Let A=pmatrix1&0&00&4&-10&12&-3pmatrix.

Question

Let $A=\begin{pmatrix}1&0&0\0&4&-1\0&12&-3\end{pmatrix}$. Then the sum of the diagonal elements of the matrix $(A+I)^{11}$ is equal to:

Accepted Answer

Direct computation shows $A^2=A$ (i.e. $A$ is idempotent), so $A^n=A$ for all $n\ge1$.

$$(A+I)^{11}=\sum_{k=0}^{11}\binom{11}{k}A^k=I+\left(\sum_{k=1}^{11}\binom{11}{k}\right)A=I+(2^{11}-1)A=I+2047A.$$

Sum of diagonal elements of $A$ is $1+4-3=2$. So $\operatorname{trace}\big((A+I)^{11}\big)=\operatorname{trace}(I)+2047\times2=3+4094=4097$.

Solutions · 1