Welcome to the **reduced row echelon form calculator** (or **rref calculator** for short), where we'll solve a system of equations of your choice using the matrix row reduction and elementary row operations. Also, we give you the option to choose whether you'd like to use the reduced version or not. Based on the choice you make, our tool can be viewed as a **Gauss-Jordan elimination calculator** (with the first variant) or a **Gauss elimination calculator**. Moreover, in case your system has an infinite number of solutions, our rref calculator will even tell you what they look like!

## What is a system of equations?

Remember all those math scenarios that try to imitate real life? Like a little girl asking you how old she is if, in ten years, her mom will be twice as old as she will be then? You know, just your **everyday conversations and everyday problems**. Well, equations are what we use to solve them.

Whenever we have some value that we don't know (like the age of the little girl), but we know that it must satisfy some property (like being twice as large as some other number), we **describe this connection using equations**. We denote the value we don't know with a symbol, which we call **a variable**. We then write what we know about it with mathematical symbols and operations, such as addition, subtraction, multiplication, or division. The resulting expression is called **an equation**.

If we have several equations and want all of them to be satisfied by the same number, then what we're dealing with is **a system of equations**. Usually, they have more than one variable in total, and the most common math problems include the same number of equations as there are variables. For example, suppose that the mother of our little girl tells us that **she's three times older than her daughter**. Now we know where that wittiness came from)... Anyway, we can translate this new mom statement into an equation as well. Together with the previous one, they would form a system of two equations with two variables: the girl's and the mother's age.

## Elementary row operations

Let's try to see **how our reduced row echelon form calculator sees a system of equations**. Take this juicy example:

Don't worry, we haven't gone back to kindergarten, we're still working with systems of equations. The above picture may not look like one, but in fact, **it is just that**. We're so used to seeing variables such as $x$x or $y$y that we tend to forget that it is just a symbol for a value we don't know. And here we have a juicy lemon, a crunchy apple, and a sweet banana, all representing numbers we don't yet know. For simplicity of notation, let's denote them by $x$x, $y$y, and $z$z, respectively. This way we can write **an equivalent system of equations**:

$\small\begin{cases}&x\!+\!y\!+\!z\! = \!32\\&y\!+\!y\!-\!x\!=\!25\\&z\!+\!z\!-\!y\!=\!16\end{cases}$⎩⎨⎧x+y+z=32y+y−x=25z+z−y=16

Now that we look at it, high school killed some of our imagination along the way, hasn't it? Still, we can now easily **simplify the system** by adding together the same symbols in consecutive equations and write the variables that appear in them **in alphabetical order**. For example, in the second equation, $y + y - x = 25$y+y−x=25, we can add the $y$y's together to obtain $2y - x = 25$2y−x=25 (since we had two copies of $y$y). Next, we flip the $x$x variable to the beginning to create an alphabetical order (remember to **bring along the number next to it**) and get $-x + 2y = 25$−x+2y=25. All in all, we obtain

$\small\begin{cases}&x\!+\!y\!+\!z\! \!= 32\\&\!-\!x\!+\!2y\!=\!25\\&\!-\!y\!+\!2z\!=\!16\end{cases}$⎩⎨⎧x+y+z=32−x+2y=25−y+2z=16

The rref calculator uses **the Gauss-Jordan elimination** and **the Gauss elimination**, and both use so-called **matrix row reduction**. This, in turn, relies on **elementary row operations**, which are:

- You can exchange any two equations.
- You can multiply any equation by a non-zero constant number.
- You can add a non-zero multiple of any equation to another equation.

What we mean by "*you can*" is that the system you'd obtain by these operations will be **equivalent** to the one you started with. This means that the two will have **exactly the same solutions**.

