Earlier, we determined this system has infinitely-many solutions.
(a)
Let \(x_1=a\) and write the solution set in the form \(\setBuilder
{
\left[\begin{array}{c} a \\ \unknown \\ \unknown \end{array}\right]
}{
a \in \IR
}
\text{.}\)
(b)
Let \(x_2=b\) and write the solution set in the form \(\setBuilder
{
\left[\begin{array}{c} \unknown \\ b \\ \unknown \end{array}\right]
}{
b \in \IR
}
\text{.}\)
(c)
Which of these was easier? What features of the RREF matrix \(\left[\begin{array}{ccc|c}
\markedPivot{1} & 2 & 0 & 4 \\
0 & 0 & \markedPivot{1} & -1
\end{array}\right]\) caused this?
Definition1.4.3.
Recall that the pivots of a matrix in \(\RREF\) form are the leading \(1\)s in each non-zero row.
The pivot columns in an augmented matrix correspond to the bound variables in the system of equations (\(x_1,x_3\) below). The remaining variables are called free variables (\(x_2\) below).
Don’t forget to correctly express the solution set of a linear system. Systems with zero or one solutions may be written by listing their elements, while systems with infinitely-many solutions may be written using set-builder notation.
Inconsistent: \(\emptyset\) or \(\{\}\)
(not \(0\) or \(\left[\begin{array}{c}0\\0\\0\end{array}\right]\))
Consistent with one solution: e.g. \(\setList{ \left[\begin{array}{c}1\\2\\3\end{array}\right] }\)
(not just \(\left[\begin{array}{c}1\\2\\3\end{array}\right]\))
Consistent with infinitely-many solutions: e.g. \(\setBuilder
{
\left[\begin{array}{c}1\\2-3a\\a\end{array}\right]
}{
a\in\IR
}\)
(not just \(\left[\begin{array}{c}1\\2-3a\\a\end{array}\right]\) )
Activity1.4.6.
Consider the following system of linear equations.
All of the following statements are not accurate or otherwise incorrect; identify what is problematic about the statements and correct them.
(a)
The matrix \(B\) is consistent with infinitely many solutions.
(b)
The solution set is given by \(\left[\begin{array}{c} a + b - 3 \\ -3 \, a - 2 \, b + 1 \\ a \\ b - 3 \\ b \end{array}\right]\text{.}\)
(c)
The variables \(x_3,x_5\) are free. Setting them equal to \(a,b\) respectively and solving for the bound variables, the solution set to the linear system is given by \(\left\{ \left[\begin{array}{c} a + b - 3 \\ -3 \, a - 2 \, b + 1 \\ b - 3 \end{array}\right] \,\middle|\, a,b \in\mathbb R \right\}\text{.}\)
Construct a system of 3 equations in 3 variables having:
0 free variables
1 free variable
2 free variables
In each case, solve the system you have created. Conjecture a relationship between the number of free variables and the type of solution set that can be obtained from a given system.
Exploration1.4.11.
For each of the following, decide if it’s true or false. If you think it’s true, can we construct a proof? If you think it’s false, can we find a counterexample?
If the coefficient matrix of a system of linear equations has a pivot in the rightmost column, then the system is inconsistent.
If a system of equations has two equations and four unknowns, then it must be consistent.
If a system of equations having four equations and three unknowns is consistent, then the solution is unique.
Suppose that a linear system has four equations and four unknowns and that the coefficient matrix has four pivots. Then the linear system is consistent and has a unique solution.
Suppose that a linear system has five equations and three unknowns and that the coefficient matrix has a pivot in every column. Then the linear system is consistent and has a unique solution.