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MATH 1554 Course Notes - Fall 2026

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Section 1.1 Systems of Linear Equations

Handout 1.1 Systems of Linear Equations

 
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Definition 1.1. A Single Linear Equation.

A linear equation has the form:
\begin{equation*} a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b \end{equation*}
where, \(a_1,\dots,a_n\) and \(b\) are the coefficients, \(x_1,\dots,x_n\) are the variables (unknowns), and \(n\) is the dimension (number of variables).
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Some Examples:
  • \(2x_1+4x_2=4\) is a line in two dimensions.
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  • \(3x_1+2x_2+x_3=6\) is a plane in three dimensions.
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  • \(\displaystyle x_1 + x_2 + x_3 = x_4 + x_5\)
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  • \(\displaystyle x_1 + \sqrt{2}x_2 -\frac{1}{7} = \pi x_3\)
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Non Examples:
  • \(\displaystyle x_1 + x_2^3 = 3\)
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  • \(\displaystyle 2x_1 - 4x_2 - \sqrt{x_3} = 4\)
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  • \(\displaystyle \dfrac{1}{x_1} + x_2 + x_3 = 4\)
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  • \(\displaystyle \sin(x_1) + x_2 = 3\)
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When we have more than one linear equation, we have a system of linear equations. For example, a system of two linear equations is:
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\begin{equation*} \begin{aligned} x_1 \amp\;+\; 1.5\,x_2 \;+\; \pi x_3 \;=\; 4 \\ 5x_1 \amp\;+\; \phantom{1.5\,x_2}\;+\; 7x_3 \;=\; 5 \end{aligned} \end{equation*}
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Definition 1.2. Solution to a Linear System.

The set of all possible values of \(x_1, x_2, \ldots, x_n\) that satisfy every equation in the system is the solution, or solution set, to the system.
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Solution.
Consider finding solutions to the system given above. At this point, we don’t have a great way of determining the solutions to a system of linear equations. We can try guessing some solutions.
  • Try \((4,0,0)\text{:}\) Note this points makes the first equation true since \(4 + 1.5(0) + \pi(0) = 4\text{,}\) but it does not make the second equation true since \(5(4) + 7(0) \neq 5\text{.}\) Therefore, \((4,0,0)\) is not a solution to the system.
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  • Try \((1,2,0)\text{:}\) Note plugging this point into the first equation yields \(1 + 1.5(2) + \pi(0) = 4\) which is a true statement. Plugging the point into the second equation yields \(5(1) + 7(0) = 5\) which is also a true statement. Since the point makes both equations true, it is a solution to the system.
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  • Can you find any more solutions?
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How many solutions can a system have?
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Example 1.3.

Consider the following systems of two linear equations in two variables and determine how many solutions there are for each.
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(a)
\(\begin{aligned} x_1 - 2x_2 \amp= -1 \\ -x_1 + 3x_2 \amp= \phantom{-}3 \end{aligned}\)
Solution.
Graphing the two lines gives the following:
Cartesian axes with two lines, showing where they intersect.
The graph of the lines \(x_1 - 2x_2 = -1\) (in red) and \(-x_1 + 3x_2 = 3\) (in blue), showing their intersection point at \((3,2)\)
Figure 1.4. Two lines in the plane with a single point of intersection.
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We see that the lines only intersect at \((3,2)\text{,}\) so this is the only solution to the system.
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(b)
\(\begin{aligned} x_1 - 2x_2 \amp= -1 \\ -x_1 + 2x_2 \amp= \phantom{-}3 \end{aligned}\)
Solution.
Graphing the two lines gives the following:
Cartesian axes with two parallel lines.
The graph of the lines \(x_1 - 2x_2 = -1\) (in red) and \(-x_1 + 2x_2 = 3\) (in blue)
Figure 1.5. Two parallel lines in the plane.
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We see that the lines are parallel, so they never intersect. Therefore, there is no solution to the system.
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(c)
\(\begin{aligned} x_1 - 2x_2 \amp= -1 \\ -x_1 + 2x_2 \amp= \phantom{-}1 \end{aligned}\)
Solution.
Graphing the two lines gives the following:
Cartesian axes with two parallel lines.
The graph of the lines \(x_1 - 2x_2 = -1\) (in red) and \(-x_1 + 2x_2 = 1\) (in blue)
Figure 1.6. Two lines lying on top of one another.
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The two linear equations give the same line. Therefore, there are infinitely many points that satisfy both equations, any point that is on the line.
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Conclusion: A system of linear equations in two variables can have \(\underline{\hspace{4in}}\) solutions.
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Answer.
zero, one, or infinitely many
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A linear equation in three variables, \(a_1x_1+a_2x_2+a_3x_3=b\text{,}\) defines a plane in \(\mathbb{R}^3\text{.}\)
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Example 1.7.

