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We are asked to solve the following linear system
$$x_1-3x_2+x_3=1$$ $$2x_1-x_2-2x_3=2$$ $$x_1+2x_2-3x_3=-1$$
by using gauss-jordan elimination method. The augmented matrix of the linear system is $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\2 & -1 & -2 & 2 \\1 & 2 & -3 & -1\end{array}\right).$$ By a series of elementary row operations, we have $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right).$$ My question is, although the question asked us to solve the linear system using gauss-jordan elimination method, can we stop immediately and conclude that the linear system is inconsistent without further apply any elementary row operation to the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ until the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ is transformed into reduced-row echelon form?
asked Dec 30, 2013 at 8:27
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You can conclude that the system is inconsistent, because the last row of your matrix implies that $0x_1+0x_2+0x_3=-2$, which cannot be satisfied.
answered Dec 30, 2013 at 8:47
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Yes, you can stop there and conclude that the system is inconsistent as $0\ne-2$. If you were to continue to reduce the matrix to reduced-row echelon form, row $3$'s inconsistency would remain unaffected. $$\left(\begin{array}{ccc|c}1 & 0 & -\frac{7}{5} & 0 \\0 & 1 & -\frac{4}{5} & 0 \\0 & 0 & 0 & 1\end{array}\right)$$ $R_3\to-\frac{1}{2}R_3$ was performed to get the new row $3$ and notice that the completely reduced-row echelon form above also has $0x_1+0x_2+0x_3=1 \implies 0=1$ which is not possible, and hence the system still maintains its inconsistency.
answered Dec 30, 2013 at 8:57
ZhoeZhoe
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$$ rank \left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\2 & -1 & -2 & 2 \\1 & 2 & -3 & -1\end{array}\right)=3$$ because $$det\left(\begin{array}{ccc}-3 & 1 & 1 \\-1 & -2 & 2 \\2 & -3 & -1\end{array}\right)\neq 0$$ and the rank of the coefficient matrix $$\left(\begin{array}{ccc}1 & -3 & 1 \\2 & -1 & -2 \\1 & 2 & -3 \end{array}\right)$$ is $2$ because its determinant is $0$ and $$det\left(\begin{array}{cc}1 & -3 \\2 & -1 \end{array}\right)\neq 0$$ Augmented matrix and coefficient matrix have different ranks then by the Rouchè-Capelli theorem the system has no solutions
answered Dec 30, 2013 at 9:10
MathemanMatheman
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Given: 2x + 3y - z = 3, - x + 4y = 9, 4x - y - z = -7
I am going to put the second equation into the first row of an augmented matrix:
#[ (-1, 4, 0,|, 9) ]#
Add the first equation as the second row in the matrix:
#[ (-1, 4, 0,|, 9), (2,3,-1,|,3) ]#
Add the third equation as the third row in the matrix:
#[ (-1, 4, 0,|, 9), (2,3,-1,|,3), (4,-1,-1,|,-7) ]#
Multiply the row 1 by 2 and add to row 2:
#[ (-1, 4, 0,|, 9), (0,11,-1,|,21), (4,-1,-1,|,-7) ]#
Multiply the row 1 by 4 and add to row 3:
#[ (-1, 4, 0,|, 9), (0,11,-1,|,21), (0,15,-1,|,29) ]#
Subtract row 3 from row 2:
#[ (-1, 4, 0,|, 9), (0,-4,0,|,-8), (0,15,-1,|,29) ]#
Add row 2 to row 1:
#[ (-1, 0, 0,|, 1), (0,-4,0,|,-8), (0,15,-1,|,29) ]#
Multiple row 1 by -1:
#[ (1, 0, 0,|, -1), (0,-4,0,|,-8), (0,15,-1,|,29) ]#
Divide row 2 by -4:
#[ (1, 0, 0,|, -1), (0,1,0,|,2), (0,15,-1,|,29) ]#
Multiply row 2 by -15 and add to row 3:
#[ (1, 0, 0,|, -1), (0,1,0,|,2), (0,0,-1,|,-1) ]#
Multiply row 3 by -1:
#[ (1, 0, 0,|, -1), (0,1,0,|,2), (0,0,1,|,1) ]#
#x = -1, y = 2 and z = 1#