For example, we could multiply the first equation by, say, $-3$−3:

$\small\begin{cases}\!-3x\!-\!3y\!-\!3z\!=\!-96\\\!-x\!+\!2y\!=\!25\\\!-y\!+\!2z\!=\!16\end{cases}$⎩⎨⎧−3x−3y−3z=−96−x+2y=25−y+2z=16

and add two copies of the second equation to the third one:

$\small\begin{cases}\!-3x\!-\!3y\!-\!3z\!=\!-96\\\!-x\!+\!2y\!=\!25\\\!-y\! + \!2z\! +\! 2\cdot (-x\! + \!2y)\! =\! 16\! +\! 2\! \cdot \!25\end{cases}$⎩⎨⎧−3x−3y−3z=−96−x+2y=25−y+2z+2⋅(−x+2y)=16+2⋅25

which is

$\small\begin{cases}\!-3x\! - \!3y \!-\! 3z \!=\! -96\\\!-x\! +\! 2y\! =\! 25\\\!-2x\! +\! 3y\! +\! 2z\! =\! 66\end{cases}$⎩⎨⎧−3x−3y−3z=−96−x+2y=25−2x+3y+2z=66

The elementary row operations **didn't change the set of solutions to our system**. Don't believe us? Go on, type the first and the last system into the reduced row echelon form calculator, and see what you get. We'll wait for you, but expect a "*we told you so*" when you get back.

## Gauss-Jordan elimination vs Gauss elimination

We can use the matrix row reduction that we've mentioned in the section above for more practical uses than just having fun with multiplying equations by random numbers. Oh come on, **we did have fun, didn't we?**

As you might have guessed, it is easier to deal with one variable than with several of them, so why not try to **eliminate some of them**? Presumably, this (but in German) was the line of thinking of **Carl Friedrich Gauss**, a mathematician behind the so-called **Gauss elimination**, but not only: meet him also at the Gauss law calculator. It is an algorithmic procedure that transforms a system of equations into a very easy to deal with form. The idea behind it is (please proceed to read the following instructions in **18th-century German accent**):

- Take an equation with the first variable in it and
**put that line as the first one in your system**. - Use elementary row operations on the first equation to
**eliminate all occurrences of the first variable**in all the other equations. - Take an equation (different from the first) with the second variable in it and
**put it as the second one in the system**. - Use elementary row operations on the second equation to
**eliminate all occurrences of the second variable**in all the later equations. **Repeat for subsequent variables**until you run out of equations, variables, or self-discipline to finish the exercise.

The system we get in the end is said to be **in row echelon form**. "*So what does the reduced in reduced row echelon form calculator stand for?*" How convenient of you to ask! That's where

**the Gauss-Jordan elimination**comes in. It is a slightly improved version of the previous algorithm, first done by

**Camille Jordan**. It took a French mathematician and a few decades to ask the fundamental question: "

*What if in the end, we divided every line by its first number?*"

**Mind = blown.**

In other words, the Gauss-Jordan elimination add-on gives us **an additional step in the algorithm**:

**Divide each equation by the coefficient of the first variable**occurring in that line.

The system we get with the upgraded version of the algorithm is said to be **in reduced row echelon form**. The advantage of that approach is that in each line the first variable will have the coefficient $1$1 in front of it instead of something complicated, like a $2$2, for example. It does, however, speed up calculations, and, as we know, every second is valuable.

It's high time for an example, wouldn't you say?

🙋 Learn everything about systems of equations with Omni's system of equations calculator!

## Example: using the reduced row echelon form calculator

**Recall the system of equations we had in** the second section, but the one right before we started playing with elementary row operations:

$\small\begin{cases}x\!+\!y\!+\!z\! \!= 32\\\!-\!x\!+\!2y\!=\!25\\\!-\!y\!+\!2z\!=\!16\end{cases}$⎩⎨⎧x+y+z=32−x+2y=25−y+2z=16

Before we move on to the step-by-step calculations, let's quickly say a few words about **how we can input such a system into our reduced row echelon form calculator**. First of all, we have three lines in the system, so we need to tell that to the calculator at the top, in the number of equations field. This will show us a symbolic picture of an arbitrary system of three linear equations.

