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Solve using reduced echelon form: 4x + 4y + z = 24 2x - 4y + 2 = 0 5x - 4y - 5z = 12
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\[\left\{x=\frac{124}{33},y=\frac{157}{66},z=-\frac{6}{11}\right\} \]\[\{3.75758,2.37879,-0.545455\} \]
what about the process of how this solution set was solve?
This link might help you how to get the reduced row echelon form: http://en.wikibooks.org/wiki/Linear_Algebra/Row_Reduction_and_Echelon_Forms In the 2nd row: Is it just 2 or 2z? For 2z in the 2nd row your RREF matrix is:\[\left[\begin{matrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 \end{matrix}\right]\]Otherwise it is:\[\left[\begin{matrix}1 & 0 & 0 & 124/33 \\ 0 & 1 & 0 & 157/66 \\ 0 & 0 & 1 & -6/11\end{matrix}\right]\]
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