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## 11.6: Solving Systems with Gaussian Elimination

## Learning Objectives

- Write the augmented matrix of a system of equations.
- Write the system of equations from an augmented matrix.
- Perform row operations on a matrix.
- Solve a system of linear equations using matrices.

## Writing the Augmented Matrix of a System of Equations

For example, consider the following \(2 × 2\) system of equations.

\[\begin{align*} 3x+4y&= 7\\ 4x-2y&= 5 \end{align*}\]

We can write this system as an augmented matrix:

\(\left[ \begin{array}{cc|c} 3&4&7\\4&-2&5\end{array} \right]\)

We can also write a matrix containing just the coefficients. This is called the coefficient matrix.

\(\begin{bmatrix}3&4\\4&−2\end{bmatrix}\)

A three-by-three system of equations such as

\[\begin{align*} 3x-y-z&= 0\\ x+y&= 5\\ 2x-3z&= 2 \end{align*}\]

\(\begin{bmatrix}3&−1&−1\\1&1&0\\2&0&−3\end{bmatrix}\)

and is represented by the augmented matrix

\(\left[ \begin{array}{ccc|c}3&−1&−1&0\\1&1&0&5\\2&0&−3&2\end{array} \right]\)

## How to: Given a system of equations, write an augmented matrix

- Write the coefficients of the \(x\)-terms as the numbers down the first column.
- Write the coefficients of the \(y\)-terms as the numbers down the second column.
- If there are \(z\)-terms, write the coefficients as the numbers down the third column.
- Draw a vertical line and write the constants to the right of the line.

## Example \(\PageIndex{1}\): Writing the Augmented Matrix for a System of Equations

Write the augmented matrix for the given system of equations.

\[\begin{align*} x+2y-z&= 3\\ 2x-y+2z&= 6\\ x-3y+3z&= 4 \end{align*}\]

\(\left[ \begin{array}{ccc|c}1&2&−1&3\\2&−1&2&6\\1&−3&3&4\end{array} \right]\)

## Exercise \(\PageIndex{1}\)

Write the augmented matrix of the given system of equations.

\[\begin{align*} 4x-3y&= 11\\ 3x+2y&= 4 \end{align*}\]

\(\left[ \begin{array}{cc|c} 4&−3&11\\3&2&4\end{array} \right]\)

## Writing a System of Equations from an Augmented Matrix

## Example \(\PageIndex{2}\): Writing a System of Equations from an Augmented Matrix Form

Find the system of equations from the augmented matrix.

\(\left[ \begin{array}{ccc|c}1&−3&−5&-2\\2&−5&−4&5\\−3&5&4&6 \end{array} \right]\)

When the columns represent the variables \(x\), \(y\), and \(z\),

## Exercise \(\PageIndex{2}\)

Write the system of equations from the augmented matrix.

\(\left[ \begin{array}{ccc|c}1&−1& 1&5\\2&−1&3&1\\0&1&1&-9\end{array}\right]\)

\(\begin{align*} x-y+z&= 5\\ 2x-y+3z&= 1\\ y+z&= -9 \end{align*}\)

## Performing Row Operations on a Matrix

Row-echelon form \(\begin{bmatrix}1&a&b\\0&1&d\\0&0&1\end{bmatrix}\)

- In any nonzero row, the first nonzero number is a \(1\). It is called a leading \(1\).
- Any all-zero rows are placed at the bottom on the matrix.
- Any leading \(1\) is below and to the right of a previous leading \(1\).
- Any column containing a leading \(1\) has zeros in all other positions in the column.

- Interchange rows. (Notation: \(R_i ↔ R_j\))
- Multiply a row by a constant. (Notation: \(cR_i\))
- Add the product of a row multiplied by a constant to another row. (Notation: \(R_i+cR_j\))

## GAUSSIAN ELIMINATION

## How to: Given an augmented matrix, perform row operations to achieve row-echelon form

- The first equation should have a leading coefficient of \(1\). Interchange rows or multiply by a constant, if necessary.
- Use row operations to obtain zeros down the first column below the first entry of \(1\).
- Use row operations to obtain a \(1\) in row 2, column 2.
- Use row operations to obtain zeros down column 2, below the entry of 1.
- Use row operations to obtain a \(1\) in row 3, column 3.
- Continue this process for all rows until there is a \(1 in every entry down the main diagonal and there are only zeros below.
- If any rows contain all zeros, place them at the bottom.

