+0

# done

+1
94
2
+82

done

Mar 17, 2020
edited by rubikx2910  Apr 15, 2020

#1
+24932
+3

Let $$A$$ be a matrix, and let $$x$$ and $$y$$ be vectors such that neither is a scalar multiple of the other and such that

$$\mathbf{A} \mathbf{x} = \mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}$$.
Then we have that $$\mathbf{A}^5 \mathbf{x} = a \mathbf{x} + b\mathbf{y}$$ for some scalars $$a$$ and $$b$$.
Find $$a$$ and $$b$$.

$$\begin{array}{|rcll|} \hline \mathbf{A} \mathbf{x} &=& \mathbf{y} \quad &| \quad \times \mathbf{A} \\ \mathbf{A}^2\mathbf{x} &=& \mathbf{A} \mathbf{y} \quad &| \quad \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y} \\ \mathbf{A}^2\mathbf{x} &=& \mathbf{x} + 2\mathbf{y} \\\\ \hline \mathbf{A}^2\mathbf{x} &=& \mathbf{x} + 2\mathbf{y} \quad &| \quad \times \mathbf{A} \\ \mathbf{A}^3\mathbf{x} &=& \mathbf{A}\mathbf{x} + 2\mathbf{A} \mathbf{y} \quad &| \quad \mathbf{A} \mathbf{x}=\mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y} \\ \mathbf{A}^3\mathbf{x} &=& \mathbf{y} + 2( \mathbf{x} + 2\mathbf{y} ) \\ \mathbf{A}^3\mathbf{x} &=& \mathbf{y} + 2\mathbf{x} + 4\mathbf{y} \\ \mathbf{A}^3\mathbf{x} &=& 2\mathbf{x} + 5\mathbf{y} \\\\ \hline \mathbf{A}^3\mathbf{x} &=& 2\mathbf{x} + 5\mathbf{y} \quad &| \quad \times \mathbf{A} \\ \mathbf{A}^4\mathbf{x} &=& 2\mathbf{A}\mathbf{x} + 5\mathbf{A}\mathbf{y} \quad &| \quad \mathbf{A} \mathbf{x}=\mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y} \\ \mathbf{A}^4\mathbf{x} &=& 2\mathbf{y} + 5( \mathbf{x} + 2\mathbf{y} ) \\ \mathbf{A}^4\mathbf{x} &=& 2\mathbf{y} + 5\mathbf{x} + 10\mathbf{y} \\ \mathbf{A}^4\mathbf{x} &=& 5\mathbf{x} + 12\mathbf{y} \\\\ \hline \mathbf{A}^4\mathbf{x} &=& 5\mathbf{x} + 12\mathbf{y} \quad &| \quad \times \mathbf{A} \\ \mathbf{A}^5\mathbf{x} &=& 5\mathbf{A}\mathbf{x} + 12\mathbf{A}\mathbf{y} \quad &| \quad \mathbf{A} \mathbf{x}=\mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y} \\ \mathbf{A}^5\mathbf{x} &=& 5\mathbf{y} + 12( \mathbf{x} + 2\mathbf{y} ) \\ \mathbf{A}^5\mathbf{x} &=& 5\mathbf{y} + 12\mathbf{x} + 24\mathbf{y} \\ \mathbf{A}^5\mathbf{x} &=& 12\mathbf{x} + 29\mathbf{y} \\ \hline \end{array}$$

$$\mathbf{a=12,\ b=29}$$

Mar 17, 2020
#2
+82
+1

done

rubikx2910  Mar 18, 2020
edited by rubikx2910  Apr 15, 2020