# 1.2: Finding solutions to systems of linear equations (2023)

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In the previous section, we looked at systems of linear equations from a graphical perspective. Since the equations had only two or three unknowns, we could study the solution spaces as the intersections of lines and planes.

Remembering that we will eventually consider many more equations and unknowns, this will, in general, not be a useful strategy. Instead, we will approach this problem algebraically and develop a technique to understand the solution spaces of general systems of linear equations.

## Gaussian elimination

We will develop an algorithm, which is usually called Gaussian elimination, that allows us to describe the solution space to a system of linear equations.

### Preview Activity 1.2.1.

Let's begin by considering some simple examples that will guide us in finding a more general approach.

1. Give a description of the solution space to the linear system:

\begin{equation*} \begin{alignedat}{3} x & & & {}={} & 2 \\ & & y & {}={} & -1. \\ \end{alignedat} \end{equation*}

2. Give a description of the solution space to the linear system:

\begin{equation*} \begin{alignedat}{4} -x & {} + {} & 2y & {}-{} & z & {}={} & -3 \\ & & 3y & {}+{} & z & {}={} & -1. \\ & & & & 2z & {}={} & 4. \\ \end{alignedat} \end{equation*}

3. Give a description of the solution space to the linear system:

\begin{equation*} \begin{alignedat}{3} x & {} + {} & 2y & {}={} & 2 \\ 2x& {}+{} & 2y & {}={} & 0. \\ \end{alignedat} \end{equation*}

4. Describe the solution space to the linear equation $$0x = 0\text{.}$$
5. Describe the solution space to the linear equation $$0x = 5\text{.}$$

As the examples in this preview activity provide some motivation for the general approach we will develop, we wish to call particular attention to two of the examples.

### Observation 1.2.1.

Let's look more carefully at two examples.

• First, finding the solution space to some systems is simple. For instance, each equation in the following system

\begin{equation*} \begin{alignedat}{3} x & & & {}={} & 2 \\ & & y & {}={} & -1. \\ \end{alignedat} \end{equation*}

has only one unknown so we can see that there is exactly one solution, which is $$(x,y) = (2,-1)\text{.}$$ We call such a system decoupled.
• Second, we may operate on a linear system transforming it into a new system that has the same solution space. For instance, given the system

\begin{equation*} \begin{alignedat}{4} -x & {} + {} & 2y & {}-{} & z & {}={} & -3 \\ & & 3y & {}+{} & z & {}={} & -1. \\ & & & & 2z & {}={} & 4, \\ \end{alignedat} \end{equation*}

we may multiply the third equation by $$1/2$$ to obtain

\begin{equation*} \begin{alignedat}{4} -x & {} + {} & 2y & {}-{} & z & {}={} & -3 \\ & & 3y & {}+{} & z & {}={} & -1. \\ & & & & z & {}={} & 2. \\ \end{alignedat} \end{equation*}

Any solution to this system of equations must then have $$z=2\text{.}$$

Once we know that, we may substitute $$z=2$$ into the first and second equation and simplify to obtain a new system of equations having the same solutions:

\begin{equation*} \begin{alignedat}{4} -x & {} + {} & 2y & {}{} & & {}={} & -1 \\ & & 3y & {}{} & & {}={} & -3. \\ & & & & z & {}={} & 2. \\ \end{alignedat} \end{equation*}

Continuing in this way, we eventually obtain a decoupled system showing that there is exactly one solution, which is $$(x,y,z)=(-1,-1,2)\text{.}$$

Our original system,

\begin{equation*} \begin{alignedat}{4} -x & {} + {} & 2y & {}-{} & z & {}={} & -3 \\ & & 3y & {}+{} & z & {}={} & -1 \\ & & & & 2z & {}={} & 4, \\ \end{alignedat} \end{equation*}

is called a triangular system due to the shape formed by the coefficients. As this example demonstrates, triangular systems are easily solved by a process called back substitution.

