An independent system of equations is a system with a unique solution. All the examples and exercises in 7.2 were examples of independent systems.
Consider the following system:
\( \mathbb{S}_1\)
\(\begin{eqnarray*} x + y + z &=& 1\\ y + z &=& 2 \end{eqnarray*}\)
Since there are fewer equations in the system than variables, there will not be a unique solution. In order to characterize the solutions of this system, we add a third equation to represent all possible values of \(z\):
\( \mathbb{S}_2\)
\(\begin{eqnarray*} x + y + z &=& 1\\ y + z &=& 2\\ z &=& t \end{eqnarray*}\)
where \(t\) can be any real number. Using back substitution, we find
\( \mathbb{S}_3\)
\(\begin{eqnarray*} x + y + z &=& 1\\ y\quad &=& 2 - t\\ z &=& t \end{eqnarray*}\)
and
\( \mathbb{S}_4\)
\(\begin{eqnarray*} x \quad\quad &=& -1\\ y\quad &=& 2 - t\\ z &=& t \end{eqnarray*}\)
So the solution is the set of all ordered triplets of the form \(( - 1, 2 - t, t )\) where \(t\) is a real number.
This is an example of a dependent system.
Dependent systems have an infinite number of solutions, since there is a different solution for each value of \(t\).
Solve the system
\( \mathbb{S}_1\)
\(\begin{eqnarray*} x + 2 y + z &=& 0\\ y - z &=& 2\\ \end{eqnarray*}\)Consider the system
\( \mathbb{S}_1\)
\(\begin{eqnarray*} x + y + z &=& 1\\ y + z &=& 2\\ y + z &=& 3 \end{eqnarray*}\)
If we attempt to triangularize the system by \(E_3 \rightarrow - E_2 + E_3\) we get the system
\( \mathbb{S}_2\)
\(\begin{eqnarray*} x + y + z &=& 1\\ y + z &=& 2\\ 0 &=& 1 \end{eqnarray*}\)
Since the last equation is false, this system has no solution.
The system is said to be inconsistent.
Solve the system
\( \mathbb{S}_2\)
\(\begin{eqnarray*} x - y + z &=& 1\\ x\quad\,\,\,\, + z &=& 1\\ x + y + z &=& 2 \end{eqnarray*}\)