## Example \(\PageIndex{3}\): Solving a \(2×2\) System by Gaussian Elimination

Solve the given system by Gaussian elimination.

\[\begin{align*} 2x+3y&= 6\\ x-y&= \dfrac{1}{2} \end{align*}\]

First, we write this as an augmented matrix.

\(\left[ \begin{array}{cc|c} 2&3&6\\1&−1&12\end{array} \right]\)

We want a \(1\) in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.

\(R_1\leftrightarrow R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\2&3&6\end{array} \right]\)

\(-2R_1+R_2=R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\0&5&5\end{array} \right]\)

We only have one more step, to multiply row 2 by \(\dfrac{1}{5}\).

\(\dfrac{1}{5}R_2=R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\0&1&1\end{array} \right]\)

\[\begin{align*} x-(1)&= \dfrac{1}{2}\\ x&= \dfrac{3}{2} \end{align*}\]

The solution is the point \(\left(\dfrac{3}{2},1\right)\).

## Exercise \(\PageIndex{3}\)

\[\begin{align*} 4x+3y&= 11\\ x-3y&= -1 \end{align*}\]

## Example \(\PageIndex{4}\): Using Gaussian Elimination to Solve a System of Equations

Use Gaussian elimination to solve the given \(2 × 2\) system of equations .

\[\begin{align*} 2x+y&= 1\\ 4x+2y&= 6 \end{align*}\]

Write the system as an augmented matrix .

\(\left[ \begin{array}{cc|c} 2&1&1\\4&2&6\end{array} \right]\)

Next, we want a \(0\) in row 2, column 1. Multiply row 1 by \(−4\) and add row 1 to row 2.

## Example \(\PageIndex{5}\): Solving a Dependent System

Solve the system of equations.

\[\begin{align*} 3x+4y&= 12\\ 6x+8y&= 24 \end{align*}\]

Perform row operations on the augmented matrix to try and achieve row-echelon form .

\(A=\left[ \begin{array}{cc|c} 3&4&12\\6&8&24\end{array} \right]\)

\(-\dfrac{1}{2}R_2+R_1=R_1\rightarrow \left[ \begin{array}{cc|c} 0&0&0\\6&8&24\end{array} \right]\)

\(R_1\leftrightarrow R_2=\left[ \begin{array}{cc|c} 6&8&24\\0&0&0\end{array} \right]\)

\[\begin{align*} 3x+4y&= 12\\ 4y&= 12-3x\\ y&= 3-\dfrac{3}{4}x \end{align*}\]

So the solution to this system is \(\left(x,3−\dfrac{3}{4}x\right)\).

## Example \(\PageIndex{6}\): Performing Row Operations on a \(3×3\) Augmented Matrix to Obtain Row-Echelon Form

Perform row operations on the given matrix to obtain row-echelon form.

\(\left[ \begin{array}{ccc|c} 1&-3&4&3\\2&-5&6&6\\-3&3&4&6\end{array} \right]\)

\(-2R_1+R_2=R_2 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\-3&3&4&6\end{array} \right]\)

Next, obtain a zero in row 3, column 1.

\(3R_1+R_3=R_3 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\0&-6&16&15\end{array} \right]\)

Next, obtain a zero in row 3, column 2.

\(6R_2+R_3=R_3 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\0&0&4&15\end{array} \right]\)

The last step is to obtain a 1 in row 3, column 3.

## Exercise \(\PageIndex{4}\)

Write the system of equations in row-echelon form.

\[\begin{align*} x−2y+3z &= 9 \\ −x+3y &= −4 \\ 2x−5y+5z &= 17 \end{align*}\]

## Solving a System of Linear Equations Using Matrices

## Example \(\PageIndex{7}\): Solving a System of Linear Equations Using Matrices

Solve the system of linear equations using matrices.

\[\begin{align*} x-y+z&= 8\\ 2x+3y-z&= -2\\ 3x-2y-9z&= 9 \end{align*}\]

First, we write the augmented matrix.

