Theorem 1.5.1. Solution Forms of Linear Systems.
Every linear system of equations has exactly one solution, infinite solutions, or no solution.
Section 1.5 Consistent and Inconsistent Systems
We’ve learned how to take a linear system, turn it into an augmented matrix, and use Gauss-Jordan elimination to put the matrix into reduced row echelon form. We used the reduced row echelon form to see what the solution was to the original linear system, and in all the examples we’ve seen so far, there was exactly one solution.
In this section, we consider different possibilities for what the solutions to a linear system can be and see how we can identify those different possibilities from the reduced row echelon form that results after performing Gauss-Jordan elimination.
Subsection Prepare
Subsubsection Types of solutions
In Activity 1.1.3, Example 1.1.7, Example 1.1.8, Example 1.3.5, and all of our other examples of linear systems, there was exactly one solution, that is, one possible assignment of values to the variables which made all of the equations in the system true at the same time. We now consider if that is always the case or not.
First, consider the following linear system with one equation:
\begin{equation*}
x-y=0\text{.}
\end{equation*}
This equation is true whenever \(x=y\text{.}\) We can picture all of the possible solutions by thinking of the graph of the equation \(y=x\text{.}\)
Each single equation with two variables can be viewed in this way, as a line in the plane. Let’s now consider systems of linear equations with two equations and two variables, so that there are two lines in the plane.
Consider the linear system
\begin{align*}
x-y\amp =0\\
x+y\amp =2\text{.}
\end{align*}
Since each equation can be viewed as a line and since the slopes of the lines are different, we know the lines will intersect somewhere, and that point of intersection is the only solution to the system.
Now consider the linear system
\begin{align*}
x-y\amp =0\\
2x-2y\amp =0\text{.}
\end{align*}
While this system has two equations, the second is a multiple of the first. The thicker line is used to represent that we’ve drawn the same line twice. In this case, we have an infinite solution set, just as if we only had the one equation \(x-y=0\text{.}\)
Finally, consider the linear system
\begin{align*}
x-y\amp =0\\
x-y\amp =2\text{.}
\end{align*}
If the difference between \(x\) and \(y\) is \(0\text{,}\) it can’t also be \(2\text{.}\) This linear system has no solution. We observe that the two lines are parallel and never intersect.
It becomes harder to visualize when we add variables, but no matter how many equations and variables we have, solutions to linear equations always come in one of three forms: exactly one solution, infinite solutions, or no solution. This is a fact that we will not prove here, but it deserves to be stated.
Visually, we think of the three options below, even though linear systems often have more than two equations and more than two variables.
Subsubsection Determining Consistency
How can we tell whether a given system of linear equations has \(0\text{,}\) \(1\text{,}\) or infinitely many solutions? The answer to this question lies with the reduced row echelon form of its corresponding augmented matrix.
Remark 1.5.3. A note on using technology.
When we are learning a new technique or procedure, it’s good to go through all the steps ourselves because understanding the process has benefits. That’s why we typically learn how to add numbers using manipulatives like blocks or fingers, with the goal of eventually being able to add faster without any aids; in fact, we eventually add numbers we never could have represented with physical manipulatives. The focus shifted from the ability to add for its own sake to solving a problem which required addition as a step.
In a similar way, once we are able to perform Gauss-Jordan elimination by hand for small systems, we move to using technology to quickly obtain the reduced row echelon form of a matrix. The focus is no longer on our ability to row reduce for its own sake but on answering a question which requires interpreting the result of row reducing a matrix.
There are many apps, websites, etc., which can produce the reduced row echelon form of a matrix very quickly. This book will use SageMath, a language which is built on Python and free. Hit the “Evaluate (Sage)” button below to see both the matrix and its reduced row echelon form.
The “QQ” in the Sage cell above is present for a technical reason, and the rest of syntax means that \(A\) is being defined as a \(2\times 3\) matrix, whose first row is \([1, 2, 3]\) and whose second row is \([4,5,6]\text{.}\)
Try modifying the contents of the above Sage cell to calculate the reduced row echelon form of a different matrix. Don’t worry about messing anything up beyond repair; refreshing the page will reset the Sage cell back to its initial state.
From now on, we will give the reduced row echelon form of the matrix without showing work.
