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I've just started my linear algebra course, and one question has me reduce an augmented matrix A of the form $A=\begin{bmatrix} n_{11} & n_{12} & n_{13} & a\\ n_{21} & n_{22} & n_{23} & b\\ n_{31} & n_{32} & n_{33} & c \end{bmatrix}$

But, after reducing this matrix to its reduced row echelon form: $rref(A)=\begin{bmatrix} 1 & 0 & 0 & a-b\\ 0 & 1 & 0 & b\\ 0 & 0 & 0 & 1 \end{bmatrix}$

Since the last row ($R_3$) is all zeros save for the rightmost column of this augmented matrix, is this matrix consistent? As in, does the system of equations represented by this matrix have solutions? Because if $R_3$ was all zeros including the last column, then you could just ignore it, but if it equals 1, does that not imply $0=1$ which would be a contradiction? Would that contradiction mean that the matrix is not consistent? Or maybe it is only consistent with infinite solutions for a certain $(a,b,c)$?

Thanks

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  • $\begingroup$ It means that $0x+0y+0z=1$... which is clearly false $\endgroup$
    – JMoravitz
    Jun 7, 2021 at 20:09

1 Answer 1

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Yes, it implies that $0=1$, which is nonsense, so the system is inconsistent. If you hypothetically had e.g. $0=a^2$ as the last row, with the freedom to choose $a=0$, then you would be able to say it has infinitely many solutions for $a=0$, but it's inconsistent for $a\neq 0$.

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