For AX = B, A is orthogonal and $$A=\begin{pmatrix}1/\sqrt2&0&1/\sqrt2\\0&1&0\\1/\sqrt2&0&-1/\sqrt2\end{pmatrix}.$$ The vectors are $$X=\begin{pmatrix}x\\y\\z\end{pmatrix},\qquad B=\begin{pmatrix}0\\1\\-\sqrt2\end{pmatrix}.$$ Find x+y+z to 1 decimal place.
Orthogonality gives $X=A^TB$. Here A is symmetric, so direct multiplication gives x = −1, y = 1, z = 1. Their sum is 1.0. This also satisfies the three original linear equations.
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