We begin by distinguishing between systems which have a solution (either one or infinitely many) and those which have no solution.
Definition 1.5.4. Consistency of Linear Systems.
A system of linear equations is consistent if it has at least one solution. A linear system is inconsistent if it does not have a solution.
What happens in the reduced row echelon form if a linear system has no solutions?
Example 1.5.5. A system with no solutions.
Find the solution to the linear system
\begin{equation*}
\begin{array}{ccccccc} x_1\amp +\amp x_2\amp +\amp x_3\amp =\amp 1\\ x_1\amp +\amp 2x_2\amp +\amp x_3\amp =\amp 2\\ 2x_1\amp +\amp 3x_2\amp +\amp 2x_3\amp =\amp 0 \end{array}
\end{equation*}
Answer: We put the corresponding augmented matrix into reduced row echelon form.
\begin{equation*}
\left[\begin{array}{cccc} 1\amp 1\amp 1\amp 1\\1\amp 2\amp 1\amp 2\\ 2\amp 3\amp 2\amp 0\\ \end{array} \right] \rrefarrow \left[\begin{array}{cccc} 1\amp 0\amp 1\amp 0\\ 0\amp 1\amp 0\amp 0\\ 0\amp 0\amp 0\amp 1\\ \end{array} \right]
\end{equation*}
Now let us take the reduced matrix and write out the corresponding equations. The first two rows give us the equations
\begin{align*}
x_1+x_3\amp =0\\
x_2 \amp = 0\text{.}
\end{align*}
So far, so good. However the last row gives us the equation
\begin{equation*}
0x_1+0x_2+0x_3 = 1\text{.}
\end{equation*}
There are no values of \(x_1, x_2\) and \(x_3\) which can make \(0=1\text{.}\) Therefore, no solution exists, and this system is inconsistent.
The key takeaway from Example 1.5.5 is: if any row has a leading \(1\) in the last column after row reducing the augmented matrix of a linear system, the system is inconsistent. Otherwise, the system is consistent, which means it has at least 1 solution.
Activity 1.5.6. Consistent or inconsistent?
For each of the following linear systems, say whether it is consistent or inconsistent.
(a)
- Consistent
- Correct!
- Inconsistent
- There is no row with a leading \(1\) in the last column. The first row says that \(x_1=-2\) and the second row says that \(x_2=1\text{.}\) If a linear system has a solution, it is consistent.
The linear system whose augmented matrix in reduced row echelon form is
\begin{equation*}
\begin{bmatrix} 1\amp 0 \amp -2\\0\amp 1 \amp 1
\end{bmatrix}\text{.}
\end{equation*}
(b)
- Consistent
- Correct!
- Inconsistent
- There is no row with a leading \(1\) in the last column. The first row says that \(x_1+x_2=-3\) and the second row says that \(0x_1+0x_2=0\text{,}\) which is true.
The linear system whose augmented matrix in reduced row echelon form is
\begin{equation*}
\begin{bmatrix} 1 \amp 1 \amp -3 \\ 0\amp 0\amp 0
\end{bmatrix}\text{.}
\end{equation*}
(c)
- Consistent
- There is a row with a leading \(1\) in the last column. The third row says that \(0x_1+0x_2=1\text{,}\) which is not true for any possible values of \(x_1\) and \(x_2\text{.}\) A system with no solutions is inconsistent.
- Inconsistent
- Correct!
The linear system whose augmented matrix has been row reduced to
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp -3 \\ 0\amp 1\amp 1 \\0\amp 0\amp 1 \\ 0 \amp 0 \amp 0
\end{bmatrix}\text{.}
\end{equation*}
(d)
- Consistent
- There is no common point of intersection. Each line is an equation, and there is no point which satisfies all three equations. The linear system pictured has no solution, which means it’s inconsistent.
- Inconsistent
- Correct!
The linear system pictured below. 
We’ve now seen a system which has no solution and systems which have exactly one solution such as Example 1.3.5, so let’s look next at a system which has infinitely many solutions.
Example 1.5.7. A system with infinitely many solutions.
