+0  
 
+1
94
2
avatar+82 

done

 Mar 17, 2020
edited by rubikx2910  Apr 15, 2020
 #1
avatar+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 \(b\).

 

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

 

\(\mathbf{a=12,\ b=29}\)

 

laugh

 Mar 17, 2020
 #2
avatar+82 
+1

done

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

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