Gauss Elimination Calculator

Gauss Elimination Calculator solve a system of three linear equations with real coefficients using Gaussian elimination algorithm. It is an online algebra tool programmed to determine an ordered triple as a solution to a system of three linear equations. Using this calculator, we will able to understand how to solve the system of linear equations using Gauss elimination algorithm.
It is necessary to follow the next steps:

1. Enter twelve coefficients of a system of linear equations in the box. These coefficients must be real numbers.
2. Press the "Generate Work" button to make the computation;
3. Gauss Elimination Calculator will give an ordered triple $(x,y,z)$ as a solution of a system of three linear equations.

Input : System of three linear equations.;
Output : Three real numbers.

How to Find Unknown Variables in the Equations by Gauss Elimination?

Gauss elimination or row reduction, is an algorithm for solving a system of linear equations. This method also called as Gauss-Jordan elimination. It is represented by a sequence of operations performed on the matrix. The method is named after Carl Friedrich Gauss (1777-1855), although it was known to Chinese mathematicians. The method of solving a system of linear equations by Gauss elimination is similar to the method of solving matrices. For instance, there is the connection between a system of three linear equations and its coefficient matrix. $$\begin{align} &a_1x+b_1y+c_1z={ d_1}\\ &a_2x+b_2y+c_2z={ d_2}\\ &a_3x+b_3y+c_3z={ d_3}\\ \end{align} \quad\longmapsto \left( \begin{array}{ccc} {a_1} & b_1 &c_1\\ {a_2} &b_2 &c_2\\ {a_3} &b_3 &c_3\\ \end{array} \right)$$ There are three types of elementary row operations:

• Replacing two rows;
• Multiplying a row by a nonzero number;
• Adding a multiple of one row to another row.

The method of Gauss elimination consists of two parts. The first part reduces a given system to \underline{row echelon form}. From the row echelon form, we can conclude whether the system has no solutions, a unique solution, or infinitely many solutions. The second part uses row operations until the solution is found. Row echelon form satisfies following properties:

• The leading coefficient of each row must be $1$;
• All elements in a column below a leading $1$ must be $0$;
• All rows that contain zeros are at the bottom of the matrix.

For example, the following matrices are in row echelon form $$\left( \begin{array}{cc} 1 & 5 \\ 0 & 1 \\ \end{array} \right), \quad \left( \begin{array}{cccc} 1 & 1 & 0 & 5 \\ 0 & 1 & 3 & 4 \\ 0&0 & 1 & 2 \\ \end{array} \right), \quad \left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 3 & 4 \\ 0&0 & 1 & 2 \\ 0&0 & 0 &0 \\ \end{array} \right)$$ A matrix is in reduced row echelon form if furthermore in every column containing a leading coefficient, all of the other entries in that column are zero. For instance, the matrices shown below are examples of matrices in reduced row echelon form. $$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right), \quad \left( \begin{array}{cccc} 1 & 0 & 0 & 7 \\ 0 & 1 & 0 & -2 \\ 0&0 & 1 & 2 \\ \end{array} \right), \quad \left( \begin{array}{cccc} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -2 \\ 0&0 & 1 & 2 \\ 0&0 & 0 &0 \\ \end{array} \right)$$ An augmented matrix is a matrix obtained by appending the columns of two given matrices. In the case of solving a system, we need to augment the coefficient matrix and the constant matrix. The vertical line indicates the separation between the coefficient matrix and the constant matrix. So, for the the system of three equations $$\begin{align} &a_1x+b_1y+c_1z={ d_1}\\ &a_2x+b_2y+c_2z={ d_2}\\ &a_3x+b_3y+c_3z={ d_3}\\ \end{align}$$ the augmented matrix is $$\left( \begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3&b_3 & c_3 & d_3 \\ \end{array} \right)$$ The number of solutions to a system depends only on the rank of the matrix representing the system and the rank of the corresponding augmented matrix. Based on the Kronecker-Capelli Theorem, any system of three linear equations has no solutions if the rank of the augmented matrix is greater than the rank of the coefficient matrix. If the ranks of these two matrices are equal, the system must have at least one solution. The solution is unique if and only if the rank equals the number of variables, in this case, if the rank is equal to $3$. For example, let us solve the solution of the system using the Gauss elimination $$\begin{align} &4x+5y+3z={ 10}\\ &3x+6y+7z={ 8}\\ &2x+3y+0z={ 8}\\ \end{align}$$ The coefficients and constant terms of the system give the matrices $$\left( \begin{array}{ccc} 4 & 5 &3\\ 3 &6 &7\\ 2 &3&0\\ \end{array} \right),\quad \left( \begin{array}{c} 10 \\ 8 \\ 8 \\ \end{array} \right)$$ The augmented matrix is $$\left( \begin{array}{ccc|c} 4 & 5 &3&10\\ 3 &6 &7&8\\ 2 &3&0&8\\ \end{array} \right)$$ To solve the system, reduce the augmented matrix to reduced row echelon form in the following way.