Find the solution to the linear system
\begin{equation*}
\begin{array}{ccccccc} \amp \amp x_2\amp -\amp x_3\amp =\amp 3\\ x_1\amp \amp \amp +\amp 2x_3\amp =\amp 2\\ \amp \amp -3x_2\amp +\amp 3x_3\amp =\amp -9 \end{array}
\end{equation*}
Answer: To find the solution, put the corresponding matrix into reduced row echelon form.
\begin{equation*}
\left[\begin{array}{cccc} 0\amp 1\amp -1\amp 3\\ 1\amp 0\amp 2\amp 2\\ 0\amp -3\amp 3\amp -9 \end{array} \right] \rrefarrow \left[\begin{array}{cccc} 1\amp 0\amp 2\amp 2\\ 0\amp 1\amp -1\amp 3\\ 0\amp 0\amp 0\amp 0 \end{array} \right]
\end{equation*}
Now convert this reduced matrix back into equations. We have
\begin{align*}
x_1 + 2x_3 \amp = 2\\
x_2-x_3\amp =3\text{.}
\end{align*}
We solve for \(x_1\) and \(x_2\) in terms of \(x_3\) and get
\begin{align*}
x_1 \amp = 2-2x_3\\
x_2\amp =3+x_3
\end{align*}
These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. There is no restriction on what \(x_3\) must be; it is free to take on the value of any real number. Since we have infinite choices for the value of \(x_3\text{,}\) we have infinitely many solutions.
In the previous example we used the word “free” to describe a certain variable. What exactly is a free variable? How do we recognize which variables are free and which are not?
Definition 1.5.8. Dependent and Independent Variables.
Consider the reduced row echelon form of an augmented matrix of a consistent system of linear equations. Then:
- a variable that corresponds to a leading \(1\) is a dependent, or determined, or basic variable, and
- a variable that does not correspond to a leading \(1\) is a free, or independent variable.
Remark 1.5.9.
Definition 1.5.8 applies only to consistent systems. If a system is inconsistent, then it’s meaningless to talk about being free (or not) to pick values for the variables because there are no values which satisfy all the equations in the system.
Definition 1.5.8 helps us understand when a consistent system of linear equations will have infinitely many solutions. If a consistent system has no free variables, then there is exactly one solution; inversely, if there are any free variables, then there are infinitely many solutions.
Analyzing the solutions of a linear system.
Consider the reduced row echelon form of the augmented matrix of a system of linear equations.
- If there is a leading \(1\) in the last column, the system has no solution.
- Otherwise, if there is a leading \(1\) for each variable, then there is exactly one solution.
- Otherwise, there are infinite solutions because there are variables which are free to take on any value.
Activity 1.5.10. Number of solutions, free and basic variables.
For each augmented matrix in reduced row echelon form, select all that is true about the corresponding linear system.
(a)
- The variable \(x_1\) is free.
- There is a leading 1 in the first column, so \(x_1\) is basic, or dependent.
- The variable \(x_2\) is free.
- Correct!
- The variable \(x_3\) is free.
- There is a leading 1 in the third column, so \(x_3\) is basic, or dependent.
- The variable \(x_4\) is free.
- Correct!
- The variable \(x_5\) is free.
- This system doesn’t have an \(x_5\text{.}\) The last column in an augmented matrix corresponds to the constants of the equations, so there are only 4 columns corresponding to variables.
\begin{equation*}
\begin{bmatrix} 1 \amp 2 \amp 0 \amp 7 \amp 6 \\ 0\amp 0 \amp 1 \amp -5 \amp -23 \end{bmatrix}
\end{equation*}
(b)
- The system has no solutions.
- There is no row with a leading 1 in the last column, so there is at least one solution.
- The system has exactly 1 solution.
- This consistent system doesn’t have a leading 1 in every column corresponding to variables, and so there is more than 1 solution.
- There are infinitely many solutions.
- Correct!
\begin{equation*}
\begin{bmatrix} 1 \amp 2 \amp 0 \amp 7 \amp 6 \\ 0\amp 0 \amp 1 \amp -5 \amp -23 \end{bmatrix}
\end{equation*}
(c)
- The variable \(x_1\) is free.
- There is a leading 1 in the first column, so \(x_1\) is basic, or dependent.
- The variable \(x_2\) is free.
- There is a leading 1 in the second column, so \(x_2\) is basic, or dependent.
- The variable \(x_3\) is free.
