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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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4y + z = 5 \\
2z - 4x = -6 \\
-x - 2y = -1
From the third equation: $x = 1 - 2y$.
2z - 4(1 - 2y) = -6 \Rightarrow 2z - 4 + 8y = -6 \Rightarrow 2z + 8y =...
Now solve the first and new second equation:
z + 4y = -1
\Rightarrow \text{Subtract: } 0 = 6 \Rightarrow \text{Contradiction}
This suggests no solution exists unless the given cross product is inc...
However, cross product outputs are orthogonal to $\mathbf{a}$. Check:

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