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Home » Let $ A = 4x $, $ B = 2x $: - Dyverse

Let $ A = 4x $, $ B = 2x $: - Dyverse

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 R\mathbf{v} = egin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} egin{pmatrix} x \ y \end{pmatrix} = egin{pmatrix} -y \ x \end{pmatrix} = egin{pmatrix} x \ y \end{pmatrix}

📌 -y = x \quad ext{and} \quad x = y

📌 From \( -y = x \) and \( x = y \), substituting: \( -x = x \Rightarrow x = 0 \), then \( y = 0 \). Only solution is \(\mathbf{v} = egin{pmatrix} 0 \ 0 \end{pmatrix}\), but \( \|\mathbf{v}\|^2 = 0

📌 Thus, no non-zero integer vector satisfies both invariance and norm condition.

📌 But wait â reconsider: âinvariant under \(90^\circ\) rotationâ means \( R\mathbf{v} = \mathbf{v} \), which we solved gives \( x = -y \) and \( x = y \Rightarrow x = 0, y = 0 \). So only trivial fixed vector.

📌 Alternatively, if the problem meant vectors that are *cyclically ordered* under symmetry, but the wording specifies âinvariant under a \(90^\circ\) rotationâ, so must satisfy \( R\mathbf{v} = \mathbf{v} \).

📌 Thus, only solution is \(\mathbf{v} = (0,0)\), which does not satisfy \(\|\mathbf{v}\|^2 = 10\).

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