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📅 May 7, 2026 👤 scraface

May 07, 2026

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From equations 2 and 3, we factor:
\( ab + bd = b(a + d) = 0 \)
\( ac + cd = c(a + d) = 0 \)
This implies either \( b = 0 \) and \( c = 0 \), or \( a + d = 0 \).
Case 1:** \( b = 0 \) and \( c = 0 \)
Then, \( a^2 = 1 \) and \( d^2 = 1 \). So, \( a = \pm 1 \) and \( d =...
Case 2:** \( a + d = 0 \)
Let \( d = -a \). Substituting into equations 1 and 4:
\( bc + a^2 = 1 \)
Both equations are identical. So, \( a^2 + bc = 1 \) holds. Thus, the...

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