§ 3.4 Matrix Solutions to Linear Systems. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.4 Solving Systems Using Matrices The following array

§ 3.4

Matrix Solutions to Linear Systems

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.4

Solving Systems Using Matrices

The following array of numbers arranged in rows and columns and placed in brackets is an example of a matrix.

12

22

5

2

8

1

1

3

3

3

1

2

The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and can be used to solve systems of equations.

A matrix gives us a shortened way of writing a system of equations.



12

22

5

2

8

1

1

3

3

3

1

2

Equations represented by matrix above

2x -3y + z = 5x + 3y + 8z = 223x – y + 2z = 12

This matrix represents a system of threeequations that are listed below.



Matrix Row Operations1) Two rows of a matrix may be interchanged. This is the same as interchanging two equations in the linear system.

2) The elements in any row may be multiplied by a nonzero number. This is the same as multiplying both sides of an equation by a nonzero number.

3) The elements in any row may be multiplied by a nonzero number, and these products may be added to the corresponding elements in any other row. This is the same as multiplying an equation by a nonzero number and then adding equations to eliminate a variable (addition method).

Two matrices are row equivalent if one can be obtained from the other by a sequence of the above row operations.



EXAMPLEEXAMPLE

Use the matrix

And perform each indicated row operation:

(a) -2R1 (b) -3R2 + R3 R3

12

22

5

2

8

1

1

3

3

3

1

2



SOLUTIONSOLUTION

(a) First, -2R1 means that we need to multiply everything in row #1 by -2.

12

22

5)2(

2

8

1)2(

1

3

)3)(2(

3

1

2)2(

12

22

10

2

8

2

1

3

6

3

1

4

CONTINUECONTINUEDD

(b) -3R2 + R3 R3 means that we need to multiply row #2 by -3, add that result to row #3 and then the result of that addition is the new row #3.



54

22

5

22

8

1

10

3

3

0

1

2

CONTINUECONTINUEDD-3R2 = -3[1 3 8 22] = [(-3)1 (-3)3 (-3)8 (-3)22] = [-3 -9 -24 -66]

ADD: R3 = [ 3 -1 2 12]

Therefore, my new row #3 is: R3 = [ 0 -10 -22 -54]

So, the matrix rewritten with its new row #3 is:



Solving Linear Systems Using Matrices1) Write the augmented matrix for the system.

2) Use matrix row operations to simplify the matrix to one with 1’s down the main diagonal from upper left to lower right, and 0’s below the 1’s.

3) Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution.



EXAMPLEEXAMPLE

Use matrices to solve the system:

SOLUTIONSOLUTION

2x - 3y + z = 5x + 3y + 8z = 223x - y + 2z = 12

12

22

5

2

8

1

1

3

3

3

1

2

1) Write the augmented matrix for the system.

2) Use matrix row operations to simplify the matrix to one with 1’s down the main diagonal from upper left to lower right, and 0’s below the 1’s.



12

5

22

2

1

8

1

3

3

3

2

1

Next we will interchange row #1 and row #2 (to facilitate calculations).

CONTINUECONTINUEDD

Now we will do -2R1 + R2 R2 (that is, I will multiply row #1 by -2, add that to row #2, and that will be my new row #2).

-2R1 = -2[1 3 8 22] = [(-2)1 (-2)3 (-2)8 (-2)22] = [-2 -6 -16 -44]

ADD: R2 = [ 2 -3 1 5]




12

39

22

2

15

8

1

9

3

3

0

1

So, the augmented matrix now looks like

CONTINUECONTINUEDD

Now we will do -3R1 + R3 R3 (that is, I will multiply row #1 by -3, add that to row #3, and that will be my new row #3).

-3R1 = -3[1 3 8 22] = [(-3)1 (-3)3 (-3)8 (-3)22] = [-3 -9 -24 -66]

ADD: R3 = [ 3 -1 2 12]


NOTE: At this point, we should have a zero in the second row, first column.



54

39

22

22

15

8

10

9

3

0

0

1


CONTINUECONTINUEDD

Now we will do -10R2 + 9R3 R3.

-10R2 = -10[0 -9 -15 -39] = [0 90 150 390]

Therefore, my new row #3 is: R3 = [0 0 -48 -96]

NOTE: At this point, we should have a zero in the third row, first column.

ADD: 9R3 = 9[0 -10 -22 -54] = [0 -90 -198 -486]



96

39

22

48

15

8

0

9

3

0

0

1


CONTINUECONTINUEDD

Now I will divide R2 by -9 and divide R3 by -48 so that the first number in each column is a 1.

NOTE: At this point, we should have a zero in the third row, second column.

The new row #2 is: (-1/9)R2 = (-1/9)[0 -9 -15 -39] = [0 1 1.7 4.3]The new row #3 is: (-1/48)R3 = (-1/48)[0 -9 -15 -39] = [0 0 1 2]



2

3.4

22

1

7.1

8

0

1

3

0

0

1


CONTINUECONTINUEDD

3) Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution. So the system of equations becomes:

NOTE: At this point, we should have a 1 in each element of the main diagonal.

x + 3y + 8z = 22y + 1.7z = 4.3

z = 2



CONTINUECONTINUEDD

Therefore, z = 2. Now I replace z with 2 into the second equation to determine y.

y + 1.7z = 4.3

y + 1.7(2) = 4.3

y + 3.4 = 4.3

y = 0.9

Replace z with 2

Multiply

Subtract 3.4 from both sides



CONTINUECONTINUEDD

x + 3y + 8z = 22

Now solve for z using the first equation.

x + 3(0.9) + 8(2) = 22

x + 2.7 + 16 = 22

x + 18.7 = 22x = 3.3

Replace z with 2 and y with 0.9

Multiply

AddSubtract 18.7 from both sides

Therefore the solution to the system is (3.3,0.9,2).

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§ 3.4 Matrix Solutions to Linear Systems. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.4 Solving Systems Using Matrices The following array