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

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 ${\mathbf{A}}^{5}x={\mathbf{A}}^{4}y$
${\mathbf{A}}^{4}y={\mathbf{A}}^{3}x+2{\mathbf{A}}^{3}y$
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Miya Swanson
You were on the right track. Here are the next steps.
${\mathbf{A}}^{3}x={\mathbf{A}}^{2}y$ and $2{\mathbf{A}}^{3}y=2{\mathbf{A}}^{2}\left(x+2y\right)$
${\mathbf{A}}^{2}y=\mathbf{A}\mathbf{A}y=\mathbf{A}\left(x+2y\right)$ and $2\mathbf{A}\mathbf{A}x+4\mathbf{A}\mathbf{A}y=2\mathbf{A}y+4\mathbf{A}\left(x+2y\right)$
Can you take it from here?
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Jase Rocha
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
$\begin{array}{rl}{A}^{2}x& =A\left(Ax\right)=Ay=x+2y\\ {A}^{2}y& =A\left(x+2y\right)=Ax+2Ay=y+2\left(x+2y\right)=2x+5y\\ {A}^{3}x& ={A}^{2}y=2x+5y\\ {A}^{4}x& ={A}^{3}y=A\left(2x+5y\right)=2y+5\left(x+2y\right)=5x+12y\end{array}$