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Assume towards contradiction that B$\mathbf x$ = $\mathbf b$ has no solutions. Therefore the augmented matrix is inconsistent.

Let $E_1$, $E_2$,..., $E_p$ be a sequence of elementary matrices representing row operations that render B row equivalent to some matrix A.

Thus we have:

A$\mathbf x$ = $E_1 \cdot E_2 \cdot ... E_p \cdot$ B = $\mathbf b$

Since we know that all elementary matrices are invertible, it follows that:

B = $(E_1 \cdot E_2 \cdot ... E_p)^{-1} \mathbf b$

At this point I feel that perhaps the proposition might be false

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Your feeling is correct.

For instance:

If $A=\left[\begin{array}{cc}1&1\\1&1 \end{array}\right]$, then $Ax=\left[\begin{array}{c}1\\1 \end{array}\right]$ has a solution.

Now $B=\left[\begin{array}{cc}1&1\\0&0 \end{array}\right]$ is row equivalent to $A$, but there is no solution to $Bx=\left[\begin{array}{c}1\\1 \end{array}\right]$

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