So far I have ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$

${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$

Camila Brandt
2022-09-27
Answered

Let $\mathbf{A}$ be a matrix, and let 𝐱 and 𝐲 be linearly independent vectors such that 𝐀𝐱=𝐲,𝐀𝐲=𝐱+2𝐲.Then we have that ${\mathbf{A}}^{5}\mathbf{x}=a\mathbf{x}+b\mathbf{y}$ for some scalars 𝑎 and 𝑏. Find the ordered pair (a, b).

So far I have ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$

${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$

So far I have ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$

${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$

You can still ask an expert for help

Miya Swanson

Answered 2022-09-28
Author has **11** answers

You were on the right track. Here are the next steps.

${\mathbf{A}}^{3}x={\mathbf{A}}^{2}y$ and $2{\mathbf{A}}^{3}y=2{\mathbf{A}}^{2}(x+2y)$

${\mathbf{A}}^{2}y=\mathbf{A}\mathbf{A}y=\mathbf{A}(x+2y)$ and $2\mathbf{A}\mathbf{A}x+4\mathbf{A}\mathbf{A}y=2\mathbf{A}y+4\mathbf{A}(x+2y)$

Can you take it from here?

${\mathbf{A}}^{3}x={\mathbf{A}}^{2}y$ and $2{\mathbf{A}}^{3}y=2{\mathbf{A}}^{2}(x+2y)$

${\mathbf{A}}^{2}y=\mathbf{A}\mathbf{A}y=\mathbf{A}(x+2y)$ and $2\mathbf{A}\mathbf{A}x+4\mathbf{A}\mathbf{A}y=2\mathbf{A}y+4\mathbf{A}(x+2y)$

Can you take it from here?

Jase Rocha

Answered 2022-09-29
Author has **2** answers

You are trying to reduce the powers, which works if you keep going. You can increase them as well. If $Ax=y$ and $Ay=x+2y$, then

$\begin{array}{rl}{A}^{2}x& =A(Ax)=Ay=x+2y\\ {A}^{2}y& =A(x+2y)=Ax+2Ay=y+2(x+2y)=2x+5y\\ {A}^{3}x& ={A}^{2}y=2x+5y\\ {A}^{4}x& ={A}^{3}y=A(2x+5y)=2y+5(x+2y)=5x+12y\end{array}$

$\begin{array}{rl}{A}^{2}x& =A(Ax)=Ay=x+2y\\ {A}^{2}y& =A(x+2y)=Ax+2Ay=y+2(x+2y)=2x+5y\\ {A}^{3}x& ={A}^{2}y=2x+5y\\ {A}^{4}x& ={A}^{3}y=A(2x+5y)=2y+5(x+2y)=5x+12y\end{array}$

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