Verified Solution • 2025–26 Session moderate

QJEEM24Ma874e: JEE Main 2024 Mathematics - Class 12 JEEM24Ma874e Solution: Let and . Then, the sum of all the elements of the matrix is:

Let A=[1amp;20amp;1]A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] and B=I+adj⁡(A)+(adj⁡A)2+…+(adj⁡A)10B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10} . Then, the sum of all the elements of the matrix BB is:

Multiple Choice Options

  • A −- 110
  • B 22
  • C −- 124
  • D −- 88

Step-by-Step Verified Solution

amp;adj⁡(A)=[1amp;−20amp;1]amp;(adj⁡A)2=[1amp;−40amp;1]amp;(adj⁡A)3=[1amp;−60amp;1]amp;(adj⁡A)4=[1amp;−80amp;1]amp;(adj⁡A)r=[1amp;(−2r)0amp;1]\begin{aligned}& \operatorname{adj}(A)=\left[\begin{array}{ll}1 & -2 \\0 & 1\end{array}\right] \\& (\operatorname{adj} A)^2=\left[\begin{array}{ll}1 & -4 \\0 & 1\end{array}\right] \\& (\operatorname{adj} A)^3=\left[\begin{array}{cc}1 & -6 \\0 & 1\end{array}\right] \\& (\operatorname{adj} A)^4=\left[\begin{array}{cc}1 & -8 \\0 & 1\end{array}\right] \\& (\operatorname{adj} A)^r=\left[\begin{array}{cc}1 & (-2 r) \\0 & 1\end{array}\right]\end{aligned}

B=∑r=010(adjA)r=[∑r=0101∑r=010(−2r)∑r=010(0)∑r=010(1)]B=[11−110011]

 Sum of elements =−110+11+11=−88\text { Sum of elements }=-110+11+11=-88

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