If the system Ax=b is consistent for every n x 1 matrix in b, then A is invertible.

Question

I have trouble understanding the proof here. I spent an hour trying to understand it but I give up. Can anyone help me with it?

Answer 1

At the beginning, they're saying that since $Ax = b$ is consistent no matter which vector $b$ is, then in particular the system \begin{equation} Ax = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \end{equation} is consistent. In other words, there exists a vector $x_1$ such that \begin{equation} Ax_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{equation}

And, since $Ax = b$ is consistent no matter what $b$ is, it must be true that the system \begin{equation} Ax = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \end{equation} is consistent. In other words, there exists a vector $x_2$ such that \begin{equation} Ax_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{equation} And so on. Now take these vectors $x_1, x_2,\ldots$ and make them the columns of a matrix $C$: \begin{equation} C = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}. \end{equation} You can check that $AC = I$, the identity matrix. So $A$ is invertible.