JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 19)

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:
$$-$$110
22
$$-$$124
$$-$$88

Explanation

$$\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{ll} 1 & -2 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^2=\left[\begin{array}{ll} 1 & -4 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^3=\left[\begin{array}{cc} 1 & -6 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^4=\left[\begin{array}{cc} 1 & -8 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^r=\left[\begin{array}{cc} 1 & (-2 r) \\ 0 & 1 \end{array}\right] \end{aligned}$$

$$\begin{aligned} & B=\sum_{r=0}^{10}(\operatorname{adj} A)^r=\left[\begin{array}{ll} \sum_\limits{r=0}^{10} 1 & \sum_\limits{r=0}^{10}(-2 r) \\ \sum_\limits{r=0}^{10}(0) & \sum_\limits{r=0}^{10}(1) \end{array}\right] \\ & B=\left[\begin{array}{cc} 11 & -110 \\ 0 & 11 \end{array}\right] \end{aligned}$$

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

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