- This system doesn’t have an \(x_3\text{.}\) The last column in an augmented matrix corresponds to the constants of the equations, so there are only 2 columns corresponding to variables.
- There are no free variables.
- Correct!
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp 2 \\ 0\amp 1 \amp -1\\ 0 \amp 0 \amp 0 \end{bmatrix}
\end{equation*}
(d)
- The system has no solutions.
- There is no row with a leading 1 in the last column, so there is at least one solution.
- The system has exactly 1 solution.
- Correct!
- There are infinitely many solutions.
- Every variable has a leading 1 in its column, so there are no free variables.
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp 2 \\ 0\amp 1 \amp -1\\ 0 \amp 0 \amp 0 \end{bmatrix}
\end{equation*}
Subsubsection More on Infinite Solutions
Systems with exactly one solution or no solution are the easiest to deal with; systems with infinite solutions are a bit harder, and we will often want to give one or two of the infinite possibilities.
Definition 1.5.11. Particular Solution.
Consider a linear system of equations with infinite solutions. A particular solution is one solution.
The easiest way to find a particular solution is to pick specific values for the free variables which then determines the values of the dependent variables.
Example 1.5.12. Infinite Solutions; Particular Solutions.
Give the solution to a linear system whose augmented matrix in reduced row echelon form is
\begin{equation*}
\left[\begin{array}{ccccc} 1\amp -1\amp 0\amp 2\amp 4\\ 0\amp 0\amp 1\amp -3\amp 7\\ 0\amp 0\amp 0\amp 0\amp 0\\ \end{array} \right]
\end{equation*}
and give two particular solutions.
Answer: We can ignore the third row as it does not give us any information about the solution. The first and second rows can be rewritten as the following equations:
\begin{align*}
x_1 - x_2 + 2x_4 \amp =4\\
x_3 - 3x_4 \amp = 7\text{.}
\end{align*}
Notice how the variables \(x_1\) and \(x_3\) correspond to the leading \(1\)’s of the given matrix. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables, in this case \(x_2\) and \(x_4\text{,}\) are free variables.
We generally write our solution with the dependent variables on the left and independent variables and constants on the right. It is also a good practice to acknowledge the fact that our free variables are, in fact, free. So all the solutions to the system would look something like
\begin{align*}
x_1 \amp = 4 +x_2 - 2x_4\\
x_2 \amp \ \ \text{ is free }\\
x_3 \amp = 7+3x_4\\
x_4 \amp \ \ \text{ is free } \text{.}
\end{align*}
To find particular solutions, choose any values you wish for the free variables.
For example, by setting \(x_2 = 0 = x_4\text{,}\) we get:
\begin{align*}
x_1 \amp = 4\\
x_2 \amp =0\\
x_3 \amp = 7\\
x_4 \amp = 0\text{.}
\end{align*}
By setting \(x_2 = 1\) and \(x_4 = -5\text{,}\) we have the solution:
\begin{align*}
x_1 \amp = 15\\
x_2 \amp =1\\
x_3 \amp = -8\\
x_4 \amp = -5\text{.}
\end{align*}
The constants and coefficients of an augmented matrix work together to determine whether a given system of linear equations has one, infinite, or no solution. The coefficients determine whether a matrix will have exactly one solution or not. In the “or not” case, the constants determine whether the system has infinite solutions or no solution.
Reading Questions Reading Questions
1. Matching Problem, Solution Types.
2.
Can there be a linear system which has exactly 3 solutions? Why or why not?
3. Reflection.
Enter a response to both of the following tasks:
- Ask a question about the material, either about something you’re not sure you fully understand, or a “what if” question.
- Give a percentage from 0 to 100 that reflects how confident you are with the material you just read about, and give one sentence as to why you feel that way. 0 means you didn’t actually do the reading and 100 means that everything makes sense so far and you think you are completely ready to engage with the material more deeply.
Worksheet Participate
Objectives
- Use the reduced row echelon form of an augmented matrix to analyze the consistency and existence of solution(s) to a linear system.
- Identify and use properties of free and basic variables to describe solutions to linear systems.
1.
Consider a linear system whose augmented matrix in reduced row echelon form is
\begin{equation*}
\begin{bmatrix} 1\amp 0\amp 0\amp 2\amp 3\\0\amp 0\amp 1\amp 4\amp 5\\ \end{bmatrix}\text{.}
\end{equation*}
(a)
Is the system consistent, inconsistent, or is there not enough information to tell?
