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residence inn phoenix desert view at mayo clinic - NBX Soluciones

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 \mathbf{M} = \sqrt{2} \cdot \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}.

📌 Thus, $ \boxed{\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}} $.

📌 Question: Planetary exploration engineers use a matrix $ \mathbf{M} $ to transform satellite coordinates. If $ \mathbf{M} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} $ and $ \mathbf{M} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix} $, find $ \mathbf{M} $.

📌 Solution: The columns of $ \mathbf{M} $ are the images of the standard basis vectors. Thus, $ \mathbf{M} = \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix} $. Verification:

📌 \mathbf{M} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 3 \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}, \quad \mathbf{M} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = 1 \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 4 \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}.

📌 The matrix is $ \boxed{\begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix}} $.

📌 Question: Compute $ (\cos 72^\circ + i \sin 72^\circ)^5 $, modeling rotational symmetry in a planetary orbit.

📌 Solution: Using De Moivre’s Theorem, $ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) $. Here, $ n = 5 $, $ \theta = 72^\circ $:

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