Q43 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
Let $A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}$. Then, the sum of all the elements of the matrix $B$ is:
- A-124
- B22
- Correct answerC-88
- D-110
Solution

$\begin{aligned} & \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -4 \\ 0 & 1\end{array}\right]+\ldots+\left[\begin{array}{cc}1 & -20 \\ 0 & 1\end{array}\right] \\ & \mathrm{B}=\left[\begin{array}{cc}11 & -110 \\ 0 & 11\end{array}\right] \Rightarrow \text { sum of elements of } \mathrm{B} \\ & =-88\end{aligned}$
Hence, the correct answer is option: 'C'.