Let's look at the process of substitution a little more carefully. A natural approach to the system

\begin{equation*} \begin{alignedat}{3} x & {} + {} & 2y & {}={} & 2 \\ 2x& {}+{} & 2y & {}={} & 0. \\ \end{alignedat} \end{equation*}

is to use the first equation to express $$x$$ in terms of $$y\text{:}$$

\begin{equation*} x = 2-2y \end{equation*}

and then substitute this into the second equation and simplify:

\begin{equation*} \begin{alignedat}{2} 2x + 2y & {}= & 0 \\ 2(2-2y) + 2y & {}={} & 0 \\ 4-4y + 2y & {}={} & 0 \\ -2y & {}={} & -4. \\ \end{alignedat} \end{equation*}

The two-step process of solving for $$x$$ and substituting into the second equation may be performed more efficiently by adding a multiple of the first equation to the second. More specifically, we multiply the first equation by -2 and add to the second equation

\begin{equation*} \begin{array}{cr} & -2(\text{equation 1}) \\ + & \text{equation 2} \\ \hline \end{array} \end{equation*}

to obtain

\begin{equation*} \begin{array}{cr} & -2(x+2y=2) \\ + & 2x+2y = 0 \\ \hline \\ \end{array} \end{equation*}

which gives us

\begin{equation*} \begin{array}{crcr} & -2x-4y & = & -4 \\ + & 2x+2y & = & 0 \\ \hline & -2y & = & -4. \\ \end{array} \end{equation*}

In this way, the system

\begin{equation*} \begin{alignedat}{3} x & {} + {} & 2y & {}={} & 2 \\ 2x& {}+{} & 2y & {}={} & 0. \\ \end{alignedat} \end{equation*}

is transformed into the system

\begin{equation*} \begin{alignedat}{3} x & {} + {} & 2y & {}={} & 2 \\ & & -2y & {}={} & -4, \\ \end{alignedat} \end{equation*}

which has the same solution space. Of course, the choice to multiply the first equation by -2 was made so that terms involving $$x$$ in the two equations will cancel when added. Notice that this operation transforms our original system into a triangular one; we may now perform back substitution to arrive at a decoupled system.

Based on these observations, we take note of three operations that transform a system of linear equations into a new system of equations having the same solution space. Our goal is to create a new system whose solution space is the same as the original system's and may be easily described.

Scaling

We may multiply one equation by a nonzero number. For instance,

\begin{equation*} 2x -4y = 6 \end{equation*}

has the same set of solutions as

\begin{equation*} \frac12(2x-4y=6) \end{equation*}

or

\begin{equation*} x-2y=3\text{.} \end{equation*}

Interchange
Interchanging equations will not change the set of solutions. For instance,

\begin{equation*} \begin{alignedat}{3} 2x & {}+{} & 4y & {}={} & 1 \\ x & {}-{} & 3y & {}={} & 0 \\ \end{alignedat} \end{equation*}

has the same set of solutions as

\begin{equation*} \begin{alignedat}{3} x & {}-{} & 3y & {}={} & 0 \\ 2x & {}+{} & 4y & {}={} & 1. \\ \end{alignedat} \end{equation*}

Replacement

As we saw above, we may multiply one equation by a real number and add it to another equation. We call this process replacement.

##### Example 1.2.2

Let's illustrate the use of these operations to find the solution space to the system of equations:

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ 2x & {}+{} & y & {}-{} & 3z & {}={} & 11 \\ -3x & {}-{} & 2y & {}+{} & z & {}={} & -10 \\ \end{alignedat} \end{equation*}

We will first transform the system into a triangular system so we start by eliminating $$x$$ from the second and third equations.

We begin with a replacement operation where we multiply the first equation by -2 and add the result to the second equation.

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & -3y & {}-{} & 3z & {}={} & 3 \\ -3x & {}-{} & 2y & {}+{} & z & {}={} & -10 \\ \end{alignedat} \end{equation*}

Scale the second equation by multiplying it by $$-1/3\text{.}$$

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & y & {}+{} & z & {}={} & -1 \\ -3x & {}-{} & 2y & {}+{} & z & {}={} & -10 \\ \end{alignedat} \end{equation*}

Another replacement operation eliminates $$x$$ from the third equation. We multiply the first equation by 3 and add to the third.

