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Test the consistency of the following equations and solve them if they are consistent $ \\ \; \\ 2x-y+z=8 \; , \; 3x-y+z=6 \;, \; 4x-y+2z=7 \;, \; -x+y-z=4 $
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The given equations can be written in matrix form AX=B as follows :

Now, $ \left[ \begin{array}{cccc} 2& -1& 1 \\ 3& -1& 1 \\ 4& -1 & 2 \\ -1& 1 & -1 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 8 \\ 6 \\ 7 \\ 4 \end{array}\right] \\ \; \\ \; \\ \; \\ R_1 \rightarrow R_1-R_2 \; , \; R_3 \rightarrow R_3 - R_2 \\ \; \\ \; \\ \therefore \left[ \begin{array}{cccc} -1& 0& 0 \\ 3& -1& 1 \\ 1& 0 & 1 \\ -1& 1 & -1 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 2 \\ 6 \\ 1 \\ 4 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3+R_1 \; , \; R_2 \rightarrow R_2 +R_4 \\ \; \\ \; \\ \therefore \left[ \begin{array}{cccc} -1& 0& 0 \\ 2& 0& 0 \\ 0& 0 & 1 \\ -1& 1 & -1 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 2 \\ 10 \\ 3 \\ 4 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2+2R_1 \; , \; R_4 \rightarrow R_4 +R_3 \\ \; \\ \; \\ \therefore \left[ \begin{array}{cccc} -1& 0& 0 \\ 0& 0& 0 \\ 0& 0 & 1 \\ -1& 1 & 0 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 2 \\ 14 \\ 3 \\ 7 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \longleftrightarrow R_4 \; , \; R_1 \rightarrow R_1 \times (-1) \\ \; \\ \; \\ \therefore \left[ \begin{array}{cccc} 1& 0& 0 \\ -1& 1 0 \\ 0& 0 & 1 \\ 0& 0 & 0 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} -2 \\ 7 \\ 3 \\ 14 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 + R_1 \\ \; \\ \; \\ \therefore \left[ \begin{array}{cccc} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0 & 1 \\ 0& 0 & 0 \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{c} -2 \\ 5 \\ 3 \\ 14 \end{array}\right] \\ \; \\ \; \\ \therefore Rank \; of \; A \; = \; 3 \\ \; \\ \; \\ \; \\ Now, \; Augmented \; Matrix \; =[A,B] \;=\; \left[ \begin{array}{cccc} 1& 0& 0 & -2 \\ 0& 1& 0 & 5 \\ 0& 0 & 1 & 3 \\ 0& 0 & 0 & 14 \end{array}\right] \\ \; \\ \; \\ $

Rank of this matrix =4

$\because$ Rank of A is not equal to rank of [A, B], the given system of equations is not consistent.

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