\(\left[ \begin{array}{ccc|c} 1&-1&1&8\\2&3&-1&-2\\3&-2&-9&9\end{array} \right]\)

Next, we perform row operations to obtain row-echelon form.

The easiest way to obtain a \(1\) in row 2 of column 1 is to interchange \(R_2\) and \(R_3\).

The last matrix represents the equivalent system.

\[\begin{align*} x−y+z &= 8 \\ y−12z &= −15 \\ z &= 1 \end{align*}\]

Using back-substitution, we obtain the solution as \((4,−3,1)\).

## Example \(\PageIndex{8}\): Solving a Dependent System of Linear Equations Using Matrices

Solve the following system of linear equations using matrices.

\[\begin{align*} −x−2y+z &= −1 \\ 2x+3y &= 2 \\ y−2z &= 0 \end{align*}\]

\(\left[ \begin{array}{ccc|c} -1&-2&1&-1\\2&3&0&2\\0&1&-2&0\end{array} \right]\)

\(-R_1\rightarrow \left[ \begin{array}{ccc|c} 1&2&-1&1\\2&3&0&2\\0&1&-2&0\end{array} \right]\)

The last matrix represents the following system.

\[\begin{align*} x+2y−z &= 1 \\ y−2z &= 0 \\ 0 &= 0 \end{align*}\]

The generic solution is \(\left(x,\dfrac{2−2x}{3},\dfrac{1−x}{3}\right)\).

## Exercise \(\PageIndex{5}\)

Solve the system using matrices.

\[\begin{align*} x+4y-z&= 4\\ 2x+5y+8z&= 1\\ 5x+3y-3z&= 1 \end{align*}\]

## Q&A: Can any system of linear equations be solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination.

## How to: Given a system of equations, solve with matrices using a calculator

- Save the augmented matrix as a matrix variable \([A], [B], [C], ….\)
- Use the ref( function in the calculator, calling up each matrix variable as needed.

## Example \(\PageIndex{9A}\): Solving Systems of Equations with Matrices Using a Calculator

\[\begin{align*} 5x+3y+9z&= -1\\ -2x+3y-z&= -2\\ -x-4y+5z&= 1 \end{align*}\]

Write the augmented matrix for the system of equations.

\(\left[ \begin{array}{ccc|c} 5&3&9&-1\\-2&3&-1&-2\\-1&-4&5&1\end{array} \right]\)

\([A]=\left[ \begin{array}{ccc|c} 5&3&9&-1\\-2&3&-1&-2\\-1&-4&5&1\end{array} \right]\)

Use the ref( function in the calculator, calling up the matrix variable \([A]\).

## Example \(\PageIndex{9B}\): Applying \(2×2\) Matrices to Finance

\[\begin{align*} x+y&= 12,000\\ 0.105x+0.12y&= 1,335 \end{align*}\]

\(\left[ \begin{array}{cc|c} 1&1&12,000\\0.105&0.12&1,335\end{array} \right]\)

Multiply row 1 by \(−0.105\) and add the result to row 2.

\(\left[ \begin{array}{cc|c} 1&1&12,000\\0&0.015&75\end{array} \right]\)

\[\begin{align*} 0.015y &= 75 \\ y &= 5,000 \end{align*}\]

Thus, \($5,000\) was invested at \(12%\) interest and \($7,000\) at \(10.5%\) interest.

## Example \(\PageIndex{10}\): Applying \(3×3\) Matrices to Finance

\[\begin{align*} x+y+z &= 10,000 \\ 0.05x+0.08y+0.09z &= 770 \\ 2x−z &= 0 \end{align*}\]

\(\left[ \begin{array}{ccc|c} 1&1&1&10,000\\0.05&0.08&0.09&770\\2&0&-1&0\end{array} \right]\)

Now, we perform Gaussian elimination to achieve row-echelon form.