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & y & {}+{} & z & {}={} & -1 \\ & & 4y & {}+{} & z & {}={} & 2 \\ \end{alignedat} \end{equation*}

Eliminate $$y$$ from the third equation by multiplying the second equation by -4 and adding it to the third.

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & y & {}+{} & z & {}={} & -1 \\ & & & & -3z & {}={} & 6 \\ \end{alignedat} \end{equation*}

After scaling the third equation by $$-1/3\text{,}$$ we have found the value for $$z\text{.}$$

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & y & {}+{} & z & {}={} & -1 \\ & & & & z & {}={} & -2 \\ \end{alignedat} \end{equation*}

The system now has a triangular form so we will begin the process of back substitution by multiplying the third equation by -1 and adding to the second.

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & & & {}={} & 4 \\ & & y & & & {}={} & 1 \\ & & & & z & {}={} & -2 \\ \end{alignedat} \end{equation*}

Finally, multiply the second equation by -2 and add to the first to obtain:

\begin{equation*} \begin{alignedat}{4} x & & & & & {}={} & 2 \\ & & y & & & {}={} & 1 \\ & & & & z & {}={} & -2 \\ \end{alignedat} \end{equation*}

Now that we have arrived at a decoupled system, we know that there is exactly one solution to our original system of equations, which is $$(x,y,z) = (2,1,-2)\text{.}$$

One could find the same result by applying a different sequence of replacement and scaling operations. However, we chose this particular sequence guided by our desire to first transform the system into a triangular one. To do this, we eliminated the first unknown $$x$$ from all but one equation and then proceeded to the next unknowns working left to right. Once we had a triangular system, we used back substitution moving through the unknowns right to left.

We call this process Gaussian elimination and note that it is our primary tool for solving systems of linear equations.

### Activity 1.2.2. Gaussian Elimination.

Use Gaussian elimination to describe the solutions to the following systems of linear equations.

1. Does the following linear system have exactly one solution, infinitely many solutions, or no solutions?

\begin{equation*} \begin{alignedat}{4} x & {}+{} & y & {}+{} & 2z & {}={} & 1 \\ 2x & {}-{} & y & {}-{} & 2z & {}={} & 2 \\ -x & {}+{} & y & {}+{} & z & {}={} & 0 \\ \end{alignedat} \end{equation*}

2. Does the following linear system have exactly one solution, infinitely many solutions, or no solutions?

\begin{equation*} \begin{alignedat}{4} -x & {}-{} & 2y & {}+{} & 2z & {}={} & -1 \\ 2x & {}+{} & 4y & {}-{} & z & {}={} & 5 \\ x & {}+{} & 2y & & & {}={} & 3 \\ \end{alignedat} \end{equation*}

3. Does the following linear system have exactly one solution, infinitely many solutions, or no solutions?

\begin{equation*} \begin{alignedat}{4} -x & {}-{} & 2y & {}+{} & 2z & {}={} & -1 \\ 2x & {}+{} & 4y & {}-{} & z & {}={} & 5 \\ x & {}+{} & 2y & & & {}={} & 2 \\ \end{alignedat} \end{equation*}

## Augmented matrices

After performing Gaussian elimination a few times, you probably noticed that you spent most of the time concentrating on the coefficients and simply recorded the unknowns as place holders. For convenience, we will therefore introduce a shorthand description of linear systems.

