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I know that a rank of a matrix is the # of rows the containing the leading $1$'s but I'm not sure how to implement that to this proof, any help is greatly appreciated.

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  • $\begingroup$ It would help if you wrote the problem out here, so it is clear what part of the proof specifically (and the proof of what) is confusing you. $\endgroup$ – Chill2Macht Apr 23 '16 at 20:24
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The system $Ax=b$ is consistent if and only if $b$ belongs to the column space of $A$ (the linear combinations of the columns of $A$), essentially by definition: if $A=[a_1\ a_2\ \dots\ a_n]$ ($a_1,\dots,a_n$ being the columns) and $[x_1\ x_2\ \dots\ x_n]^T$ is a solution, then $$ b=x_1a_1+x_2a_2+\dots+x_na_n $$ The converse is similarly clear.

In particular, if the system is consistent, the column space of $[A\mid b]$ is the same as the column space of $A$. Since the rank of a matrix is the dimension of the column space, we have that if the system is consistent, then $\operatorname{rank}A=\operatorname{rank}[A\mid b]$.

Conversely, if the ranks are equal, then the system is consistent. Indeed the column space of $A$ is a subspace of the column space of $[A\mid b]$ and so, if they have the same dimension, they are equal. So $b$ belongs to the column space of $A$.

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  • $\begingroup$ But the proof asks for inconsistent , Rank (A)<Rank[A|b] $\endgroup$ – Peter B. Apr 23 '16 at 21:03
  • $\begingroup$ @PeterB. If the ranks are different, then the system must be inconsistent! Proof by contrapositive. $\endgroup$ – egreg Apr 23 '16 at 21:06
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Well the augmented matrix is in general an n by (n+1) matrix, so if it has strictly higher rank than the original matrix, then the original matrix A had rank strictly less than n.

This is because an n by (n+1) matrix can have rank no greater than n.

Thus at least one of the n equations (for the homogeneous system defined by A) is linearly dependent of the others.

This means that there is not enough information to solve the system, since we basically have the equivalent of n-1 or fewer equations.

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