The third row tells us \(−\dfrac{1}{3}z=−2,000\); thus \(z=6,000\).

The second row tells us \(y+\dfrac{4}{3}z=9,000\). Substituting \(z=6,000\),we get

\[\begin{align*} y+\dfrac{4}{3}(6,000) &= 9,000 \\ y+8,000 &= 9,000 \\ y &= 1,000 \end{align*}\]

The first row tells us \(x+y+z=10,000\). Substituting \(y=1,000\) and \(z=6,000\),we get

\[\begin{align*} x+1,000+6,000 &= 10,000 \\ x &= 3,000 \end{align*}\]

## Exercise \(\PageIndex{6}\)

\($150,000\) at \(7%\), \($750,000\) at \(8%\), \($600,000\) at \(10%\)

- Solve a System of Two Equations Using an Augmented Matrix
- Solve a System of Three Equations Using an Augmented Matrix
- Augmented Matrices on the Calculator

## Key Concepts

- An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example \(\PageIndex{1}\).
- A matrix augmented with the constant column can be represented as the original system of equations. See Example \(\PageIndex{2}\).
- Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows.
- We can use Gaussian elimination to solve a system of equations. See Example \(\PageIndex{3}\), Example \(\PageIndex{4}\), and Example \(\PageIndex{5}\).
- Row operations are performed on matrices to obtain row-echelon form. See Example \(\PageIndex{6}\).
- To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions. See Example \(\PageIndex{7}\) and Example \(\PageIndex{8}\).
- A calculator can be used to solve systems of equations using matrices. See Example \(\PageIndex{9}\).
- Many real-world problems can be solved using augmented matrices. See Example \(\PageIndex{10}\) and Example \(\PageIndex{11}\).

## Gauss Elimination Method

This can be summarized in a table as given below:

## What is the Gauss Elimination Method?

- Swapping two rows and this can be expressed using the notation ↔, for example, R 2 ↔ R 3
- Multiplying a row by a nonzero number, for example, R 1 → kR 2 where k is some nonzero number
- Adding a multiple of one row to another row, for example, R 2 → R 2 + 3R 1

Learn more about the elementary operations of a matrix here.

## Gauss Elimination Method with Example

Let’s have a look at the gauss elimination method example with a solution.

Solve the following system of equations:

Given system of equations are:

Let us write these equations in matrix form.

Subtracting R 1 from R 2 to get the new elements of R 2 , i.e. R 2 → R 2 – R 1 .

Let us make another operation as R 3 → R 3 – 2R 1

Subtract R 2 from R 1 to get the new elements of R 1 , i.e. R 1 → R 1 – R 2 .

Now, subtract R 2 from R 3 to get the new elements of R 3 , i.e. R 3 → R 3 – R 2 .

That means, there is no solution for the given system of equations.

## Gauss Elimination Method Problems

1. Solve the following system of equations using Gauss elimination method.

2. Solve the following linear system using the Gaussian elimination method.

3. Using Gauss elimination method, solve:

## Frequently Asked Questions on Gauss Elimination Method

Why gauss elimination method is used.

## What type of method is the Gauss elimination method?

## What are the steps of the Gauss elimination method?

## How Gauss-Jordan Method is different from the Gauss-Elimination method?

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Example 1 : Solve this system:

Example 2 : Solve this system:

The first step is to write the coefficients of the unknowns in a matrix:

Then, perform a sequence of elementary row operations , which are any of the following:

Type 1. Interchange any two rows.

Type 2. Multiply a row by a nonzero constant.

Type 3. Add a multiple of one row to another row.

Example 3 : Solve the following system using Gaussian elimination:

The augmented matrix which represents this system is

Now, add −5 times the second row to the third row:

Example 4 : Solve the following system using Gaussian elimination:

For this system, the augmented matrix (vertical line omitted) is

Interchanging the second and third rows then gives the desired upper‐triangular coefficient matrix:

Therefore, the goal is solve the system

The augmented matrix for this system is reduced as follows:

This final matrix immediately gives the solution: a = −5, b = 10, and c = 2.

