+0  
 
0
121
2
avatar+78 

done

 Mar 18, 2020
edited by rubikx2910  Apr 15, 2020
 #1
avatar+25243 
+2

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 \(\mathbf{A}^{-1} \mathbf{x} = a \mathbf{x} + b\mathbf{y}\) for some scalars \(a\) and \(b\).
Find \(a\) and \(b\).

 

\(\begin{array}{|rcll|} \hline \mathbf{A} \mathbf{x} &=& \mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{x} &=& \mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{x} &=& \mathbf{A}^{-1}\mathbf{y} \\\\ \hline \mathbf{A} \mathbf{y} &=& \mathbf{x} + 2\mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{y} = \mathbf{x} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{x} \\ \mathbf{A}^{-1}\mathbf{x} &=& -2\mathbf{x}+\mathbf{y} \\ \hline \end{array}\)

 

\(\mathbf{a=-2,\ b=1} \)

 

laugh

 Mar 18, 2020
edited by heureka  Mar 18, 2020
 #2
avatar+78 
+1

done

rubikx2910  Mar 19, 2020
edited by rubikx2910  Apr 15, 2020

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