# System of linear equations with a unique solution

Suppose the system of linear equations $$AX=B$$ has a unique solution for some $$B$$ . Prove that rref of $$A$$ is the same as $$I_n$$. ($$A$$ is a square matrix)

My try : Because the system has a unique solution therefore it's possible to write $$x_1 = a , x_2 = b , \dots , x_n = z$$ . In the matrix form, for the coefficient matrix, we have :

$$\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}$$ I don't know whether that's enough for proving that statement or not .

• If $\mathbf{x}_0$ is a solution to $AX=0$ and $\mathbf{x}_B$ is a solution to $AX=B$, then $\mathbf{x}_B+\mathbf{x}_0$ is a solution to $AX=B$. So if $AX=B$ has a unique solution, then $AX=0$ must also have a unique solution. Do you know anything about when homogeneous systems have unique solutions? – Arturo Magidin Jun 6 at 20:26
• @ArturoMagidin Unfortunately no . – S.H.W Jun 6 at 20:32
• Are you familiar with the concepts of rank and nullity? – paulinho Jun 6 at 20:48
• @paulinho I'm only familiar with the rank. It is the number of leading 1's in ref form. – S.H.W Jun 6 at 20:50
• Counterexample: $\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}X=\begin{bmatrix}1\\1\\2\end{bmatrix}$. This has a unique solution, but the coefficient matrix isn’t square, so its rref can’t possibly equal the identity matrix. Seems like there are some missing conditions from the claim you’re trying to prove. – amd Jun 7 at 0:47