Question

Solve the equation by reduction method. X+2y+3z=9. 2x+3y+z=4. 4x+5y-15+4z= 0

Answer 1

$$\begin{gathered} x+2y+3z=9 \\ 2x+3y+z=4 \\ 4x+5y+4z=15 \\ Inmatrixformwecanwriteitas \\ \left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 4& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 4\\ 15\end{array}\right] \\ {R}_2\to R_2-2R_1 \\ \left[\begin{array}{ccc}1& 2& 3\\ 0& -1& -5\\ 4& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ -14\\ 15\end{array}\right] \\ {R}_3\to R_3-4R_1 \\ \left[\begin{array}{ccc}1& 2& 3\\ 0& -1& -5\\ 0& -3& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ -14\\ -21\end{array}\right] \\ {R}_1\to R_1+2R_2 \\ \left[\begin{array}{ccc}1& 0& -7\\ 0& -1& -5\\ 0& -3& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ -14\\ -21\end{array}\right] \\ {R}_3\to R_3-3R_2 \\ \left[\begin{array}{ccc}1& 0& -7\\ 0& -1& -5\\ 0& 0& 7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ -14\\ 21\end{array}\right] \\ {R}_1\to R_1+R_3 \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& -1& -5\\ 0& 0& 7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -14\\ 21\end{array}\right] \\ {R}_2\to R_2+\frac{5}{7}{R}_3 \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 21\end{array}\right] \\ {R}_2\to -R_2 \\ {R}_3\to \frac{1}{7}{R}_3 \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right] \\ Regards \end{gathered}$$