Theorem:Given a system of linear equations Ax=b whereA∈Mm×n(R),x∈Rcoln,b∈Rcolm.Deduce that a solution x exists if and only if rank(A∣b)=rank(A) where A∣b is the augmented coefficient matrix of this systemI am having trouble proving the above theorem from my Linear Algebra course, I understand that A|b must reduce under elementary row operations to a form which is consistent but I don't understand exactly why the matrix A|b need have the same rank as A for this to happen.Please correct me if I am mistaken

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Theorem:Given a system of linear equations Ax=b whereA∈Mm×n(R),x∈Rcoln,b∈Rcolm.Deduce that a solution x exists if and only if rank(A∣b)=rank(A) where A∣b is the augmented coefficient matrix of this systemI am having trouble proving the above theorem from my Linear Algebra course, I understand that A|b must reduce under elementary row operations to a form which is consistent but I don&#039;t understand exactly why the matrix A|b need have the same rank as A for this to happen.Please correct me if I am mistaken

Josef Beil · Accepted Answer

To see this note that rank of the matrix is dimension of the span of columns of the matrix. Now if Ax=b has solution, then it means that some linear combination of columns of A gives us b, which implies that b lies in span(A) and so rank(A∣b)=rank(A). You can argue similarly in the reverse direction.

Ulgelmorgs · Accepted Answer

To get started,  If the system contains a row such that [ 0 0 0 0 | b ] with b=/=0 then the system is inconsistent and has no solution. The rank is the number of pivots matrix X has in echelon form, whereby b is the pivot in this row.  Let M=[A ,B], the augmented matrix, where A is the original matrix.  The system has a solution if and only if rank(A)=rank(M).  If b is a pivot of M then the solution does not exist. Hence rank(m)&amp;gt;rank(a).

Theorem: Given a system of linear equations Ax=b where A \in M_{m \times n}(\ma

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