Consider the system \(Ax = b\) defined over the real numbers where \(A=\begin{bmatrix} 1 & 2 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 3 \end{bmatrix}\), \(x = \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}\), and \(b = \begin{bmatrix} 4\\5\\6\end{bmatrix}\). Note that \(A\) is in reduced row-echelon form with no zero row. So the system has at least one solution. As there are two free variables \(x_2\) and \(x_5\), the system has infinitely many solutions. Is there a succint way to describe all the solutions?
The key is to set each free variable to a parameter. Here, setting \(x_2 = s\) and \(x_5 = t\), the system gives \begin{eqnarray} x_1 + 2s - t & = & 4 \\ x_3 + 2t & = & 5 \\ x_4 + 3t & = & 6 \\ \end{eqnarray} Solving for \(x_1\), \(x_3\), and \(x_4\) gives \(x_1 = 4 - 2s + t\), \(x_3 = 5-2t\), \(x_4 = 6 - 3t\).
So the solutions are \(\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix} 4-2s+t\\s\\5-2t\\6-3t\\t\end{bmatrix} = \begin{bmatrix} 4\\0\\5\\6\\0\end{bmatrix} +s\begin{bmatrix} -2\\1\\0\\0\\0\end{bmatrix} +t\begin{bmatrix} 1\\0\\-2\\-3\\1\end{bmatrix}\) where \(s\) and \(t\) are arbitrary numbers.
Consider the following system defined over \(GF(2)\). \begin{eqnarray} x_1 + x_3 & = & 1\\ x_2 + x_3 & = & 0\\ \end{eqnarray} Here, \(x_3\) is a free variable. So we can set \(x_3\) to a paramter and write \(x_1\) and \(x_2\) in terms of \(x_3\). As \(0\) and \(1\) are the only elements in \(GF(2)\), there are only two solutions.
Setting \(x_3 = 0\) gives the solution \(\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 1\\0\\0\end{bmatrix}\).
Setting \(x_3 = 1\) gives the solution \(\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 0\\1\\1\end{bmatrix}\).
For each of the following system, give a succinct description of the set of solutions.
\(\begin{array}{r} x -y + z = 1 \\ y - 2z = 2 \end{array}\)
\(3x - 4y + z= 12\)
\(\begin{array}{r} x_1 - x_3 + x_4 = 0 \\ x_2 - 2x_4 = 0 \end{array}\)