(b)
Which variables are free and which are basic?
(c)
Solve the system. If there are infinitely many solutions, give two particular solutions.
2.
Consider a linear system whose augmented matrix in reduced row echelon form is
\begin{equation*}
\begin{bmatrix} 1\amp 0\amp 1\amp 4\\0\amp 1\amp 0\amp 1\\ \end{bmatrix}\text{.}
\end{equation*}
(a)
Is the system consistent, inconsistent, or is there not enough information to tell?
(b)
Which variables are free and which are basic?
(c)
Solve the system. If there are infinitely many solutions, give two particular solutions.
3.
Construct two different inconsistent linear systems with 3 variables. Use Sage or another tool to calculate the reduced row echelon form of the augmented matrix of your linear systems.
4.
Construct a linear system with 5 variables that has infinitely many solutions. Use Sage or another tool to calculate the reduced row echelon form of the augmented matrix of your linear system.
Analyzing the Effect of Different Coefficient and Constant Values.
For what values of \(k\) will the given system have exactly one solution, infinite solutions, or no solution?
5.
\begin{align*}
x_1 + 2x_2\amp =3\\
3x_1 + kx_2\amp = 9
\end{align*}
6.
\begin{align*}
x_1 + 2x_2\amp =3\\
3x_1 + kx_2\amp = 10
\end{align*}
7.
\begin{align*}
x_1 + 2x_2\amp =3\\
3x_1 + 6x_2\amp = k
\end{align*}
8.
\begin{align*}
x_1 + 2x_2\amp =3\\
3x_1 + 7x_2\amp = k
\end{align*}
Summary.
- Every linear system has either no solutions, one solution, or infinitely many solutions. We call a linear system which has at least one solution consistent, and we call a linear system with no solutions inconsistent.
- To analyze how many solutions a given linear system has, we examine the reduced row echelon form of the system’s augmented matrix. We look for number and location of leading \(1\)’s to see whether the system is inconsistent, or in the case that it is consistent, to see which variables are free or independent and which variables are basic or dependent.
- If a linear system has infinitely many solutions, we can describe all the possible solutions by solving for the dependent variables in terms of the free variables, and we can also give particular solutions by choosing specific values for the free variables.
Subsection Practice
Exercise 1.5.1.
The reduced row-echelon forms of the augmented matrices of four systems are given below. How many solutions does each system have?
\(\left\lbrack \begin{array}{rrr|r} 1 \amp 0 \amp -2 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right\rbrack\)
- No solutions
- Infinitely many solutions
- Unique solution
- None of the above
\(\left\lbrack \begin{array}{rrr|r} 0 \amp 1 \amp 0 \amp -16 \\ 0 \amp 0 \amp 1 \amp 5 \end{array} \right\rbrack\)
- Infinitely many solutions
- No solutions
- Unique solution
- None of the above
\(\left\lbrack \begin{array}{rr|r} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp 11 \\ 0 \amp 0 \amp 0 \end{array} \right\rbrack\)
- Unique solution
- No solutions
- Infinitely many solutions
- None of the above
\(\left\lbrack \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp 13 \\ 0 \amp 0 \amp 1 \amp -15 \end{array} \right\rbrack\)
- Infinitely many solutions
- No solutions
- Unique solution
- None of the above
Exercise 1.5.2.
How many free variables does each augmented matrix have?
- Choose
- None
- One
- Two
- Three
- Choose
- None
- One
- Two
- Three
- Choose
- None
- One
- Two
- Three
- Choose
- None
- One
- Two
- Three
Exercise 1.5.3.
How many determined (basic) variables does each augmented matrix have?
- Choose
- Zero
- One
- Two
- Three
- Four
- Choose
- Zero
- One
- Two
- Three
- Four
- Choose
- Zero
- One
- Two
- Three
- Four
- Choose
- Zero
- One
- Two
- Three
- Four
Exercise 1.5.4.
