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Home » The bird travels this displacement in 5 hours, so speed = \( \frac2\sqrt135 \). - Dachbleche24

The bird travels this displacement in 5 hours, so speed = \( \frac2\sqrt135 \). - Dachbleche24

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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 $:

📌 The result is $ \boxed{1} $.

📌 Question: In a city map, vectors $ \overrightarrow{OA} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $ and $ \overrightarrow{OB} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} $ represent street segments. Find $ \overrightarrow{OC} $ such that $ \overrightarrow{OC} = 2\overrightarrow{OA} - 3\overrightarrow{OB} $.

📌 Solution: Compute $ 2\overrightarrow{OA} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} $ and $ 3\overrightarrow{OB} = \begin{bmatrix} -3 \\ 9 \end{bmatrix} $. Subtract:

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