When writing a linear system, we always write the unknowns in the same order in each equation. We then construct an augmented matrix by simply forgetting about the unknowns and recording the numerical data in a rectangular array. For instance, the system of equations below has the following augmented matrix

\begin{equation*} \begin{alignedat}{4} -x & {}-{} & 2y & {}+{} & 2z & {}={} & -1 \\ 2x & {}+{} & 4y & {}-{} & z & {}={} & 5 \\ x & {}+{} & 2y & & & {}={} & 3 \\ \end{alignedat} \end{equation*}

\begin{equation*} \left[ \begin{array}{rrr|r} -1 & -2 & 2 & -1 \\ 2 & 4 & -1 & 5 \\ 1 & 2 & 0 & 3 \\ \end{array} \right]. \end{equation*}

The vertical line reminds us where the equals signs appear in the equations. Entries to the left corresponds to coefficients of the equations. We will sometimes choose to focus only on the coefficients of the system in which we case we write the coefficient matrix as

\begin{equation*} \left[ \begin{array}{rrr} -1 & -2 & 2 \\ 2 & 4 & -1 \\ 1 & 2 & 0 \\ \end{array} \right]. \end{equation*}

The three operations we perform on systems of equations translate naturally into operations on matrices. For instance, the replacement operation that multiplies the first equation by 2 and adds it to the second may be recorded as

\begin{equation*} \left[ \begin{array}{rrr|r} -1 & -2 & 2 & -1 \\ 2 & 4 & -1 & 5 \\ 1 & 2 & 0 & 3 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} -1 & -2 & 2 & -1 \\ 0 & 0 & 3 & 3 \\ 1 & 2 & 0 & 3 \\ \end{array} \right]. \end{equation*}

The symbol $$\sim$$ between the matrices indicates that the two matrices are related by a sequence of scaling, interchange, and replacement operations. Since these operations act on the rows of the matrices, we say that the matrices are row equivalent.

### Activity 1.2.3. Augmented matrices and solution spaces.

1. Write the augmented matrix for the system of equations

\begin{equation*} \begin{alignedat}{4} x & {}+{} & 2y & {}-{} & z & {}={} & 1 \\ 3x & {}+{} & 2y & {}+{} & 2z & {}={} & 7 \\ -x & & & {}+{} & 4z & {}={} & -3 \\ \end{alignedat} \end{equation*}

and perform Gaussian elimination to describe the solution space of the system of equations in as much detail as you can.

2. Suppose that you have a system of linear equations in the unknowns $$x$$ and $$y$$ whose augmented matrix is row equivalent to

\begin{equation*} \left[ \begin{array}{rr|r} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right]. \end{equation*}

Write the system of linear equations corresponding to the augmented matrix. Then describe the solution set of the system of equations in as much detail as you can.

3. Suppose that you have a system of linear equations in the unknowns $$x$$ and $$y$$ whose augmented matrix is row equivalent to

\begin{equation*} \left[ \begin{array}{rr|r} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]. \end{equation*}

Write the system of linear equations corresponding to the augmented matrix. Then describe the solution set of the system of equations in as much detail as you can.

4. Suppose that the augmented matrix of a system of linear equations has the following shape where $$*$$ could be any real number.

\begin{equation*} \left[ \begin{array}{rrrrr|r} * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \\ \end{array} \right]. \end{equation*}

1. How many equations are there in this system and how many unknowns?
2. Based on our earlier discussion in Section 1.1, do you think it's possible that this system has exactly one solution, infinitely many solutions, or no solutions?
3. Suppose that this augmented matrix is row equivalent to

\begin{equation*} \left[ \begin{array}{rrrrr|r} 1 & 2 & 0 & 0 & 3 & 2 \\ 0 & 0 & 1 & 2 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{equation*}

Make a choice for the names of the unknowns and write the corresponding system of linear equations. Does the system have exactly one solution, infinitely many solutions, or no solutions?

## Reduced row echelon form

There is a special class of matrices whose form makes it especially easy to describe the solution space of the corresponding linear system. As we describe the properties of this class of matrices, it may be helpful to consider an example, such as the following matrix.

\begin{equation*} \left[ \begin{array}{rrrrrr} 1 & * & 0 & * & 0 & * \\ 0 & 0 & 1 & * & 0 & * \\ 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{equation*}

##### Definition 1.2.3

We say that a matrix is in reduced row echelon form if the following properties are satisfied.

