Question Number 198293 by Mingma last updated on 16/Oct/23
Answered by MM42 last updated on 17/Oct/23
![B^n = [(((−1)^n 0)),(( 0 2^n )) ] & (ABA^(−1) )^n =AB^n A^(−1) ⇒(ABA^(−1) )^n = [(( 2 5)),((−1 −3)) ] [(((−1)^n 0)),(( 0 2^n )) ] [(( 3 5)),((−1 −2)) ]](https://www.tinkutara.com/question/Q198300.png)
$${B}^{{n}} =\begin{bmatrix}{\left(−\mathrm{1}\right)^{{n}} \:\:\:\:\mathrm{0}}\\{\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}} }\end{bmatrix}\:\&\:\:\left({ABA}^{−\mathrm{1}} \right)^{{n}} ={AB}^{{n}} {A}^{−\mathrm{1}} \\ $$$$\Rightarrow\left({ABA}^{−\mathrm{1}} \right)^{{n}} =\begin{bmatrix}{\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:−\mathrm{3}}\end{bmatrix}\begin{bmatrix}{\left(−\mathrm{1}\right)^{{n}} \:\:\:\mathrm{0}}\\{\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}} }\end{bmatrix}\begin{bmatrix}{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:−\mathrm{2}}\end{bmatrix} \\ $$$$ \\ $$
Commented by Mingma last updated on 17/Oct/23
Perfect