Convert the system
\begin{equation*}
\begin{array}{rcrcrcr}-4 x_1 \amp - \amp 7 x_{2} \amp = \amp 2 \\4 x_1 \amp + \amp 9 x_{2} \amp = \amp -6 \\ x_1 \amp + \amp 2 x_{2} \amp = \amp -1\end{array}
\end{equation*}
to an augmented matrix. Then reduce the system to echelon form and determine if the system is consistent.
Augmented matrix: (3 × 3 array)
Echelon form: (3 × 3 array)
Is the system consistent?
- select
- yes
- no
Exercise 1.5.5.
Solve the system by finding the reduced row-echelon form of the augmented matrix.
\begin{equation*}
\left\{
\begin{array}{ll}
x+ y+ 3 z \amp = -5 \\
2 x+ 3 y+ 7 z \amp = -6 \\
-3 x- y - 7 z \amp = 23
\end{array}\right.
\end{equation*}
reduced row-echelon form:
How many solutions are there to this system?
- None
- Exactly 1
- Exactly 2
- Exactly 3
- Infinitely many
- None of the above
If there is one solution, give its coordinates in the answer spaces below.
If there are infinitely many solutions, enter z in the answer blank for \(z\text{,}\) enter a formula for \(y\) in terms of \(z\) in the answer blank for \(y\) and enter a formula for \(x\) in terms of \(z\) in the answer blank for \(x\text{.}\)
If there are no solutions, leave the answer blanks for \(x\text{,}\) \(y\) and \(z\) empty.
\(x =\)
\(y =\)
\(z =\)
Exercises Additional Practice
Exercise Group.
In the following exercises, find the solution to the given linear system. If the system has infinite solutions, give \(2\) particular solutions.
1.
\begin{align*}
2x_1 + 4x_2\amp = 2\\
x_1 + 2x_2\amp = 1
\end{align*}
Answer.
\(x_1=1-2x_2\text{;}\) \(x_2\) is free. Two possible particular solutions: \(x_1=1\text{,}\) \(x_2=0\) or \(x_1=-1\text{,}\) \(x_2=1\text{.}\)
2.
\begin{align*}
-x_1 + 5x_2\amp = 3\\
2x_1 - 10x_2\amp = -6
\end{align*}
Answer.
\(x_1=-3+5x_2\text{;}\) \(x_2\) is free. Two possible particular solutions: \(x_1 = 3\text{,}\) \(x_2=0\) or \(x_1 = -8\text{,}\) \(x_2 = -1\)
3.
\begin{align*}
x_1 + x_2\amp = 3\\
2x_1 + x_2\amp = 4
\end{align*}
Answer.
\(x_1=1\text{;}\) \(x_2=2\)
4.
\begin{align*}
-3x_1 + 7x_2\amp = -7\\
2x_1 - 8x_2\amp = 8
\end{align*}
Answer.
\(x_1=0\text{;}\) \(x_2=-1\)
5.
\begin{align*}
2x_1 + 3x_2\amp = 1\\
-2x_1 - 3x_2\amp = 1
\end{align*}
Answer.
No solution; the system is inconsistent.
6.
\begin{align*}
x_1+ 2x_2\amp = 1\\
-x_1- 2x_2\amp = 5
\end{align*}
Answer.
No solution; the system is inconsistent.
7.
\begin{align*}
-2x_1+ 4x_2+ 4x_3\amp = 6\\
x_1- 3x_2+ 2x_3\amp = 1
\end{align*}
Answer.
\(x_1=-11+10x_3\text{;}\) \(x_2=-4+4x_3\text{;}\) \(x_3\) is free. Two possible particular solutions: \(x_1=-11\text{,}\) \(x_2 = -4\text{,}\) \(x_3=0\) or \(x_1 = -1\text{,}\) \(x_2 = 0\) and \(x_3 = 1\text{.}\)
8.
\begin{align*}
-x_1+ 2x_2+ 2x_3\amp = 2\\
2x_1+ 5x_2+ x_3\amp = 2
\end{align*}
Answer.
\(x_1=-\frac23+\frac89x_3\text{;}\) \(x_2=\frac23-\frac59x_3\text{;}\) \(x_3\) is free. Two possible particular solutions: \(x_1 = -\frac23\text{,}\) \(x_2 = \frac23\text{,}\) \(x_3 = 0\) and \(x_1 = \frac49\text{,}\) \(x_2 = -\frac19\text{,}\) \(x_3 = 1\)
9.
\begin{align*}
-x_1-x_2+x_3+x_4\amp = 0\\
-2x_1-2x_2+x_3\amp = -1
\end{align*}
Answer.
