91Ó°ÊÓ

First, let's assume that the given linear system \(A\mathbf{x}=\mathbf{b}\) is consistent. This means that there exists a solution \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{b}\). Now consider the augmented matrix \((A | \mathbf{b})\). We will apply Gaussian elimination or row reduction to this matrix to obtain an equivalent matrix in row echelon form, which we will call \((A' | \mathbf{b}')\). Since the linear system is consistent, there will be no rows in \(A'\) of the form \([0 \dots 0 | 1]\), because such a row would indicate an inconsistent system. Therefore, the number of nonzero rows in \(A'\) would be the same as the number of nonzero rows in \(A\), after applying the same row reduction process to \(A\) alone. The rank of a matrix is defined as the number of nonzero rows in its row echelon form, so in this case, the rank of \(A\) is equal to the rank of \((A | \mathbf{b})\).

Now, let's assume that the rank of \((A | \mathbf{b})\) equals the rank of \(A\). Apply Gaussian elimination to both matrices separately to obtain their row echelon forms, denoted as \(A'\) and \((A' | \mathbf{b}')\) respectively. Since the ranks are equal, this means the number of nonzero rows in both \(A'\) and \((A' | \mathbf{b}')\) are the same. This implies that there cannot be any rows in \((A' | \mathbf{b}')\) of the form \([0 \dots 0 | 1]\), because, if such a row existed, the rank of \((A | \mathbf{b})\) would be greater than the rank of \(A\). Since the row echelon form of the augmented matrix does not contain any inconsistent rows, the linear system \(A\mathbf{x}=\mathbf{b}\) must be consistent.

We have shown that, if the linear system is consistent, the ranks of \(A\) and \((A | \mathbf{b})\) are equal and that if the ranks of \(A\) and \((A | \mathbf{b})\) are equal, the linear system is consistent. Thus, the given linear system \(A\mathbf{x}=\mathbf{b}\) is consistent if and only if the rank of \((A | \mathbf{b})\) equals the rank of \(A\).