Example 6 : Solve the following system using Gaussian elimination:

The augmented matrix for this system is

Next, −1 times the second row is added to the third row:

Example 7 : Solve the following system using Gaussian elimination:

Back substituting z = t and y = 6 + 5 t into the first row ( x + y − 3 z = 4) determines x :

Therefore, every solution of the system has the form

Example 8 : Find all solutions to the system

Thus, the solutions of the system have the form

where t 1 t 2 are allowed to take on any real values.

Example 9 : Let b = ( b 1 , b 2 , b 3 ) T and let A be the matrix

For what values of b 1 , b 2 , and b 3 will the system A x = b be consistent?

The augmented matrix for the system A x = b reads

which Gaussian eliminatin reduces as follows:

Example 10 : Solve the following system (compare to Example 12):

and back‐substitution of z = t and y = 5 t into the first row ( x + y − 3 z = 0) determines x :

general solution to Ax = 0 : ( x, y, z ) = (−2 t , 5 t , t )

general solution to Ax = b : ( x, y, z ) = (−2 t , 5 t , t ) + (−2, 6, 0)

Example 11 : Determine all solutions of the system

Write down the augmented matrix and perform the following sequence of operations:

Therefore, every solution of this system has the form

where t 1 and t 2 are any real numbers. Another way to write the solution is as follows:

Example 12 : Determine the general solution of

which is the homogeneous system corresponding to the nonhomoeneous one in Example 11 above.

Since the solution to the nonhomogeneous system in Example 11 is

Therefore, x 1 + t ( x 1 − x 2 ) is indeed a solution of A x = b , and the theorem is proved.

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## Gaussian Elimination

Gaussian elimination is a method for solving matrix equations of the form

To perform Gaussian elimination starting with the system of equations

compose the " augmented matrix equation"

A matrix that has undergone Gaussian elimination is said to be in echelon form .

For example, consider the matrix equation

In augmented form, this becomes

Subtracting 9 times the first row from the third row gives

Subtracting 4 times the first row from the second row gives

Restoring the transformed matrix equation gives

## Explore with Wolfram|Alpha

## Referenced on Wolfram|Alpha

## Subject classifications

## Augmented Matrix

## What is Gaussian Elimination?

- to represent a system of linear equations in an augmented matrix form
- then performing the $ 3 $ row operations on it until the reduced row echelon form (RREF) is achieved
- Lastly, we can easily recognize the solutions from the RREF

A system of linear equations is shown below:

$ \begin{align*} 2x + 3y &= \,7 \\ x – y &= 4 \end{align*} $

$ \left[ \begin{array}{ r r | r } 2 & 3 & 7 \\ 1 & -1 & 4 \end{array} \right] $

An example using $ 3 $ simultaneous equations is shown below:

$ \begin{align*} 2x + y + z &= \,10 \\ x + 2y + 3z &= 1 \\ – x – y – z &= 2 \end{align*} $

Representing this system as an augmented matrix:

- Interchange $ 2 $ rows
- Multiply a row by a non-zero ($ \neq 0 $) scalar
- Add or subtract the scalar multiple of one row to another row.

- Rows with zero entries (all elements of that row are $ 0 $s) are at the matrix’s bottom.
- The leading entry (the first non-zero entry in a row) of each non-zero row is to the right of the row’s leading entry directly above it.
- The leading entry in any non-zero row is $ 1 $.
- All entries in the column containing the leading entry ($ 1 $) are zeroes.

## How to do Gauss Jordan Elimination

- Swap rows so that all rows with zero entries are on the bottom of the matrix.
- Swap rows so that the row with the largest left-most digit is on the top of the matrix.
- Multiply the top row by a scalar that converts the top row’s leading entry into $ 1 $ (If the leading entry of the top row is $ a $, then multiply it by $ \frac{ 1 }{ a } $ to get $ 1 $ ).
- Add or subtract multiples of the top row to the other rows so that the entry’s in the column of the top row’s leading entry are all zeroes.
- Perform Steps $ 2 – 4 $ for the next leftmost non-zero entry until all the leading entries of each row are $ 1 $.
- Swap the rows so that the leading entry of each nonzero row is to the right of the leading entry of the row directly above it