• Any rows in which all the entries are zero are at the bottom of the matrix.
• If we move across a row from left to right, the first nonzero entry we encounter is 1. We call this entry the leading entry in the row.
• The leading entry in one row is to the right of the leading entry in any row above.
• A leading entry is the only nonzero entry in its column.

We call a matrix in reduced row echelon form a reduced row echelon matrix.

We have been intentionally vague about whether the matrix we are considering is an augmented matrix corresponding to a linear system or a coefficient matrix since we will eventually consider both possibilities.

### Activity 1.2.4. Identifying reduced row echelon matrices.

Consider each of the following augmented matrices. Determine if the matrix is in reduced row echelon form. If it is not, perform a sequence of scaling, interchange, and replacement operations to obtain a row equivalent matrix that is in reduced row echelon form. Then use the reduced row echelon matrix to describe the solution space.

1. $$\displaystyle \left[ \begin{array}{rrr|r} 2 & 0 & 4 & -8 \\ 0 & 1 & 3 & 2 \\ \end{array} \right].$$
2. $$\displaystyle \left[ \begin{array}{rrr|r} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \\ \end{array} \right].$$
3. $$\displaystyle \left[ \begin{array}{rrr|r} 1 & 0 & 4 & 2 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 1 \\ \end{array} \right].$$
4. $$\displaystyle \left[ \begin{array}{rrr|r} 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 4 & 2 \\ \end{array} \right].$$
5. $$\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 2 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 1 \\ \end{array} \right].$$

If we are given a matrix, the examples in the previous activity indicate that there is a sequence of row operations that produces a matrix in reduced row echelon form. Moreover, the conditions that define reduced row echelon matrices guarantee that this matrix is unique.

### Theorem 1.2.4.

Given a matrix, there is exactly one reduced row echelon matrix to which it is row equivalent.

Once we have this reduced row echelon matrix, we may describe the set of solutions to the corresponding linear system with relative ease.

##### Example 1.2.5. Describing the solution space from a reduced row echelon matrix
1. Consider the row reduced echelon matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 & 0 & 2 & -1 \\ 0 & 1 & 1 & 2 \\ \end{array} \right]. \end{equation*}

Its corresponding linear system may be written as

\begin{equation*} \begin{alignedat}{4} x & & & {}+{} & 2z & {}={} & -1 \\ & & y & {}+{} & z & {}={} & 2. \\ \end{alignedat} \end{equation*}

Let's rewrite the equations as

\begin{equation*} \begin{alignedat}{2} x & {}={} & -1 -2z\\ y & {}={} & 2-z. \\ \end{alignedat} \end{equation*}

From this description, it is clear that we obtain a solution for any value of the variable $$z\text{.}$$ For instance, if $$z=2\text{,}$$ then $$x = -5$$ and $$y=0$$ so that $$(x,y,z) = (-5,0,2)$$ is a solution. Similarly, if $$z=0\text{,}$$ we see that $$(x,y,z) = (-1,2,0)$$ is also a solution.

Because there is no restriction on the value of $$z\text{,}$$ we call it a free variable, and note that the linear system has infinitely many solutions. The variables $$x$$ and $$y$$ are called basic variables as they are determined once we make a choice of the free variable.

We will call this description of the solution space, in which the basic variables are written in terms of the free variables, a parametric description of the solution space.

2. Consider the matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{equation*}

The last equation gives

\begin{equation*} 0x +0y+0z = 0\text{,} \end{equation*}

which is true for any $$(x,y,z)\text{.}$$ We may safely ignore this equation since it does not provide a restriction on the choice of $$(x,y,z)\text{.}$$ We then see that there is a unique solution $$(x,y,z) = (4,-3,1)\text{.}$$

3. Consider the matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 & 0 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right]. \end{equation*}

Beginning with the last equation, we see that

\begin{equation*} 0x +0y+0z = 0 = 1\text{,} \end{equation*}

which is not true for any $$(x,y,z)\text{.}$$ There is no solution to this particular equation and therefore no solution to the system of equations.

