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$$ A = \left( \begin{array}{ccc} 10 & 29 & 41 \\ 23 & 27 & 42 \\ 24 & 28 & 48 \\ \end{array} \right) $$ $\det (A) = -1748$.

Now $B$ is formed when the second column is multiplied by $15$ and added to the first column. $$ B = \left( \begin{array}{ccc} 445 & 29 & 41 \\ 428 & 27 & 42 \\ 444 & 28 & 48 \\ \end{array} \right) $$ $\det (B) = -1748$.

When i reduced these two matrices the final row is $[0,0,1]$, which means there are no solutions.

Why is that? If my system is inconsistent then what does it mean?

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First of all, if your final row is $[0,0,1]$ it is not a row of "zeros". Secondly, if you have pivots (leading ones) in every column then there is a solution. In this case, since $\det(A) = \det(B) \neq 0$ the matrix is invertible and your solution is unique. Third of all, I think you meant $\det(B)$ after you define $B$, not $\det(A)$. – user139388 Apr 6 '14 at 2:01
1) Corrected. 2) I didn't get that statement. 3) Corrected as well. – Miodrag Apr 6 '14 at 2:06
What makes you think that there are no solutions when you get a final row of $[0,0,1]$? The only way for there to be no solutions is if your system is $Ax=b$ with $b \neq 0$, and such that when you reduce to the RREF of $[A \mid b ]$ you have a leading $1$ in the last column. – user139388 Apr 6 '14 at 2:12
@Amzoti No i don't have that system. I simply used my graphics calculator to get that final row using the function RREF. – Miodrag Apr 6 '14 at 2:24
up vote 0 down vote accepted

You are confusing reducing the matrix itself with reducing the augmented matrix --- the original matrix with the column of constants appended as the last column. If this augmented matrix reduces to a matrix with a row of zeros except in the last column, then the system is inconsistent and has no solutions.

In the case you've given, the matrix of coefficients $A$ has nonzero determinant, so you know (depending on what you've learned so far) that the system has a unique solution.

As far as what an inconsistent system means, consider the case of two equations in two unknowns. Each of those equations represents a line in the $xy$-plane, and a solution to the system is an intersection point of those lines. If the lines are nonparallel, they intersect in exactly one point and the system has a unique solution. If they coincide, there are an infinite number of solutions. But if they are parallel and not the same line, they do not intersect, so the system has no solutions and is inconsistent. The situation with more variables and more equations is more complicated when you try to visualize it, but is conceptually the same: if the surfaces defined have a unique intersection point, the system has a unique solution, and so on.

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So what can i comment on this two matrices? Why didn't the determinant change? – Miodrag Apr 6 '14 at 2:29
Because you performed the elementary operation of adding a multiple of one row to another row. That does not change the determinant of a matrix. – rogerl Apr 6 '14 at 3:14
So that is it. Do i need to add anything more? Any more mathematical statements to solidify my working? Thanks for the answer though. – Miodrag Apr 6 '14 at 3:18

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