\(x_1=1-x_2-x_4\text{;}\) \(x_2\) is free; \(x_3=1-2x_4\text{;}\) \(x_4\) is free. Two possible solutions: \(x_1 = 1\text{,}\) \(x_2 = 0\text{,}\) \(x_3 = 1\text{,}\) \(x_4 = 0\) or \(x_1 = -2\text{,}\) \(x_2 = 1\text{,}\) \(x_3 = -3\text{,}\) \(x_4=2\)
10.
\begin{align*}
x_1+x_2+6x_3+9x_4\amp = 0\\
-x_1-x_3-2x_4\amp = -3
\end{align*}
Answer.
\(x_1=3-x_3-2x_4\text{;}\) \(x_2=-3-5x_3-7x_4\text{;}\) \(x_3\) is free; \(x_4\) is free. Two possible solutions: \(x_1 =3\text{,}\) \(x_2 = -3\text{,}\) \(x_3=0\text{,}\) \(x_4=0\) or \(x_1 = 0\text{,}\) \(x_2 = -5\text{,}\) \(x_3 =-1\text{,}\) \(x_4=1\)
11.
\begin{align*}
2x_1+ x_2+ 2x_3\amp = 0\\
x_1+ x_2+ 3x_3\amp = 1\\
3x_1+ 2x_2+ 5x_3\amp = 3
\end{align*}
Answer.
No solution; the system is inconsistent.
12.
\begin{align*}
x_1+ 3x_2+ 3x_3\amp = 1\\
2x_1- x_2+ 2x_3\amp = -1\\
4x_1+ 5x_2+ 8x_3\amp = 2
\end{align*}
Answer.
No solution; the system is inconsistent.
13.
\begin{align*}
x_1+ 2x_2+ 2x_3\amp =1\\
2x_1+ x_2+ 3x_3\amp = 1\\
3x_1+ 3x_2+ 5x_3\amp = 2
\end{align*}
Answer.
\(x_1=\frac13-\frac43x_3\text{;}\) \(x_2=\frac13-\frac13x_3\text{;}\) \(x_3\) is free. Two possible solutions: \(x_1 = \frac13\text{,}\) \(x_2=\frac13\text{,}\) \(x_3=0\) or \(x_1 = -1\text{,}\) \(x_2 = 0\text{,}\) \(x_3=1\)
14.
\begin{align*}
2x_1+ 4x_2+ 6x_3\amp = 2\\
1x_1+ 2x_2+ 3x_3\amp = 1\\
-3x_1- 6x_2- 9x_3\amp = -3
\end{align*}
Answer.
\(x_1=1-2x_2-3x_3\text{;}\) \(x_2\) is free; \(x_3\) is free. Two possible solutions: \(x_1=1\text{,}\) \(x_2=0\text{,}\) \(x_3=0\) or \(x_1=8\text{,}\) \(x_2=1\text{,}\) \(x_3 = -3\)
Exercise Group.
In the following exercises, state for which values of \(k\) the given system will have exactly \(1\) solution, infinite solutions, or no solution.
15.
\begin{align*}
x_1+ 2x_2\amp = 1\\
2x_1+ 4x_2\amp = k
\end{align*}
Answer.
Never exactly \(1\) solution; infinite solutions if \(k=2\text{;}\) no solution if \(k\neq 2\text{.}\)
16.
\begin{align*}
x_1+ 2x_2\amp = 1\\
x_1+ kx_2\amp = 1
\end{align*}
Answer.
Exactly \(1\) solution if \(k\neq 2\text{;}\) infinite solutions if \(k=2\text{;}\) never no solution.
17.
\begin{align*}
x_1+ 2x_2\amp = 1\\
x_1+ kx_2\amp = 2
\end{align*}
Answer.
Exactly \(1\) solution if \(k\neq 2\text{;}\) no solution if \(k=2\text{;}\) never infinite solutions.
18.
\begin{align*}
x_1+ 2x_2\amp = 1\\
x_1+ 3x_2\amp = k
\end{align*}
Answer.
Exactly \(1\) solution for all \(k\text{.}\)