## Summary

We saw several importants concepts in this chapter.

• We can describe the solution space to a linear system by transforming it into a new linear system through a sequence of scaling, interchange, and replacement operations.
• We represented a system of linear equations by an augmented matrix. Using scaling, interchange, and replacement operations, the augmented matrix is row equivalent to exactly one reduced row echelon matrix.
• The reduced row echelon matrix allows us to easily describe the solution space to a system of linear equations.

## Exercises 1.2.5Exercises

##### 1

For each of the linear systems below, write the associated augmented matrix and find the reduced row echelon matrix that is row equivalent to it. Identify the basic and free variables and then describe the solution space of the original linear system using a parametric description, if appropriate.

1. \begin{equation*} \begin{alignedat}{3} 2x & {}+{} & y & {}={} & 0 \\ x & {}+{} & 2y & {}={} & 3 \\ -2x & {}+{} & 2y & {}={} & 6 \\ \end{alignedat} \end{equation*}

2. \begin{equation*} \begin{alignedat}{5} -x_1 & {}+{} & 2x_2 & & & {}+{} & x_3 & {}={} & 2 \\ 3x_1 & & & & & {}+{} & 2x_3 & {}={} & -1 \\ -x_1 & {}-{} & x_2 & & & {}+{} & x_3 & {}={} & 2 \\ \end{alignedat} \end{equation*}

3. \begin{equation*} \begin{alignedat}{5} x_1 & {}+{} & 2x_2 & {}-{} & 5x_3 & {}-{} & x_4 & {}={} & -3 \\ -2x_1 & {}-{} & 2x_2 & {}+{} & 6x_3 & {}-{} & 2x_4 & {}={} & 4 \\ x_1 & & & {}-{} & x_3 & {}+{} & 9x_4 & {}={} & 7 \\ & & -x_2 & {}+{} & 2x_3 & {}-{} & x_4 & {}={} & 4 \\ \end{alignedat} \end{equation*}

##### 2

Consider each matrix below and determine if it is in reduced row echelon form. If not, indicate the reason and apply a sequence of row operations to find its reduced row echelon matrix. For each matrix, indicate whether the linear system has infinitely many solutions, exactly one solution, or no solutions.

1. \begin{equation*} \left[ \begin{array}{rrrr|r} 1 & 1 & 0 & 3 & 3 \\ 0 & 1 & 0 & -2 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ \end{array} \right] \end{equation*}

2. \begin{equation*} \left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & -3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \end{equation*}

3. \begin{equation*} \left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 3 & 3 \\ 0 & 1 & 0 & -2 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 3 & 3 \\ \end{array} \right] \end{equation*}

4. \begin{equation*} \left[ \begin{array}{rrrr|r} 0 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 0 & 3 \\ 1 & 1 & 1 & 1 & 2 \\ \end{array} \right] \end{equation*}

##### 3

Give an example of a reduced row echelon matrix that describes a linear system having the stated properties. If it is not possible to find such an example, explain why not.

1. Write a reduced row echelon matrix for a linear system having five equations and three unknowns and having exactly one solution.
2. Write a reduced row echelon matrix for a linear system having three equations and three unknowns and having no solution.
3. Write a reduced row echelon matrix for a linear system having three equations and five unknowns and having infinitely many solutions.
4. Write a reduced row echelon matrix for a linear system having three equations and four unknowns and having exactly one solution.
5. Write a reduced row echelon matrix for a linear system having four equations and four unknowns and having exactly one solution.
##### 4

For each of the questions below, provide a justification for your response.

1. What does the presence of a row whose entries are all zero in an augmented matrix tell us about the solution space of the linear system?
2. How can you determine if a linear system has no solutions directly from its reduced row echelon matrix?
3. How can you determine if a linear system has infinitely many solutions directly from its reduced row echelon matrix?
4. What can you say the solution space of a linear system if there are more unknowns than equations and at least one solution exists?
##### 5

Determine whether the following statements are true or false and explain your reasoning.