Solve the system shown below using the Gauss Jordan Elimination method:

$ \begin{align*} { – x } + 2y &= \, { – 6 } \\ { 3x } – 4y &= { 14 } \end{align*} $

The first step is to write the augmented matrix of the system. We show this below:

$ \left[ \begin{array}{ r r | r } – 1 & 2 & – 6 \\ 3 & -4 & 14 \end{array} \right] $

The augmented matrix that we have is:

$ \left[ \begin{array}{ r r | r } – 1 & 2 & – 6 \\ 3 & – 4 & 14 \end{array} \right] $

We can multiply the first row by $ – 1 $ to make the leading entry $ 1 $. Shown below:

$ \left[ \begin{array}{ r r | r } 1 & – 2 & 6 \\ 3 & – 4 & 14 \end{array} \right] $

We can now multiply the first row by $ 3 $ and subtract it from the second row. Shown below:

$ = \left[ \begin{array}{ r r | r } 1 & – 2 & 6 \\ 0 & 2 & – 4 \end{array} \right] $

We have a $ 0 $ as the first entry of the second row.

$ = \left[ \begin{array}{ r r | r } 1 & – 2 & 6 \\ 0 & 1 & – 2 \end{array} \right] $

$ = \left[ \begin{array}{ r r | r } 1 & 0 & 2 \\ 0 & 1 & – 2 \end{array} \right] $

$ \begin{align*} x + 0y &= \, 2 \\ 0x + y &= -2 \end{align*} $

$ \begin{align*} x &= \, 2 \\ y &= – 2 \end{align*} $

Thus, the solution of the system of equations is $ x = 2 $ and $ y = – 2 $.

$ \begin{align*} x + 2y &= \, 4 \\ x – 2y &= 6 \end{align*} $

Let’s write the augmented matrix of the system of equations:

$ \left[ \begin{array}{ r r | r } 1 & 2 & 4 \\ 1 & – 2 & 6 \end{array} \right] $

$ \left[ \begin{array}{ r r | r } 1 & 2 & 4 \\ 1 – 1 & – 2 – 2 & 6 – 4 \end{array} \right] $

$ =\left[ \begin{array}{ r r | r } 1 & 2 & 4 \\ 0 & – 4 & 2 \end{array} \right] $

We multiply the second row by $ -\frac{ 1 }{ 4 }$ to make the second entry of the row, $ 1 $:

$ =\left[ \begin{array}{ r r | r } 1 & 2 & 4 \\ 0 & 1 & -\frac{ 1 }{ 2 } \end{array} \right] $

$=\left[ \begin{array}{ r r | r } 1 & 0 & 5 \\ 0 & 1 & -\frac{ 1 }{ 2 } \end{array} \right] $

$ \begin{align*} x + 0y &= \, 5 \\ 0x+ y &= -\frac{ 1 }{ 2 } \end{align*} $

$ \begin{align*} x &= \, 5 \\ y &= -\frac{ 1 }{ 2 } \end{align*} $

Thus, the solution of the system of equations is $ x = 5 $ and $ y = -\frac{ 1 }{ 2 } $.

$ \begin{align*} 2x + y &= \, – 3 \\ – x – y &= 2 \end{align*} $

$ \begin{align*} x + 5y &= \, 15 \\ – x + 5y &= 25 \end{align*} $

We start off by writing the augmented matrix of the system of equations:

$ \left[ \begin{array}{r r | r} 2 & 1 & – 3 \\ – 1 & – 1 & 2 \end{array} \right] $

Now, we do the elementary row operations to arrive at our solution.

From this augmented matrix , we can write two equations (solutions):

$ \begin{align*} x + 0y &= \, – 1 \\ 0x+ y &= – 1 \end{align*} $

$ \begin{align*} x &= \, – 1 \\ y &= – 1 \end{align*} $

Thus, the solution of the system of equations is $ x = – 1 $ and $ y = – 1 $.

$ \begin{align*} x &= \, – 5 \\ y &= 4 \end{align*} $

Thus, the solution of the system of equations is $ x = – 5 $ and $ y = 4 $.

