# 11.6 Solving systems with gaussian elimination  (Page 3/13)

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## Solving a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ System by gaussian elimination

Solve the given system by Gaussian elimination.

First, we write this as an augmented matrix.

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

${R}_{1}↔{R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 2& \hfill 3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 6\end{array}\right]$

We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by $\text{\hspace{0.17em}}-2,$ and then adding the result to row 2.

$-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 5& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 5\end{array}\right]$

We only have one more step, to multiply row 2 by $\text{\hspace{0.17em}}\frac{1}{5}.$

$\frac{1}{5}{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 1& \hfill \end{array}|\begin{array}{cc}& \frac{1}{2}\\ & 1\end{array}\right]$

Use back-substitution. The second row of the matrix represents $\text{\hspace{0.17em}}y=1.\text{\hspace{0.17em}}$ Back-substitute $\text{\hspace{0.17em}}y=1\text{\hspace{0.17em}}$ into the first equation.

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

Solve the given system by Gaussian elimination.

$\left(2,\text{\hspace{0.17em}}1\right)$

## Using gaussian elimination to solve a system of equations

Use Gaussian elimination    to solve the given $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ system of equations .

Write the system as an augmented matrix    .

Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by $\text{\hspace{0.17em}}\frac{1}{2}.$

Next, we want a 0 in row 2, column 1. Multiply row 1 by $\text{\hspace{0.17em}}-4\text{\hspace{0.17em}}$ and add row 1 to row 2.

The second row represents the equation $\text{\hspace{0.17em}}0=4.\text{\hspace{0.17em}}$ Therefore, the system is inconsistent and has no solution.

## Solving a dependent system

Solve the system of equations.

$\begin{array}{l}3x+4y=12\\ 6x+8y=24\end{array}$

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

$A=\left[\begin{array}{llll}3\hfill & \hfill & 4\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 12\hfill \\ \hfill & 24\hfill \end{array}\right]$
$\begin{array}{l}\hfill \\ \begin{array}{l}-\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{llll}0\hfill & \hfill & 0\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill \\ \hfill & 24\hfill \end{array}\right]\hfill \\ {R}_{1}↔{R}_{2}\to \left[\begin{array}{llll}6\hfill & \hfill & 8\hfill & \hfill \\ 0\hfill & \hfill & 0\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 24\hfill \\ \hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill \end{array}\right]\hfill \end{array}\hfill \end{array}$

The matrix ends up with all zeros in the last row: $\text{\hspace{0.17em}}0y=0.\text{\hspace{0.17em}}$ Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for $\text{\hspace{0.17em}}y.$

So the solution to this system is $\text{\hspace{0.17em}}\left(x,3-\frac{3}{4}x\right).$

## 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.

The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add it to row 2. Then replace row 2 with the result.

$-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill -3& \hfill & \hfill 3& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 6\end{array}\right]$

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

$3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -6& \hfill & \hfill 16& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]$

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

$6{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]$

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

Write the system of equations in row-echelon form.

$\left[\text{\hspace{0.17em}}\begin{array}{ccc}\text{\hspace{0.17em}}1& -\frac{5}{2}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{2}\\ \text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 5\\ \text{\hspace{0.17em}}0& \text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\text{\hspace{0.17em}}|\begin{array}{c}\frac{17}{2}\\ 9\\ 2\end{array}\right]$

## Solving a system of linear equations using matrices

We have seen how to write a system of equations with an augmented matrix    , and then how to use row operations and back-substitution to obtain row-echelon form    . Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.

## Solving a system of linear equations using matrices

Solve the system of linear equations using matrices.

$\begin{array}{c}\begin{array}{l}\hfill \\ \hfill \\ x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}z=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\hfill \end{array}\\ 2x\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}z=-2\\ 3x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2y\text{\hspace{0.17em}}\text{\hspace{0.17em}}-9z=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9\end{array}$

First, we write the augmented matrix.

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

$\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 3& \hfill & \hfill -2& \hfill & \hfill -9& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill -15\end{array}\right]\end{array}$

The easiest way to obtain a 1 in row 2 of column 1 is to interchange $\text{\hspace{0.17em}}{R}_{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{R}_{3}.$

$\text{Interchange}\text{\hspace{0.17em}}{R}_{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{R}_{3}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill & \hfill 8\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill & \hfill -15\\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill & \hfill -18\end{array}\right]$

Then

$\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 57& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 1\end{array}\right]\end{array}\end{array}$

The last matrix represents the equivalent system.

Using back-substitution, we obtain the solution as $\text{\hspace{0.17em}}\left(4,-3,1\right).$

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