# If A and B are row equivalent, and A$\mathbf x$ = $\mathbf b$ has a solution, then B$\mathbf x$ = $\mathbf b$ has a solution

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

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 $$Bx=\left[\begin{array}{c}1\\1 \end{array}\right]$$