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

 Page 3 / 13

## 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 $\text{\hspace{0.17em}}\left(4,-3,1\right).$

#### Questions & Answers

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
Kaitlyn Reply
The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
give me the waec 2019 questions
Aaron Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

 By By By By Edward Biton By