Question

System of Linear Equations In Exercises 25–38, solve the system using ei ther Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Answer 1

Hi, you haven't provided the system of linear equations that you need to solve. Therefore, I'll just explain how to use Gauss-Jordan in a system of equations and you can apply the same method to the system of equations you have.

Answer with explanation and step by step solution:

1. For the system of equations:

$4X_{1} + 8X_{2} + 12X_{3} = 36\\8X_{1} + 10X_{2} + 12X_{3} = 48\\4X_{1} + 14X_{2} + 24X_{3} = 60$

2. We can represent it as a matrix by placing every number of the equation as follow:  

$\left[\begin{array}{ccccc}4&8&12&|&36\\4&5&6&|&24\\2&7&12&|&30\end{array}\right]$

3. As you can see all the coefficients in the equation are divisible by two, so we can express the system of equations as follow:  

 $\left[\begin{array}{ccccc}2&4&6&|&18\\4&5&6&|&24\\2&7&12&|&30\end{array}\right]$

4. Gauss-Jordan method solves the system of equations by applying simple operations to the Matrix: Multiplication by non-zero numbers, adding a multiple of one row to another and swapping rows.

Step by step solution:  

Divide both sides of equation one by two:

$\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\2&7&12&|&30\end{array}\right]$

 

Subtract two times the equation two to the equation three:  

$\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\-6&-3&0&|&-18\end{array}\right]$

Divide equation number three by minus three and subtract two times the equation one to equation two:

$\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\2&1&0&|&6\end{array}\right]$

 

Subtract the equation two to the equation three:  

$\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\0&0&0&|&0\end{array}\right]$

Because now we have two equations for three unknown values X1, X2 and X3 the system has an infinite number of solutions.  

Equivalente system (From matrix to equation notation):  

$1X_{1} + 2X_{2} + 3X_{3} = 9\\2X_{1} + 1X_{2} = 6$

Conclusion:  

For whatever system you have you need to convert the system into a matrix notation and using the basic operations, described here, reduce the complexity of the system until:  

You have a solution, you discover that the system has an infinite number of solutions or the system of equation is inconsistent.  

Example of inconsistency

If after making the basic operations to your system you get a result like this

$\left[\begin{array}{ccccc}7&0&4&|&9\\2&1&0&|&6\\0&0&0&|&-1\end{array}\right]$  

You can say that the system is inconsistent because zero is not equal to minus one.  

Example of solution  

If after making the basic operations to your system you get a result like this

$\left[\begin{array}{ccccc}1&0&0&|&9\\0&1&0&|&-6\\0&0&1&|&-1\end{array}\right]$

You can say that the system have a solution in which X1 = 9, X2 = -6 and X3 = -1