We need to **identify which number corresponds to which symbol** from the rref calculator. In the picture, the first equation has symbols $a_1$a1, $b_1$b1, $c_1$c1, and $d_1$d1, which are respectively next to $x$x, to $y$y, to $z$z, and to the right hand side of the $=$= sign. These are the numbers that we're looking for in our system. Looking at the first of our equations, we identify that $a_1 = 1$a1=1, $b_1= 1$b1=1, $c_1 = 1$c1=1, and $d_1 = 32$d1=32 (**remember that no number in front of a variable means that the coefficient is equal to** $1$1).

Similarly, for the next two lines we get $a_2 = -1$a2=−1, $b_2 = 2$b2=2, $c_2 = 0$c2=0, $d_2 = 25$d2=25, and $a_3 = 0$a3=0, $b_3 = -1$b3=−1, $c_3 = 2$c3=2, $d_3 = 16$d3=16 (**remember that if an equation doesn't have some variable, then the coefficient next to that variable is** $0$0). If you input all this data into the reduced row echelon form calculator, **you'll get a spoiler of what the answer is**. Also, note here that our rref calculator **doesn't allow non-linear** (e.g., quadratic) **equations**. If you need to solve individual equations of this type, visit the quadratic formula calculator!

We'll now follow the instructions on matrix row reduction given by **the Gauss elimination** to transform it into a row echelon form. Lastly, we'll do the extra step from **the Gauss-Jordan elimination** to make it into the reduced version, which is used by default in the rref calculator.

According to the algorithm, we start by **choosing an equation with the first variable** (in our case, it's $x$x) and putting it in the top line. Note, that our system is already in this form, so we don't have to change anything. Next, **we use the first equation to eliminate the** $x$x**'s from the other two lines**. Observe that we only have to deal with the second one since the third equation has no $x$x. To get rid of the $-x$−x in the middle line, we need to add to that equation a multiple of the first equation so that the $x$x's will cancel each other out. Since $-x + x = 0$−x+x=0, we need to have $x$x with coefficient $1$1 in what we add to the second line. Fortunately, this is exactly what we have in the top equation. Therefore, **we add the first line to the second** to obtain:

$-x\! +\! 2y\! +\! (x\! +\! y\! +\! z) \!= \!25 \!+ \!32$−x+2y+(x+y+z)=25+32

Which is:

$3y \!+\! z\! =\! 57$3y+z=57

Together with the other two equations, this gives:

$\begin{cases}\!x\! +\! y \!+\! z \!=\! 32\\\!3y\! +\! z\! =\! 57\\\!-y\! +\! 2z\! =\! 16\end{cases}$⎩⎨⎧x+y+z=323y+z=57−y+2z=16

Great! We now have the **two last lines with no** $x$x**'s in them**. True, the second equation gained a $z$z that was not there before, but that's just a price we have to pay.

Now we need to do something about the $y$y in the last equation, and we'll use the second line for it. However, **it's not going to be as easy as last time** - we have $3y$3y at our disposal and $-y$−y to deal with. Well, the tools they gave us will have to do.

To eliminate $-y$−y from the third equation, we'll need to get $y$y (i.e., $y$y with coefficient $1$1) from the second one since $-y + y = 0$−y+y=0. To get it from the $3y$3y, it's enough to divide it by $3$3. In other words, in the language of matrix row reduction, **we'll add a multiple of** $1/3$1/3 (equivalent to dividing by $3$3) **of the second equation to the bottom line**. This gives:

$-y\! +\! 2z\! +\!\frac{1}{3}\!\cdot\! (3y\! +\! z)\! =\! 16\! +\! \frac{1}{3}\! \cdot \!57$−y+2z+31⋅(3y+z)=16+31⋅57

