MCQ
Let $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $\mathrm{B}=\mathrm{I}+\operatorname{adj}(\mathrm{A})+(\operatorname{adj} \mathrm{A})^2+\ldots+$ $(\operatorname{adj} \mathrm{A})^{10}$. Then, the sum of all the elements of the matrix $B$ is :
  • A
    $-110$
  • B
    $22$
  • $-88$
  • D
    $-124$

Answer

Correct option: C.
$-88$
c
$\begin{aligned} & \operatorname{Adj}(\mathrm{A})=\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right] \\ & (\operatorname{AdjA})^2=\left[\begin{array}{cc}1 & -4 \\ 0 & 1\end{array}\right] \\ & (\operatorname{AdjA})^{10}=\left[\begin{array}{cc}1 & -20 \\ 0 & 1\end{array}\right] \\ & \mathrm{B}=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -4 \\ 0 & 1\end{array}\right]+\ldots+\left[\begin{array}{cc}1 & -20 \\ 0 & 1\end{array}\right] \\ & \mathrm{B}=\left[\begin{array}{cc}11 & -110 \\ 0 & 11\end{array}\right] \Rightarrow \text { sum of elements of } \mathrm{B} \\ & =-88\end{aligned}$

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