Let [A=[[arrayll 1 & 2 0 & 1 array]]] and [B=I+adj(A)+(adj A)^2++(adj A)^10].

Question

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:

Accepted Answer

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

$$\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}\end{aligned}$$\right] \
\& B=\left[$$\begin{array}{cc} 11 & -110 \ 0 & 11 \end{array}$$\right]
\]

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

Solutions · 1