Note how the $1/3$1/3 **also appeared on the right side** with the $57$57. After simplification, this gives:

$\frac{7}{3}\!\cdot\!z=35$37⋅z=35

Which, together with the other two equations, is:

$\begin{cases}\!x\! +\! y \!+\! z \!=\! 32\\\!3y\! +\! z\! =\! 57\\\frac{7}{3}\!\cdot\!z=35\end{cases}$⎩⎨⎧x+y+z=323y+z=5737⋅z=35

**Voilà!** That is **the row echelon form given by the Gauss elimination**. Note, that such systems are obtained in our rref calculator by answering "*No*" to the question of whether to show the reduced form at the top of the calculator.

🙋 In your calculations, you may need to make your equations simpler: learn a way to do so with, for example, the simplify fractions calculator at Omni!

To obtain the reduced row echelon form, we follow the sixth step mentioned in the section above - **we divide each equation by the coefficient of its first variable**. This means that we need to divide the first line by $1$1 (the coefficient of $x$x), the second by $3$3 (the coefficient of $y$y), and the third by $7/3$7/3 (the coefficient of $z$z). This gives

$\begin{cases}\!x\! +\! y \!+\! z \!=\! 32\\\!y \!+\! \frac{1}{3}\!\cdot \!z\! =\! 19\\\!z\!=\!15\end{cases}$⎩⎨⎧x+y+z=32y+31⋅z=19z=15

and **marks an end of the Gauss-Jordan elimination algorithm**. We can get such systems in our reduced row echelon form calculator by answering "*Yes*" to the top question (as is done by default).

Observe that **now it is easy to find the solution to our system**. From the last line, we know that $z = 15$z=15 so we can substitute it in the second equation to get:

$y \!+\! \frac{1}{3}\!\cdot \!15\! = \!19$y+31⋅15=19

From that we get $y = 14$y=14, and we can substitute that and $z = 15$z=15 into the first line to obtain:

$x \!+ \!14 \!+ \!15 \!= \!32$x+14+15=32

Which gives $x = 3$x=3. Coming back to the picture we started with, this means that **the lemon is equal to** $3$3, **the apple to** $14$14, **and the banana to** $15$15. Now that we know our fruits, we can chop them up and have them with some pancakes. **We deserve it**.

## FAQs

### How do you calculate row reduced echelon? ›

...

**To change X to its reduced row echelon form, we take the following steps:**

- Interchange Rows 1 and 2, producing X
_{1}. - In X
_{1}, multiply Row 2 by -5 and add it to Row 3, producing X_{2}. - In X
_{2}, multiply Row 2 by -2 and add it to Row 1, producing X_{rref}.

**What is echelon formula? ›**

**A system of linear equations is said to be in row echelon form if its augmented matrix is in row echelon form**. Similarly, a system of linear equations is said to be in reduced row echelon form or in canonical form if its augmented matrix is in reduced row echelon form.

**How do you calculate a row reduction? ›**

To row reduce a matrix: **Perform elementary row operations to yield a "1" in the first row, first column**. Create zeros in all the rows of the first column except the first row by adding the first row times a constant to each other row. Perform elementary row operations to yield a "1" in the second row, second column.

**What is the command to reduced echelon form of a matrix? ›**

**R = rref( A )** returns the reduced row echelon form of A using Gauss-Jordan elimination with partial pivoting. R = rref( A , tol ) specifies a pivot tolerance that the algorithm uses to determine negligible columns.

**What is meant by reduced echelon form? ›**

Definition of reduced row echelon form

Definition We say that **a matrix is in reduced row echelon form if and only if it is in row echelon form, all its pivots are equal to 1 and the pivots are the only non-zero entries of the basic columns**.

**Is Gauss-Jordan reduced row echelon form? ›**

**The purpose of Gauss-Jordan Elimination is to use the three elementary row operations to convert a matrix into reduced-row echelon form**. A matrix is in reduced-row echelon form, also known as row canonical form, if the following conditions are satisfied: All rows with only zero entries are at the bottom of the matrix.