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## Gaussian Elimination Calculator

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## What Is Gaussian Elimination Algorithm?

In the light of mathematical analysis:

## How to Apply Gaussian Elimination Algorithm?

Find solution of the following system of equations as under:

The equivalent augmented matrix form of the above equations are as follows:

$$ \begin{bmatrix} 3&6&23 \\ 6&2&34 \\\end{bmatrix} $$

$$ \left[\begin{array}{cc|c}1&2& \frac{23}{3} \\6&2&34 \\\end{array}\right] $$

Multiply the first row by 6 and then subtract it from the zeroth row.

$$ \left[\begin{array}{cc|c}1&2&\frac{23}{3} \\0&-10&-12 \\\end{array}\right] $$

Go for dividing the first row by -10.

$$ \left[\begin{array}{cc|c}1&2&\frac{23}{3} \\0&1&\frac{6}{5}\\\end{array}\right] $$

So the final results are as follows:

The same results can also be verified by using outer free gauss elimination calculator.

## How Gaussian Elimination Method Calculator Works?

- First, set up the order of the matrix from drop-down lists
- After you do that, click the “Set Matrices” button to get the desired matrix format
- Now fetch the numbers in their fields
- After you are done with the stuff, hit the calculate button

The best gauss jordan elimination calculator with steps does the following calculations:

## References:

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## Keyboard Shortcuts

- Swap the positions of two of the rows
- Multiply one of the rows by a nonzero scalar.
- Add or subtract the scalar multiple of one row to another row.

For an example of the first elementary row operation, swap the positions of the 1st and 3rd row.

For an example of the second elementary row operation, multiply the second row by 3.

For an example of the third elementary row operation, add twice the 1st row to the 2nd row.

## Reduced-row echelon form

- All rows with only zero entries are at the bottom of the matrix
- The first nonzero entry in a row, called the leading entry or the pivot , of each nonzero row is to the right of the leading entry of the row above it.
- The leading entry, also known as the pivot, in any nonzero row is 1.
- All other entries in the column containing a leading 1 are zeroes.

## Steps for Gauss-Jordan Elimination

To perform Gauss-Jordan Elimination:

- Swap the rows so that all rows with all zero entries are on the bottom
- Swap the rows so that the row with the largest, leftmost nonzero entry is on top.
- Multiply the top row by a scalar so that top row's leading entry becomes 1.
- Add/subtract multiples of the top row to the other rows so that all other entries in the column containing the top row's leading entry are all zero.
- Repeat steps 2-4 for the next leftmost nonzero entry until all the leading entries are 1.
- Swap the rows so that the leading entry of each nonzero row is to the right of the leading entry of the row above it.

Selected video examples are shown below:

- Gauss-Jordan Elimination - Jonathan Mitchell (YouTube)
- Using Gauss-Jordan to Solve a System of Three Linear Equations - Example 1 - patrickJMT (YouTube)
- Algebra - Matrices - Gauss Jordan Method Part 1 Augmented Matrix - IntuitiveMath (YouTube)
- Gaussian Elimination - patrickJMT (YouTube)

To obtain the inverse of a n × n matrix A :

- Create the partitioned matrix \(( A | I )\) , where I is the identity matrix.
- Perform Gauss-Jordan Elimination on the partitioned matrix with the objective of converting the first part of the matrix to reduced-row echelon form.
- If done correctly, the resulting partitioned matrix will take the form \(( I | A^{-1} )\)
- Double-check your work by making sure that \(AA^{-1} = I\).

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## Gaussian Elimination technique by matlab

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Solve the given system by Gaussian elimination. 4x + 3y = 11 x − 3y = − 1 Answer Example 9.6.4: Using Gaussian Elimination to Solve a System of Equations Use Gaussian elimination to solve the given 2 × 2 system of equations. 2x + y = 1 4x + 2y = 6 Solution Write the system as an augmented matrix. [2 1 1 4 2 6] Obtain a 1 in row 1, column 1.

351K views 5 years ago New Precalculus Video Playlist This precalculus video tutorial provides a basic introduction into the gaussian elimination with 4 variables using elementary row...

