Let
\(A = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 2 & 1\\ 0 & 1 & 0 \end{bmatrix}\),
\(b = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}\),
\(b' = \begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix}\), and
\(b'' = \begin{bmatrix} 3 \\ 2 \\ 1\end{bmatrix}\) be defined over the real
numbers.
Does each of the systems \(Ax = b\), \(Ax = b'\), and \(Ax = b''\) have
a solution?
The answer is “No”.
The extended augmented matrix is
\(\left[\begin{array}{ccc|ccc}
1 & 2 & -1 & 3 & 2 & 3 \\
-1 & 2 & 1 & 1 & -2 & 2 \\
0 & 1 & 0 & 1 & 0 & 1
\end{array}\right]\).
Performing \(R_2 \leftarrow R_2 + R_1\), we get
\(\left[\begin{array}{ccc|ccc}
1 & 2 & -1 & 3 & 2 & 3 \\
0 & 4 & 0 & 4 & 0 & 5 \\
0 & 1 & 0 & 1 & 0 & 1
\end{array}\right]\).
Performing \(R_2 \leftarrow R_2 - 4R_3\), we get
\(\left[\begin{array}{ccc|ccc}
1 & 2 & -1 & 3 & 2 & 3 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 1
\end{array}\right]\).
Performing \(R_1 \leftarrow R_1 - 2R_3\), we get
\(\left[\begin{array}{ccc|ccc}
1 & 0 & -1 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 1
\end{array}\right]\).
Swapping rows 2 and 3, we get
\(\left[\begin{array}{ccc|ccc}
1 & 0 & -1 & 1 & 2 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 1
\end{array}\right]\).
From this, we see that \(Ax = b\) and \(Ax = b'\) have solutions but
\(Ax = b''\) does not.