1. If every variable is basic, then the linear system has exactly one solution.
2. If two augmented matrices are row equivalent to one another, then they describe two linear systems having the same solution spaces.
3. The presence of a free variable indicates that there are no solutions to the linear system.
4. If a linear system has exactly one solution, then it must have the same number of equations as unknowns.
5. If a linear system has the same number of equations as unknowns, then it has exactly one solution.

## FAQs

### Can a solution to a system of linear equations have 1 or 2 solutions? ›

Number of Solutions

Most linear systems you will encounter will have exactly one solution. However, it is possible that there are no solutions, or infinitely many. (It is not possible that there are exactly two solutions.) The word unique in this context means there is a solution, and it's the only one.

How to find number of solutions of a linear equation in two variables? ›

A linear equation in two variables is an equation of the form ax + by + c = 0 where a, b, c ∈ R, a, and b ≠ 0. When we consider a system of linear equations, we can find the number of solutions by comparing the coefficients of the variables of the equations.

How do you know a linear system will have 1 solution? ›

A system of linear equations has one solution when the graphs intersect at a point.

What are the 3 solutions of systems of linear equation? ›

A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation.

How do you solve linear systems step by step? ›

1. Step 1 : First, solve one linear equation for y in terms of x .
2. Step 2 : Then substitute that expression for y in the other linear equation. ...
3. Step 3 : Solve this, and you have the x -coordinate of the intersection.
4. Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.

How do you know if a system of equations has no solution? ›

When two equations have the same slope but different y-axis, they are parallel. Since there are no intersection points, the system has no solutions.

How many solutions for a linear equation in two variables is 1? ›

A linear equation in two variables has only 1 solution.

What is an example of a system of linear equations with one solution? ›

The linear equations in one variable is an equation which is expressed in the form of ax+b = 0, where a and b are two integers, and x is a variable and has only one solution. For example, 2x+3=8 is a linear equation having a single variable in it. Therefore, this equation has only one solution, which is x = 5/2.

How many solutions can a system of 2 linear equations in 2 variables have? ›

there can be either no solution, exactly one solution, or an infinite number of solutions. If you are dealing with two lines then the lines will either never intersect, intersect at only one point, or be on top of each other.

How do you find the solution of an equation? ›

Substitute the number for the variable in the equation. Simplify the expressions on both sides of the equation. Determine whether the resulting equation is true. If it is true, the number is a solution.

### How do you tell if a system of equations has one solution or infinitely many? ›

If the two lines have the same y-intercept and the slope, they are actually in the same exact line. In other words, when the two lines are the same line, then the system should have infinite solutions.

How do you tell if a system of equations has no solution or infinitely many? ›

If you get a unique solution for each variable, there is one solution. If you get a contradiction like 0 = 1, then there is no solution. If you get an equation that is always true, such as 0 = 0, then there are infinite solutions.

When a linear system has no solution? ›

A system of linear equations that has no solution is called an inconsistent pair of linear equations.

What are all solutions of a linear system? ›

A solution of a linear system is an assignment of values to the variables x1, x2, ..., xn such that each of the equations is satisfied. The set of all possible solutions is called the solution set. A linear system may behave in any one of three possible ways: The system has infinitely many solutions.

What are the 3 types of solutions to a system of equations? ›

There are three types of systems of linear equations in two variables, and three types of solutions.
• An independent system has exactly one solution pair (x,y). The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. ...
• A dependent system has infinitely many solutions.
May 1, 2022

What are 3 examples of system of linear equations in three variables? ›

Examples. 3x + 4y - 7z = 2, -2x + y - z = -6, x - 17z = 4, 4y = 0, and x + y + z = 2 are all linear equations in three variables.

What are 4 methods of solving linear systems? ›

There are a few different methods of solving systems of linear equations:
• The Graphing Method . ...
• The Substitution Method . ...
• The Linear Combination Method , aka The Addition Method , aka The Elimination Method. ...
• The Matrix Method .

