Let A=pmatrix3&-41&-1pmatrix and B be such that A^100=100B+I.

Question

Let $A=\begin{pmatrix}3&-4\1&-1\end{pmatrix}$ and $B$ be such that $A^{100}=100B+I$. Then the sum of all the elements of $B^{100}$ is \_\_\_\_.

Answer: 0

Accepted Answer

Write $A=I+C$ where $C=\begin{pmatrix}2&-4\1&-2\end{pmatrix}$. Direct computation gives $C^2=O$, so $C^n=O$ for all $n\ge2$.

By the binomial theorem, $A^{100}=(I+C)^{100}=I+100C$ (all higher terms vanish since $C^2=O$).

Comparing with $A^{100}=100B+I$ gives $100B=100C$, so $B=C$.

Since $C^2=O$, $B^{100}=C^{100}=O$ (the zero matrix, as $100\ge2$). The sum of all elements of the zero matrix is $0$.

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