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Solving a dependent system of linear equations using matrices

Solve the following system of linear equations using matrices.

x −2 y + z = −1   2 x + 3 y = 2 y −2 z = 0

Write the augmented matrix.

[ −1 −2 1 2 3 0 0 1 −2    |    −1 2 0 ]

First, multiply row 1 by −1 to get a 1 in row 1, column 1. Then, perform row operations    to obtain row-echelon form.

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

The last matrix represents the following system.

  x + 2 y z = 1         y 2 z = 0                0 = 0

We see by the identity 0 = 0 that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for y and substituting it into the first equation we can solve for z in terms of x .

      x + 2 y z = 1                    y = 2 z x + 2 ( 2 z ) z = 1             x + 3 z = 1                    z = 1 x 3

Now we substitute the expression for z into the second equation to solve for y in terms of x .

           y 2 z = 0                 z = 1 x 3   y 2 ( 1 x 3 ) = 0                   y = 2 2 x 3

The generic solution is ( x , 2 −2 x 3 , 1 x 3 ) .

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Solve the system using matrices.

x + 4 y z = 4 2 x + 5 y + 8 z = 15 x + 3 y −3 z = 1

( 1 , 1 , 1 )

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Can any system of linear equations be solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination.

Given a system of equations, solve with matrices using a calculator.

  1. Save the augmented matrix as a matrix variable [ A ] , [ B ] , [ C ] ,   .
  2. Use the ref( function in the calculator, calling up each matrix variable as needed.

Solving systems of equations with matrices using a calculator

Solve the system of equations.

  5 x + 3 y + 9 z = −1 −2 x + 3 y z = −2 x −4 y + 5 z = 1

Write the augmented matrix for the system of equations.

[ 5 3 9 −2 3 −1 −1 −4 5    |    5 −2 −1 ]

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [ A ] .

[ A ] = [ 5 3 9 −1 −2 3 −1 −2 −1 −4 5 1 ]

Use the ref( function in the calculator, calling up the matrix variable [ A ] .

ref ( [ A ] )

Evaluate.

[ 1 3 5 9 5 1 5 0 1 13 21 4 7 0 0 1 24 187 ] x + 3 5 y + 9 5 z = 1 5         y + 13 21 z = 4 7                    z = 24 187

Using back-substitution, the solution is ( 61 187 , 92 187 , 24 187 ) .

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Applying 2 × 2 matrices to finance

Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?

We have a system of two equations in two variables. Let x = the amount invested at 10.5% interest, and y = the amount invested at 12% interest.

                 x + y = 12,000 0.105 x + 0.12 y = 1,335

As a matrix, we have

[ 1 1 0.105 0.12    |    12,000 1,335 ]

Multiply row 1 by −0.105 and add the result to row 2.

[ 1 1 0 0.015    |    12,000 75 ]

Then,

0.015 y = 75           y = 5,000

So 12,000 −5,000 = 7,000.

Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.

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Applying 3 × 3 matrices to finance

Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?

We have a system of three equations in three variables. Let x be the amount invested at 5% interest, let y be the amount invested at 8% interest, and let z be the amount invested at 9% interest. Thus,

                      x + y + z = 10 , 000 0.05 x + 0.08 y + 0.09 z = 770                           2 x z = 0

As a matrix, we have

[ 1 1 1 0.05 0.08 0.09 2 0 −1    |    10 , 000 770 0 ]

Now, we perform Gaussian elimination to achieve row-echelon form.

−0.05 R 1 + R 2 = R 2 [ 1 1 1 0 0.03 0.04 2 0 −1 | 10,000 270 0 ] −2 R 1 + R 3 = R 3 [ 1 1 1 0 0.03 0.04 0 −2 −3 | 10,000 270 −20,000 ] 1 0.03 R 2 = R 2 [ 0 1 1 0 1 4 3 0 −2 −3 | 10,000 9,000 −20,000 ] 2 R 2 + R 3 = R 3 [ 1 1 1 0 1 4 3 0 0 1 3 | 10,000 9,000 −2,000 ]

The third row tells us 1 3 z = −2,000 ; thus z = 6,000.

The second row tells us y + 4 3 z = 9,000. Substituting z = 6,000 , we get

y + 4 3 ( 6,000 ) = 9,000 y + 8,000 = 9,000 y = 1,000

The first row tells us x + y + z = 10 , 000. Substituting y = 1 , 000 and z = 6 , 000 , we get

x + 1 , 000 + 6 , 000 = 10,000                                x = 3,000    

The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.

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Practice Key Terms 7

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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