0
2.6kviews
Solve the following system of equations : $2x-2y-5z=0, \; 4x-y+z=0, \; 3x-2y+3z=0, \; x-3y+7z=0$
1 Answer
0
167views

Now, $ \left[ \begin{array}{ccc} 2& -2& -5\\ 4& -1& 1\\ 3& -2& 3 \\ 1 & -3 & 7 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_1 \rightarrow R_1 - R_4 \; , \; R_2 \rightarrow R_2 - 4R_4 \; , \; R_3 \rightarrow R_3 - 4R_4 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 11& -27\\ 0& 7& -18 \\ 1 & -3 & 7 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_4 \rightarrow R_4 - R_1 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 11& -27\\ 0& 7& -18 \\ 0 & -4 & 19 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 - R_3 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 4& -9\\ 0& 7& -18 \\ 0 & -4 & 19 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_4 \rightarrow R_4 + R_2 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 4& -9\\ 0& 7& -18 \\ 0 & 0 & 10 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3 - R_2 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 4& -9\\ 0& 3& -9 \\ 0 & 0 & 10 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 - R_3 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 1& 0\\ 0& 3& -9 \\ 0 & 0 & 10 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3 - 3R_2 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 1& 0\\ 0& 0& -9 \\ 0 & 0 & 10 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3 + R_4 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 1& 0\\ 0& 0& 1 \\ 0 & 0 & 10 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_4 \rightarrow R_4 -10 R_3 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& -12\\ 0& 1& 0\\ 0& 0& 1 \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} x\\ y\\ z \end{array}\right] \; =\; \left[ \begin{array}{ccc} 0\\ 0\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ $

Hence, Rank of Matrix = 3 = No. of unknowns.

$ \\ $

Thus, the system has trivial solution, i.e. x=y=z=0

Please log in to add an answer.