What are the 4 steps in solving linear functions? ›

Step 1: Substitute the value of f(x) into the problem. Step 2: Isolate the variable. Step 3: Continue to isolate the variable. Step 4: Confirming the answer.

How do you find the solution to a system of equations without graphing? ›

To solve a system of linear equations without graphing, you can use the substitution method. This method works by solving one of the linear equations for one of the variables, then substituting this value for the same variable in the other linear equation and solving for the other variable.

What is an example of a system of linear equations with no solution? ›

As an example, consider 3x + 5 = 3x - 5. This equation has no solution. There is no value that will ever satisfy this type of equation.

### What is an example of a linear system with no solution? ›

An example of an equation with no solution is x = x + 5. 2x + 6y = 7.

What are examples of linear equations? ›

Some of the examples of linear equations are 2x – 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, 3x – y + z = 3. In this article, we are going to discuss the definition of linear equations, standard form for linear equation in one variable, two variables, three variables and their examples with complete explanation.

How many solutions does an equation with one variable have? ›

A linear equation in one variable cannot have more than one solution.

Can a linear equation have exactly two solutions? ›

No, a system of linear equations cannot have exactly two solutions.

How many solutions can there be to a system of equations that has one linear equation and one quadratic equation? ›

The number of solutions for a system with one linear equation and one quadratic equation is either (never intersect), (intersect in one place), or (intersect in two places). Solve the system by graphing the equations and . Graph each equation and locate the points of intersection. This system has two solutions.

How can you tell if a point is a solution to a system? ›

To verify whether a point is a solution to a system or not, we will either determine whether it is the point of intersection of two lines on a graph or we will determine whether or not the point lies on both lines algebraically.

What is the number of solutions of a linear system? ›

A system of linear equations can have no solutions, exactly one solution, or infinitely many solutions. If the system has two or more distinct solutions, it must have infinitely many solutions.

How many solutions are possible for a system of two linear equations? ›

Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts.

How do you find how many solutions does this equation have? ›

If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.

Why can't a linear system have 2 solutions? ›

For linear equations, no, that's it. Because through any two points is a line. And then they'll have all the same points so infinite solutions. You can get more solutions with different types of functions but that's it for linear functions.

### When can a system of equations have 2 solutions? ›

If the graphs of the equations do not intersect, then there are no solutions that are true for both equations. If the line intersects the parabola in two places, then there are two solutions that are true for both equations.

How many solutions can an equation have? ›

An equation can have infinitely many solutions when it should satisfy some conditions. The system of an equation has infinitely many solutions when the lines are coincident, and they have the same y-intercept. If the two lines have the same y-intercept and the slope, they are actually in the same exact line.

How do you tell if an equation has no solution or infinitely many? ›

If you get a unique solution for each variable, there is one solution. If you get a contradiction like 0 = 1, then there is no solution. If you get an equation that is always true, such as 0 = 0, then there are infinite solutions.

How many solutions will there by if two equations are the same line? ›

If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations. If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

How many solutions can a consistent system of linear equations have? ›

A system of two linear equations can have one solution, an infinite number of solutions, or no solution. Systems of equations can be classified by the number of solutions. If a system has at least one solution, it is said to be consistent . If a consistent system has exactly one solution, it is independent .

What are solutions of two linear equations? ›

The solution of a linear equation is defined as the points, in which the lines represent the intersection of two linear equations. In other words, the solution set of the system of linear equations is the set of all possible values to the variables that satisfies the given linear equation.

Can linear systems be hard to learn? ›

Statistically hard to learn linear system classes have worst- case sample complexity that is at least exponential with the system dimension, regardless of the identification algorithm. Using tools from minimax theory, we show that classes of linear systems can be hard to learn.

How many solutions can a system of 2 equations have? ›

Under normal circumstances a system of two linear equations can have 0 , 1 or infinitely many solutions.

Which equation has infinitely many solutions? ›

An infinite solution has both sides equal. For example, 6x + 2y - 8 = 12x +4y - 16. If you simplify the equation using an infinite solutions formula or method, you'll get both sides equal, hence, it is an infinite solution.

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