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Solving a 2 × 2 System by gaussian elimination

Solve the given system by Gaussian elimination.

2 x + 3 y = 6      x y = 1 2

First, we write this as an augmented matrix.

[ 2 3 1 −1    |    6 1 2 ]

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

R 1 R 2 [ 1 −1 2 3 | 1 2 6 ]

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 −2 , and then adding the result to row 2.

−2 R 1 + R 2 = R 2 [ 1 −1 0 5 | 1 2 5 ]

We only have one more step, to multiply row 2 by 1 5 .

1 5 R 2 = R 2 [ 1 −1 0 1 | 1 2 1 ]

Use back-substitution. The second row of the matrix represents y = 1. Back-substitute y = 1 into the first equation.

x ( 1 ) = 1 2           x = 3 2

The solution is the point ( 3 2 , 1 ) .

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Solve the given system by Gaussian elimination.

4 x + 3 y = 11    x −3 y = −1

( 2 , 1 )

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Using gaussian elimination to solve a system of equations

Use Gaussian elimination    to solve the given 2 × 2 system of equations .

   2 x + y = 1 4 x + 2 y = 6

Write the system as an augmented matrix    .

[ 2 1 4 2    |    1 6 ]

Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by 1 2 .

1 2 R 1 = R 1 [ 1 1 2 4 2    |    1 2 6 ]

Next, we want a 0 in row 2, column 1. Multiply row 1 by −4 and add row 1 to row 2.

−4 R 1 + R 2 = R 2 [ 1 1 2 0 0    |    1 2 4 ]

The second row represents the equation 0 = 4. Therefore, the system is inconsistent and has no solution.

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Solving a dependent system

Solve the system of equations.

3 x + 4 y = 12 6 x + 8 y = 24

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

A = [ 3 4 6 8 | 12 24 ]
1 2 R 2 + R 1 = R 1 [ 0 0 6 8 | 0 24 ] R 1 R 2 [ 6 8 0 0 | 24 0 ]

The matrix ends up with all zeros in the last row: 0 y = 0. 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 y .

3 x + 4 y = 12           4 y = 12 −3 x             y = 3 3 4 x

So the solution to this system is ( x , 3 3 4 x ) .

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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.

[ 1 −3 4 2 −5 6 −3 3 4    |    3 6 6 ]

The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by −2 and add it to row 2. Then replace row 2 with the result.

−2 R 1 + R 2 = R 2 [ 1 −3 4 0 1 −2 −3 3 4 | 3 0 6 ]

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

3 R 1 + R 3 = R 3 [ 1 −3 4 0 1 −2 0 −6 16 | 3 0 15 ]

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

6 R 2 + R 3 = R 3 [ 1 −3 4 0 1 −2 0 0 4 | 3 0 15 ]

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

1 2 R 3 = R 3 [ 1 −3 4 0 1 −2 0 0 1    |    3 −6 21 2 ]
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Write the system of equations in row-echelon form.

   x 2 y + 3 z = 9       x + 3 y = 4 2 x 5 y + 5 z = 17

[ 1 5 2 5 2 0 1 5 0 0 1 | 17 2 9 2 ]

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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.

x y + z = 8 2 x + 3 y z = −2 3 x 2 y 9 z = 9

First, we write the augmented matrix.

[ 1 1 1 2 3 1 3 2 9     |    8 2 9 ]

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

2 R 1 + R 2 = R 2 [ 1 1 1 0 5 3 3 2 9 | 8 18 9 ] 3 R 1 + R 3 = R 3 [ 1 1 1 0 5 3 0 1 12 | 8 18 15 ]

The easiest way to obtain a 1 in row 2 of column 1 is to interchange R 2 and R 3 .

Interchange R 2 and R 3 [ 1 −1 1 8 0 1 −12 −15 0 5 −3 −18 ]

Then

−5 R 2 + R 3 = R 3 [ 1 −1 1 0 1 −12 0 0 57 | 8 −15 57 ] 1 57 R 3 = R 3 [ 1 −1 1 0 1 −12 0 0 1 | 8 −15 1 ]

The last matrix represents the equivalent system.

  x y + z = 8     y 12 z = −15              z = 1

Using back-substitution, we obtain the solution as ( 4 , −3 , 1 ) .

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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.
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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