**What is echelon form of matrix examples? ›**

A rectangular matrix is in echelon form if it has the three properties: **All non-zero rows are above any rows of all zeros**. Each leading entry of a row is in a column to right of the leading entry of the row above it. All entries in a column below a leading entry are zeros.

**Why echelon form is used? ›**

The major application of row echelon form is Gaussian elimination. In linear algebra, Gaussian elimination is a method used on coefficent matrices **to solve systems of linear equations**. By transforming matrices into row echelon form, the values of the variables given the coefficients becomes evident.

**What are the rules for row echelon form? ›**

**Echelon Form**

- All zero rows are at the bottom of the matrix.
- The leading entry of each nonzero row after the first occurs to the right of the leading entry of the previous row.
- The leading entry in any nonzero row is 1.
- All entries in the column above and below a leading 1 are zero.

**What are the possible reduced row echelon forms of 3x3 matrices? ›**

For a 3×3 matrix in reduced row echelon form to have rank 1, it must have 2 rows which are all 0s. The non-zero row must be the first row, and it must have a leading 1. These conditions imply that the matrix must be of one of the following forms: **[1ab000000],[01c000000], or [001000000]**.

### How do you find a 3x3 matrix? ›

To find determinant of 3x3 matrix, you first take the first element of the first row and multiply it by a secondary 2x2 matrix which comes from the elements remaining in the 3x3 matrix that do not belong to the row or column to which your first selected element belongs.

**Is there only one reduced row echelon form? ›**

Any nonzero matrix may be row reduced into more than one matrix in echelon form, by using different sequences of row operations. However, no matter how one gets to it, **the reduced row echelon form of every matrix is unique**.

**How many types of 2 2 matrices in reduced row echelon form are there? ›**

There are **4** types of 2x2 matrices in rref: ( %), ( 0).

**Can every matrix be reduced to row echelon form? ›**

As we have seen in earlier sections, we know that **every matrix can be brought into reduced row-echelon form by a sequence of elementary row operations**.

**What is the difference between row echelon form and reduced row echelon form? ›**

Echelon Form vs Reduced Echelon Form

Continuing the elimination process gives a matrix with all the other terms of a column containing a 1 is zero. A matrix in that form is said to be in the reduced row echelon form. But the above condition restricts the possibility of having columns with values except 1 and zero.

**Is rref same as Gauss-Jordan? ›**

**Gauss-Jordan elimination is a mechanical procedure for transforming a given system of linear equations to Rx=d with R in RREF using only elementary row operations**. In casual terms, the process of transforming a matrix into RREF is called row reduction.

**What is the difference between Gaussian and Gauss-Jordan? ›**

Difference between gaussian elimination and gauss jordan elimination. The difference between Gaussian elimination and the Gaussian Jordan elimination is that **one produces a matrix in row echelon form while the other produces a matrix in row reduced echelon form**.

**Is zero matrix reduced row echelon? ›**

**The zero matrix is vacuously in reduced row echelon form** as it satisfies: All zero rows are at the bottom of the matrix. The leading entry of each nonzero row subsequently to the first is right of the leading entry of the preceding row. The leading entry in any nonzero row is a 1.

**Which matrix is in row echelon form? ›**

A matrix is in row echelon form (ref) when it satisfies the following conditions. The first non-zero element in each row, called the leading entry, is 1. Each leading entry is in a column to the right of the leading entry in the previous row.

**Is null matrix A echelon form? ›**

Hint: Here, we have given a null matrix and we should know that **a null matrix is a zero matrix**. We should also know that the zero matrix has no non-zero rows or columns in the echelon form to find the required value.

### Why is reduced echelon form useful? ›

So echelon form **expedites solving system of equations**. There are other benefits as well such as discerning the rank, dimension of various subspaces, etc.

