrm{B} \\ & =-88\end{aligned}
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