How many solutions can a system of two linear equations in three variables have?
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Hint.
Play around with drawing two planes in 3-space and seeing how they intersect. Can they never intersect? What about intersect in a finite number of points? What about intersect in infinitely many points?
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Answer.
zero or infinitely many. It is impossible to have exactly one solution in this case.
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Example 1.8.

How many solutions can a system of three linear equations in three variables have?
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Hint.
Play around with drawing three planes in 3-space and seeing how they intersect. Can they never intersect? What about intersect in a finite number of points? What about intersect in infinitely many points?
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Answer.
zero, one, or infinitely many
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Conclusion: Every system of linear equations, no matter the number of equations and number of variables has \(\underline{\hspace{4in}}\) solutions.
Answer.
zero, one, or infinitely many
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As we saw in the previous examples, one way to solve a system of linear equations is by graphing. This is not practical for most examples, especially if we are working with four or more varialbes. A more useful approach to solving a linear system is to manipulate the equations through the use of row operations.
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  1. Replacement/Addition: add a multiple of one row to another.
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  2. Interchange: swap two rows.
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  3. Scaling: multiply a row by a nonzero scalar.
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Example 1.9.

Apply row operations to systematically simplify the system below to an equivalent (easier to read) form in order to more easily identify the solution(s).
\begin{alignat*}{4} x_1 \amp {}-{} \amp 2x_2 \amp {}+{} \amp x_3 \amp {}={} \amp 0 \\ \amp \amp 2x_2 \amp {}-{} \amp 8x_3 \amp {}={} \amp 8 \\ 5x_1 \amp \amp \amp {}-{} \amp 5x_3 \amp {}={} \amp 10 \end{alignat*}
Solution.
We will simplify the system using appropriate row operations. I am going to go ahead and use augmented matrices since that is what we will do from here on out.
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\(\begin{eqnarray*}\begin{bmatrix} 1 \amp -2 \amp 1 \amp \rvert \amp 0 \\ 0 \amp 2 \amp -8 \amp \rvert \amp 8 \\ 5 \amp 0 \amp -5 \amp \rvert \amp 10 \end{bmatrix} \amp \overset{R_3-5R_1 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp -2 \amp 1 \amp \rvert \amp 0 \\ 0 \amp 2 \amp -8 \amp \rvert \amp 8 \\ 0 \amp 10 \amp -10 \amp \rvert \amp 10 \end{bmatrix} \\ \amp \overset{\frac{1}{2}R_2 \rightarrow R_2}{\underset{\frac{1}{10}R_3 \rightarrow R_3}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp -2 \amp 1 \amp \rvert \amp 0 \\ 0 \amp 1 \amp -4 \amp \rvert \amp 4 \\ 0 \amp 1 \amp -1 \amp \rvert \amp 1\end{bmatrix} \\ \amp \overset{R_1 + 2R_2 \rightarrow R_1}{\underset{R_3 - R_2 \rightarrow R_3}{\Huge \sim}} \amp \begin{bmatrix}1 \amp 0 \amp -7 \amp \rvert \amp 8 \\ 0 \amp 1 \amp -4 \amp \rvert \amp 4 \\ 0 \amp 0 \amp 3 \amp \rvert \amp -3 \end{bmatrix} \\ \amp \overset{\frac{1}{3}R_3 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 0 \amp -7 \amp \rvert \amp 8 \\ 0 \amp 1 \amp -4 \amp \rvert \amp 4 \\ 0 \amp 0 \amp 1 \amp \rvert \amp -1 \end{bmatrix} \\ \amp \overset{R_1 + 7R_3 \rightarrow R_1}{\underset{R_2 + 4R_3 \rightarrow R_2}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 0 \amp 0 \amp \rvert \amp 1 \\ 0 \amp 1 \amp 0 \amp \rvert \amp 0 \\ 0 \amp 0 \amp 1 \amp \rvert \amp -1 \end{bmatrix} \end{eqnarray*}\)
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Therefore, we have that the only solution is \((1,0,-1)\text{.}\)
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It’s redundant to write variables repeatedly, so we use a augmented matrix to simplify the notation for the system. For example, the previous system
\begin{alignat*}{4} x_1 \amp {}-{} \amp 2x_2 \amp {}+{} \amp x_3 \amp {}={} \amp 0 \\ \amp \amp 2x_2 \amp {}-{} \amp 8x_3 \amp {}={} \amp 8 \\ 5x_1 \amp \amp \amp {}-{} \amp 5x_3 \amp {}={} \amp 10 \end{alignat*}
Can be writen as the augmented matrix
\begin{equation*} \left[\begin{array}{rrr|r} 1 \amp -2 \amp 1 \amp 0 \\ 0 \amp 2 \amp -8 \amp 8 \\ 5 \amp 0 \amp -5 \amp 10 \end{array}\right] \end{equation*}
The vertical bar separates the coefficient matrix (for \(x_1,x_2,x_3\)) from the constants.
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Example 1.10.