• Divide row $1$ by $4$ ($R_1=\frac {R_1}4)$, to get $$\left( \begin{array}{ccc|c} 1 & \frac 54 &\frac 34&\frac{5}2\\ 3 &6 &7&8\\ 2 &3&0&8\\ \end{array} \right)$$
• Subtract row $1$ multiplied by $3$ from row $2$ ($R_2=R_2-3R_1$), to get $$\left( \begin{array}{ccc|c} 1 & \frac 54 &\frac 34&\frac{5}2\\ 0 &\frac 94 &\frac{19}4&\frac 12\\ 2 &3&0&8\\ \end{array} \right)$$
• Subtract row $1$ multiplied by $2$ from row $3$ ($R_3=R_3-2R_1$), to get $$\left( \begin{array}{ccc|c} 1 & \frac 54 &\frac 34&\frac{5}2\\ 0 &\frac 94 &\frac{19}4&\frac 12\\ 0 &\frac12 &-\frac 32&3\\ \end{array} \right)$$
• Multiply row $2$ by $\frac 49$ ($R_2=\frac49 R_2$), to get $$\left( \begin{array}{ccc|c} 1 & \frac 54 &\frac 34&\frac{5}2\\ 0 &1 &\frac{19}9&\frac 29\\\ 0 &\frac12 &-\frac 32&3\\ \end{array} \right)$$
• Subtract row $2$ multiplied by $\frac 54$ from row $1$ ($R_1=R_1-\frac54 R_2$), to get $$\left( \begin{array}{ccc|c} 1 & 0 &-\frac {17}9&\frac{20}9\\ 0 &1 &\frac{19}9&\frac 29\\ 0 &\frac12 &-\frac 32&3\\ \end{array} \right)$$
• Subtract row $2$ multiplied by $\frac 12$ from row $3$ ($R_3=R_3-\frac12R_2$), to get $$\left( \begin{array}{ccc|c} 1 & 0 &-\frac {17}9&\frac{20}9\\ 0 &1 &\frac{19}9&\frac 29\\ 0 &0&-\frac{23}9&\frac{26}9\\ \end{array} \right)$$
• Multiply row $3$ by $-\frac9{23}$ ($R_3=-\frac9{23}R_3$), to get $$\left( \begin{array}{ccc|c} 1 & 0 &-\frac {17}9&\frac{20}9\\ 0 &1 &\frac{19}9&\frac 29\\ 0 &0&1&-\frac{26}{23}\\ \end{array} \right)$$
• Add row $3$ multiplied by $\frac{17}9$ to row $1$ ($R_1=R_1+\frac{17}9R_3$), to get $$\left( \begin{array}{ccc|c} 1 & 0 &0&\frac2{23}\\ 0 &1 &\frac{19}9&\frac 29\\ 0 &0&1&-\frac{26}{23}\\ \end{array} \right)$$
• Subtract row $3$ multiplied by $\frac {19}9$ from row $2$ ($R_2=R_2-\frac{19}9R_3$), to obtain $$\left( \begin{array}{ccc|c} 1 & 0 &0&\frac2{23}\\ 0 &1 &0&\frac {60}{23}\\ 0 &0&1&-\frac{26}{23}\\ \end{array} \right)$$ So the solution of the system is $(x, y, z) = (\frac{2}{23},\frac{60}{23}, -\frac{26}{23})$.

The Gauss Elimination work with steps shows the complete step-by-step calculation for finding a solution of a linear system of three equations using the Gauss elimination method. For any other system, just supply twelve real numbers as coefficients of linear equations and click on the Generate Work button. The grade school students use this Gauss Elimination Calculator to generate the work, verify the results of solving systems of linear equations derived by hand or do their homework problems efficiently. In many applications, it is necessary to calculate the matrix elimination where this online Matrix Gauss Elimination calculator can assist to effortlessly make calculations easy for the respective inputs.

Real World Problems Using Gauss elimination

Gaussian elimination algorithm is useful for determining the rank of a matrix (an important property of each matrix). This method can also help us to find the inverse of a matrix. In Geometry, the equation $Ax+By+Cz=D$ defines a plane in the three-dimensional coordinate system. If we consider a system of three variables, we can think about the points of intersection of planes. Hence, we can determine whether planes are parallel, intersect each other or coincide.

Gauss Elimination Practice Problems

Practice Problem 1:
Using the Gauss elimination, solve the system of equations $$\begin{align} &2x+4y-z=-1\\ &x+3y+7z=2\\ &x+2y+z=-5\\ \end{align} $$ Practice Problem 2:
A math library wants to purchase $25$ books for $\$2,800$. Three different types of books are available: a geometry with a price of $\$35$, an algebra with a price of $\$70$, a statistics with a price of $\$140$. How many of each type of books should the library purchase?

The Gauss Elimination Calculator, formula, example calculation (work with steps) and practice problems would be very useful for grade school students of K-12 education to understand the concept of solving systems of linear equations. This concept is conceived in almost all areas of science, so it will be helpful in solving more complex problems.