In the given figure, the vectors u and v are related as \( Au = v \) by a transformation matrix A. The correct choice of matrix A is
% Vector details
u = (4,3), v = (5,0)
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A transformation that preserves the length of a vector is a rotation. The matrix for a counter-clockwise rotation by angle \( \theta \) is \( \begin{pmatrix} \cos\theta & -\sin\theta \sin\theta & \cos\theta \end{pmatrix} \).
Step 1: Analyze the vectors \( u \) and \( v \). The initial vector is \( u = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \). The transformed vector is \( v = \begin{pmatrix} 5 \\ 0 \end{pmatrix} \). Let's calculate their magnitudes (lengths): \( ||u|| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). \( ||v|| = \sqrt{5^2 + 0^2} = 5 \). Since the magnitudes are equal, the transformation is a rotation.
Step 2: Determine the angle of rotation. The vector \(u\) makes an angle \( \theta \) with the positive x-axis, where \( \cos\theta = \frac{4}{5} \) and \( \sin\theta = \frac{3}{5} \). The vector \(v\) lies on the positive x-axis, so its angle is 0. To transform \(u\) to \(v\), we need to rotate \(u\) by an angle of \( -\theta \).
Step 3: Construct the rotation matrix. The general rotation matrix for an angle \( \alpha \) is \( R(\alpha) = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} \). We need to use the angle \( \alpha = -\theta \). We know that \( \cos(-\theta) = \cos\theta = \frac{4}{5} \) and \( \sin(-\theta) = -\sin\theta = -\frac{3}{5} \). Substituting these into the rotation matrix formula: \[ A = R(-\theta) = \begin{pmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{pmatrix} = \begin{pmatrix} \frac{4}{5} & -(-\frac{3}{5}) \\ -\frac{3}{5} & \frac{4}{5} \end{pmatrix} = \begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5} \end{pmatrix} \] Step 4: Verify the result. Let's multiply the matrix \( A \) with vector \( u \): \[ Au = \begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5} \end{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} \left(\frac{4}{5}\right)(4) + \left(\frac{3}{5}\right)(3) \\ \left(-\frac{3}{5}\right)(4) + \left(\frac{4}{5}\right)(3) \end{pmatrix} = \begin{pmatrix} \frac{16+9}{5} \\ \frac{-12+12}{5} \end{pmatrix} = \begin{pmatrix} \frac{25}{5} \\ 0 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} = v \] The result matches, so the matrix is correct.