Determine if the following system has a solution.
\begin{alignat*}{4} \amp \amp x_2 \amp {}-{} \amp 4x_3 \amp {}={} \amp 8 \\ 2x_1 \amp {}-{} \amp 3x_2 \amp {}+{} \amp 2x_3 \amp {}={} \amp 1 \\ 4x_1 \amp {}-{} \amp 8x_2 \amp {}+{} \amp 12x_3 \amp {}={} \amp 1 \end{alignat*}
Solution.
Note the following:
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\(\begin{eqnarray*} \begin{bmatrix} 0 \amp 1 \amp -4 \amp \rvert \amp 8 \\ 2 \amp -3 \amp 2 \amp \rvert \amp 1 \\ 4 \amp -8 \amp 12 \amp \rvert \amp 1 \end{bmatrix} \amp \overset{\frac{1}{4}R_3 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 0 \amp 1 \amp -4 \amp \rvert \amp 8 \\ 2 \amp -3 \amp 2 \amp \rvert \amp 1 \\ 1 \amp -2 \amp 3 \amp \rvert \amp \frac{1}{4} \end{bmatrix} \\[1em] \amp \overset{\text{swap } R_1 \text{ and } R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp -2 \amp 3 \amp \rvert \amp \frac{1}{4} \\ 2 \amp -3 \amp 2 \amp \rvert \amp 1 \\ 0 \amp 1 \amp -4 \amp \rvert \amp 8 \end{bmatrix} \\[1em] \amp \overset{R_2 - 2R_1 \rightarrow R_2}{\Huge \sim} \amp \begin{bmatrix} 1 \amp -2 \amp 3 \amp \rvert \amp \frac{1}{4} \\ 0 \amp 1 \amp -4 \amp \rvert \amp \frac{1}{2} \\ 0 \amp 1 \amp -4 \amp \rvert \amp 8 \end{bmatrix} \\[1em] \amp \overset{R_3 - R_2 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp -2 \amp 3 \amp \rvert \amp \frac{1}{4} \\ 0 \amp 1 \amp -4 \amp \rvert \amp \frac{1}{2} \\ 0 \amp 0 \amp 0 \amp \rvert \amp \frac{15}{2} \end{bmatrix} \end{eqnarray*}\)
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Note that the final row suggests that \(0 = \frac{15}{2}\) which is impossible, no matter the choice of \(x_1, x_2, x_3\text{.}\) Therefore, there is no solution. So the system is inconsistent.
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Definition 1.11. Consistent.

A linear system is consistent if it has at least one solution. Otherwise, it is inconsistent.
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Definition 1.12. Row Equivalence.

Two matrices are row equivalent if a sequence of row operations transforms one matrix into the other.
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Note: If the augmented matrices of two linear systems are row equivalent, they have the same solution set.
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Fundamental Questions: Two questions that we will revisit many times throughout the course:
  1. Does a given linear system have a solution? In other words, is it consistent?
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  2. If it is consistent, is the solution unique?
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