Let \(A\) be an \(m \times n\) matrix over the field \(F\), and consider the system of equations \(AX = Y\). Prove that this system of equations has a solution if and only if the row rank of \(A\) is equal to the row rank of the augmented matrix of the system.
Solution
Let \([A\mid Y]\) denote the augmented matrix.
If \(AX=Y\) has a solution, then \(Y\) is a linear combination of the columns of \(A\). Hence adjoining \(Y\) as an extra column does not enlarge the column space, so \[ \operatorname{rank}[A\mid Y]=\operatorname{rank}A. \] Since row rank equals column rank for every matrix, the row ranks are equal as well.
Conversely, suppose \[ \operatorname{rowrank}[A\mid Y]=\operatorname{rowrank}A. \] By equality of row and column rank, \[ \operatorname{colrank}[A\mid Y]=\operatorname{colrank}A. \] Thus adjoining \(Y\) does not enlarge the column space of \(A\), so \(Y\) lies in the span of the columns of \(A\). Therefore there is a column vector \(X\) with \[ AX=Y. \] Hence the system is solvable if and only if the row ranks of \(A\) and its augmented matrix are equal.