We then find the generic solution. By solving the second equation for y and substituting it into the first equation we can solve for z in terms of x. x + 2y − z = 1 y = 2z x + 2(2z) − z = 1 x + 3z = 1 z = 1 − x 3. Now we substitute the expression for z into the second equation to solve for y in terms of x.

Gauss Elimination Method Problems 1. Solve the following system of equations using Gauss elimination method. x + y + z = 9 2x + 5y + 7z = 52 2x + y - z = 0 2. Solve the following linear system using the Gaussian elimination method. 4x - 5y = -6 2x - 2y = 1 3. Using Gauss elimination method, solve: 2x - y + 3z = 9 x + y + z = 6 x - y + z = 2

Gauss-Jordan is augmented by an n x n identity matrix, which will yield the inverse of the original matrix as the original matrix is manipulated into the identity matrix. In the case that Sal is discussing above, we are augmenting with the linear "answers", and solving for the variables (in this case, x_1, x_2, x_3, x_4) when we get to row ...

Example 4: Solve the following system using Gaussian elimination: For this system, the augmented matrix (vertical line omitted) is First, multiply row 1 by 1/2: Now, adding −1 times the first row to the second row yields zeros below the first entry in the first column:

Gaussian elimination is a method for solving matrix equations of the form. (1) To perform Gaussian elimination starting with the system of equations. (2) compose the " augmented matrix equation". (3) Here, the column vector in the variables is carried along for labeling the matrix rows. Now, perform elementary row operations to put the ...

In general, when the process of Gaussian elimination without pivoting is applied to solving a linear system Ax= b,weobtainA= LUwith Land Uconstructed as above. For the case in which partial pivoting is used, we ob-tain the slightly modiﬁed result LU= PA where Land Uare constructed as before and Pis a permutation matrix. For example, consider P=

http://www.greenemath.com/http://www.facebook.com/mathematicsbyjgreeneIn this video, we will look at one example of how to solve a four variable linear syste...

Let's take a few examples to elucidate the process of solving a system of linear equations via the Gauss Jordan Elimination Method. Example 1 Solve the system shown below using the Gauss Jordan Elimination method: - x + 2 y = - 6 3 x - 4 y = 14 Solution The first step is to write the augmented matrix of the system. We show this below:

Gaussian elimination is the process of using valid row operations on a matrix until it is in reduced row echelon form. There are three types of valid row operations that may be performed on a...

Gauss-Jordan elimination is a lot faster but only for certain matrices--if the inverse matrix ends up having loads of fractions in it, then it's too hard to see the next step for Gauss-Jordan and the determinant/adjugate method is the only way I can solve the problem without pulling my hair out.

Example # 01: Find solution of the following system of equations as under: 3x1 + 6x2 = 23 6x1 + 2x2 = 34 Solution: No doubt our widely used gaussian elimination calculator with steps will show detailed calculations to simplify these equations, but we need to analyse the scenario manually.

Steps for Gauss-Jordan Elimination. To perform Gauss-Jordan Elimination: Swap the rows so that all rows with all zero entries are on the bottom. Swap the rows so that the row with the largest, leftmost nonzero entry is on top. Multiply the top row by a scalar so that top row's leading entry becomes 1. Add/subtract multiples of the top row to ...

Hello every body , i am trying to solve an (nxn) system equations by Gaussian Elimination method using Matlab , for example the system below : Theme Copy x1 + 2x2 - x3 = 3 2x1 + x2 - 2x3 = 3 -3x1 + x2 + x3 = -6 C = [ 1 2 -1 ; 2 1 -2 ; -3 1 1 ] b= [ 3 3 -6 ] by using this code : Theme Copy % Matlab Program to solve (nxn) system equation

A familiar 3 4 Example 2 Ignoring the rst row and column, ... We'll apply the Gauss-Jordan elimination algorithm to (!:. A. Havens The Gauss-Jordan Elimination Algorithm. 1: 1. =: and;:; (;; = = : The Gauss-Jordan Elimination Algorithm - Solving Systems of Real Linear Equations ...