Let \(A\in M_{m\times n}(F)\) and \(b\in F^m\). Let \([A\mid b]\) be the augmented matrix.
- Show that \(Ax=b\) has a solution if and only if \[ \operatorname{rank}A=\operatorname{rank}[A\mid b]. \]
- Assuming the system is consistent, show that the solution is unique if and only if \[ \operatorname{rank}A=n. \]
Solution
1. The system is consistent iff \(b\) lies in the column space of \(A\).
Writing the columns of \(A\) as \(a_1,\ldots,a_n\), the equation \[ Ax=b \] means exactly that \[ b=x_1a_1+\cdots+x_na_n. \] Thus a solution exists iff \(b\in\operatorname{col}(A)\). This is equivalent to adjoining \(b\) as an extra column without increasing the column-space dimension, namely \[ \operatorname{rank}[A\mid b]=\operatorname{rank}A. \]
2. A consistent system has a unique solution iff \(\ker A=0\).
Fix one solution \(x_0\). Then every solution is of the form \[ x_0+v, \qquad v\in\ker A, \] because \[ A(x_0+v)=b \iff Av=0. \] Hence the solution is unique iff \(\ker A=0\). By rank-nullity, \[ n=\operatorname{rank}A+\dim\ker A, \] so \(\ker A=0\) iff \(\operatorname{rank}A=n\).