**Why is reduced echelon form used? ›**

Reduced row-echelon form of a matrix is used **to solve the system of linear equations**. Matrix is said to be in reduced row-echelon form or reduced echelon form if: The number 1 is the first non-zero integer in each row, i.e., the leading entry.

**Does row echelon form have to start with 1? ›**

Row echelon form must meet three requirements: **The leading coefficient of each row must be a one**. All entries in a column below a leading one must be zero. All rows that just contain zeros are at the bottom of the matrix.

**How do you know if a matrix is in reduced row echelon form? ›**

**A matrix is in reduced row-echelon form if it satisfies the following:**

- In each row, the left-most nonzero entry is 1 and the column that contains this 1 has all other entries equal to 0. ...
- The leading 1 in the second row or beyond is to the right of the leading 1 in the row just above.

**What is Cramer's rule 3x3? ›**

Cramer's Rule **requires us to find the determinant of 2 x 2 and 3 x 3 matrices** (depends on your linear system). However, this rule can only be used if you have the same number of equations and variables. If you have a different number of equations and variables, then finding the determinant will be impossible.

**Is Row reduction only for augmented matrices? ›**

**The row reduction algorithm applies only to augmented matrices for a linear system**. Answer: False. Any matrix can be reduced.

**How do you solve a system of equations with an augmented matrix? ›**

Step 1: Translate the system of linear equations into an augmented matrix. Step 2: Use elementary row operations to get a leading 1 in the first row. Step 3: Use elementary row operations to get zeroes below the leading 1 . Step 4: Use elementary row operations to get a leading 1 in the second row, second column.

**How do you find the inverse of a matrix using an augmented matrix? ›**

The Method for Finding the Inverse of a Matrix

**Write the augmented matrix in step 1 in reduced row echelon form.** If the reduced row echelon form in 2 is [In|B], then B is the inverse of A. If the left side of the row reduced echelon is not an identity matrix, the inverse does not exist.

**Can there be multiple reduced row echelon forms? ›**

**Any nonzero matrix may be row reduced into more than one matrix in echelon form**, by using different sequences of row operations. However, no matter how one gets to it, the reduced row echelon form of every matrix is unique.

**Why do we want a matrix in row reduced form? ›**

The main point of row operations is that they do not change the solution set of the underlying linear system. So when you take a system of linear equations, write down its (augmented) coefficient matrix, and row reduce that matrix, you get a new system of equations that has the same solutions as the original system.

### What is the difference between matrix and augmented matrix? ›

Solution: A coefficient matrix is a matrix made up of the coefficients from a system of linear equations. An augmented matrix is similar in that it, too, is a coefficient matrix, but in addition it is augmented with a column consisting of the values on the right-hand side of the equations of the linear system.

**What is the fastest way to find the inverse of a matrix? ›**

We can find the matrix inverse only for square matrices, whose number of rows and columns are equal such as 2 × 2, 3 × 3, etc. In simple words, inverse matrix is obtained by **dividing the adjugate of the given matrix by the determinant of the given matrix**.

**Can you find the inverse of a 3x3 matrix? ›**

What is the Inverse of 3x3 Matrix? The inverse of a 3x3 matrix, say A, is **a matrix of the same order denoted by A ^{-}^{1} where AA^{-}^{1} = A^{-}^{1}A = I, where I is the identity matrix of order 3x3**. i.e., I = ⎡⎢⎣100010010⎤⎥⎦ [ 1 0 0 0 1 0 0 1 0 ] .

**What is augmented matrix example? ›**

Examples on Augmented Matrix

Example 1: Represent the equations **3x + 2y + z = 8, 4x - 3y + 3z = 7, and x + 5y - 3z = 11**, as an augmented matrix. Solution: The three given equations are: 3x + 2y + z = 8.

**Why do we use augmented matrix? ›**

In linear algebra, an augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices. This is **useful when solving systems of linear equations**.