27 Systems of Linear Equations

Linear algebra introduces methods that we can use to solve systems that involve hundreds of equations. Systems of this magnitude are most easily solved technologically, but how do computers go about solving this type of problem? In this section, we will introduce some terminology and demonstrate a process of solving systems of equations using matrices. We will be using only a few equations, but this process can be expanded to solve systems with hundreds more. The steps come from Larson and Falvo and are as follows [16].

  1. Convert the system of linear equations into an augmented matrix.
  2. Find the row-echelon form of the matrix using Gaussian elimination.
  3. Convert the matrix back into a system of linear equations and back-substitute.

First, convert the system into an augmented matrix. An augmented matrix is a matrix with a row for each equation and a column for each variable with an additional column for the constant terms of the system. The entries of every column but the last one are the coefficients of the system. The first row is made up of the coefficients and constant term of the first equation, the second row, the coefficients and constant term of the second equation, and so on. The first column is made up of the coefficients of the first variable, the second, the coefficients of the second, and so on for all columns but the last one, which is made up of the constant terms.

Second, find the row-echelon form using Gaussian elimination. Gaussian elimination is performed by using elementary row operations. These operations are as follows.

  • Switch two rows
  • Multiply a row by a nonzero constant
  • Add a multiple of a row to another row

These are the only operations we can use to transform the matrix into row-echelon form, which is described by Larson and Falvo as follows [16].




Definition VI.1

A matrix in row-echelon form has the following properties.


    

  1.  All rows consisting entirely of zeros occur at the bottom of the matrix.
    2. 

  2.  For each row that does not consist entirely of zeros, the first nonzero entry is 1 (called a leading 1).
    3. 

  3.  For two successive (nonzero) rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row.
    4. 





Our final step is to convert back to a system of linear equations by doing the opposite of what was done in the first step and back-substitute to get the final solution. 


Let’s go through an example to have a better idea of what all of this is saying. This example is a homework problem I completed for Elementary Linear Algebra and comes from Larson and Falvo [16].





Example 61

Solve the following system using Gaussian elimination.  


      \begin{align*} 1x_1&+1x_2-5x_3=3\\ 1x_1&+0x_2-2x_3=1\\ 2x_1&-1x_2-1x_3=0 \end{align*}


Solution

The augmented matrix is \left[      \begin{array}{@{}*{7}{r}@{}}        1    &  1&-5&3   \\         1     & 0&-2&1  \\         2      & -1&-1&0 \\           \end{array}      \right].


Using Gaussian elimination, we can arrive at the row-echelon form as follows.


      \begin{align*} &\left[      \begin{array}{@{}*{7}{r}@{}}        1    &  1&-5&3   \\         1     & 0&-2&1  \\         2      & -1&-1&0 \\           \end{array}      \right] &&\text{$-R_1+R_2\rightarrow R_2$ and $-2R_1+R_3\rightarrow R_3$}\\ &\left[      \begin{array}{@{}*{7}{r}@{}}        1    &  1&-5&3   \\         0     & -1&3&-2  \\         0      & -3&9&-6 \\           \end{array}      \right]             &&\text{$-3R_2+R_3\rightarrow R_3$}        \\      &\left[      \begin{array}{@{}*{7}{r}@{}}        1    &  1&-5&3   \\         0     & -1&3&-2  \\         0     & 0&0&0 \\           \end{array}      \right]   &&\text{$-R_2\rightarrow R_2$}\\ &\left[      \begin{array}{@{}*{7}{r}@{}}        1    &  1&-5&3   \\         0     & 1&-3&2 \\         0      & 0&0&0 \\           \end{array}      \right]  \\ \end{align*}


Converting back into a system of equations, we get


      \begin{align*} x_1+x_2-5x_3&=3\\ x_2-3x_3&=2\\ \end{align*}


Because this system has more variables than it does equations, we know that there will be an infinite number of solutions.  Let x_3=t where t\in\mathbb{R}. By back-substitution, 


      \begin{align*} x_3&=t\\ x_2&=2+3t\\ x_1&=3-2-3t+5t\\ &=1+2t\\  \end{align*}




Next, we are going to look at a characteristic specific to a square matrix that has many applications and can give us much information about the matrix itself.  This characteristic is known as the determinant.





					
		
					

						

							

								
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