In earlier lectures, we discussed systems of linear equations and their solutions. A solution of a linear system is a value for each variable that makes every equation in the system true. We also discussed the row-reduction process, which allows us to manipulate the augmented matrix of a linear system and put it into echelon form or reduced echelon form. Importantly, these row operations are reversible, which means that row operations preserve the solutions of the original equations.
Definition. Given a matrix, any matrix that can be obtained from that matrix through row operations is said to be row-equivalent to the original matrix.
We will illustrate how to use echelon form to find solutions to a linear system using an example.
Example 1. Analyze the solutions of this linear system: \[ \begin{eqnarray*} x_1 + x_2 - 6x_3 + x_4 && = && -4 \\ x_1 - 9x_3 + x_4 && = && -3 \\ 2x_2 + 6x_3 + x_4 && = && -9 \end{eqnarray*} \]
We first set up the corresponding augmented matrix for this system: \[ \begin{bmatrix} 1 & 1 & -6 & 1 & -4 \\ 1 & 0 & -9 & 1 & -3 \\ 0 & 2 & 6 & 1 & 9 \end{bmatrix} \]
Next, we row reduce the matrix using the process we learned in Lecture 4: \[ \begin{bmatrix} 1 & 1 & -6 & 1 & -4 \\ 1 & 0 & -9 & 1 & -3 \\ 0 & 2 & 6 & 1 & 9 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & -9 & 0 & 4 \\ 0 & 1 & 3 & 0 & -1 \\ 0 & 0 & 0 & 1 & -7 \end{bmatrix} \]
Now, take the reduced matrix and write the corresponding equations: \begin{eqnarray*} x_1 - 9x_3 && = && 4 \\ x_2 + 3x_3 && = && -1 \\ x_4 && = && -7 \end{eqnarray*}
The variables corresponding to pivot columns are called basic variables, and the other variables are called free variables. In this example, our reduced matrix has pivots in the first, second, and fourth columns, so the basic variables are \( x_1 \), \( x_2 \), and \( x_4 \). Since there is no pivot in the third column, \( x_3 \) is a free variable. Note that the fifth column also does not have a pivot, but we have no "\( x_5 \)" in this system. The fifth column corresponds to the numbers on the right-hand sides of our equations!
Each of these "reduced" equations contains exactly one basic variable, and each basic variable appears in a different equation. Any free variables can appear multiple times in different equations. These facts derive from the criteria for echelon form that we learned in Lecture 3.
We now solve each equation for the basic variable, and indicate any variables that are free: \[ \begin{array}{ll} x_1 && = 9x_3+4 \\ x_2 && = -3x_3-1 \\ x_3 && \mbox{is free} \\ x_4 && = -7 \end{array} \]
Each value of \( x_3 \) gives a new solution of the original system. For example, if \( x_3 = 2 \), this generates the solution \( (22, -7, 2, -7) \). The original system of equations has infinitely many solutions, one for each value of \( x_3 \). \( \Box \)
Here is the general process for solving a system of linear equations:
We now have a process for explicitly finding the solutions of a given system of linear equations. However, sometimes we will only need to characterize the solutions of a system. That is, we will only want to know whether a particular system of equations has zero, one, or infinitely many solutions. For this, echelon form (rather than reduced echelon form) is sufficient.
Definition. A system of linear equations is consistent if it has one or more solutions. The system is inconsistent if it has no solutions.
Example 2. Determine whether the following system of equation is consistent or inconsistent: \begin{eqnarray*} 2x_1 + 6x_3 + 7x_4 && = && 7 \\ 6x_1 + 18x_3 + 24x_4 && = && 6 \\ 2x_1 - x_2 + 5x_4 && = && 14 \\ 2x_2 + 12x_3 + 4x_4 && = && -10 \end{eqnarray*}
We begin by setting up the augmented matrix for this system and row-reducing. While we could go all the way to reduced echelon form, we can get to echelon form with fewer steps. This can be helpful if the matrix is particularly large, or if we are solving this problem without the aid of technology. Suppose that we reduce our augmented matrix to the following echelon form: \[ \begin{bmatrix} 2 & -1 & 0 & 5 & 10 \\ 0 & 1 & 6 & 2 & -7 \\ 0 & 0 & 0 & 1 & -5 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix} \]
This echelon form is sufficient for us to see the pivot positions in our augmented matrix. Specifically, we see that there is a pivot in the last column. The row containing that pivot corresponds to the equation \( 0 = 4 \), which is always false no matter what values we use for our variables. This means that the system of equations has no solutions because there is no way to make all of the equations true. \( \Box \)
In order to characterize the solutions of a system of linear equations, it is enough to know the pivot positions of its corresponding augmented matrix: