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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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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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