9.3 Matrices and Systems of Equations

  • Solve systems of equations using matrices.

Matrices and Row-Equivalent Operations

In this section, we consider additional techniques for solving systems of equations. You have probably observed that when we solve a system of equations, we perform computations with the coefficients and the constants and continually rewrite the variables. We can streamline the solution process by omitting the variables until a solution is found. For example, the system

2x-3y=7,x+4y=−2
2xx−+3y4y==7,−2

can be written more simply as

[2−314|7−2].
[21−34∣∣∣7−2].

The vertical line replaces the equals signs.

A rectangular array of numbers like the one above is called a matrix (pl., matrices). The matrix above is called an augmented matrix for the given system of equations, because it contains not only the coefficients but also the constant terms. The matrix

[2−314]
[21−34]

is called the coefficient matrix of the system.

The rows of a matrix are horizontal, and the columns are vertical. The augmented matrix above has 2 rows and 3 columns, and the coefficient matrix has 2 rows and 2 columns. A matrix with m rows and n columns is said to be of order m×nm×n. Thus the order of the augmented matrix above is 2×32×3, and the order of the coefficient matrix is 2×2.2×2. When m=nm=n, a matrix is said to be square. The coefficient matrix above is a square matrix. The numbers 2 and 4 lie on the main diagonal of the coefficient matrix. The numbers in a matrix are called entries, or elements.

Gaussian Elimination with Matrices

In Section 9.2, we described a series of operations that can be used to transform a system of equations to an equivalent system. Each of these operations corresponds to one that can be used to produce row-equivalent matrices.

We can use these operations on the augmented matrix of a system of equations to solve the system.

Example 1

Solve the following system:

2x−y+4z=−3,x−2y−10z=−6,3x+4z=7.
2xx3x−−y2y+−+4z10z4z===−3,−6,7.

Solution

First, we write the augmented matrix, writing 0 for the missing y-term in the last equation:

[2−141−2−10304|−3−67].
⎡⎣⎢213−1−204−104∣∣∣∣−3−67⎤⎦⎥.

Our goal is to find a row-equivalent matrix of the form

[1ab01d001|cef].
⎡⎣⎢100a10bd1∣∣∣∣cef⎤⎦⎥.

The variables can then be reinserted to form equations from which we can complete the solution. This is done by working from the bottom equation to the top and using back-substitution.

The first step is to multiply and/or interchange rows so that each number in the first column below the first number is a multiple of that number. In this case, we interchange the first and second rows to obtain a 1 in the upper left-hand corner.

[1−2−102−14304|−6−37]     New row 1 = row 2    New row 2 = row 1
⎡⎣⎢123−2−10−1044∣∣∣∣−6−37⎤⎦⎥     New row 1 = row 2    New row 2 = row 1

Next, we multiply the first row by −2−2 and add it to the second row. We also multiply the first row by −3−3 and add it to the third row.

[1−2−100  3240  634|−6   925]   Row 1 is unchanged.    New row 2 = −2(row 1) + row 2   New row 3 = −3(row 1) + row 3
⎡⎣⎢100−2  3  6−102434∣∣∣∣−6   925⎤⎦⎥   Row 1 is unchanged.    New row 2 = −2(row 1) + row 2   New row 3 = −3(row 1) + row 3

Now we multiply the second row by 1313 to get a 1 in the second row, second column.

[1−2−100180634|−6   325]    New row 2 = 13(row 2)   
⎡⎣⎢100−216−10834∣∣∣∣−6   325⎤⎦⎥    New row 2 = 13(row 2)   

Then we multiply the second row by −6−6 and add it to the third row.

[1−2−100180  0−14|−637]     New row 3 = −6(row 2) + row 3  
⎡⎣⎢100−21  0−108−14∣∣∣∣−637⎤⎦⎥     New row 3 = −6(row 2) + row 3  

Finally, we multiply the third row by −114−114 to get a 1 in the third row, third column.

[1−2−10018001|−6   3−12]      New row 3 = −114(row 3)  
⎡⎣⎢100−210−1081∣∣∣∣∣−6   3−12⎤⎦⎥      New row 3 = −114(row 3)  

Now we can write the system of equations that corresponds to the last matrix above:

x−2y−10z=−6,(1)y+8z=3,(2)z=−12.(3)
x−2yy−+10z8zz===−6,3,−12.(1)(2)(3)

We back-substitute −12−12 for z in equation (2) and solve for y:

y+8(−12)=3y−4=3y=7.
y+8(−12)y−4y===337.

Next, we back-substitute 7 for y and −12−12 for z in equation (1) and solve for x:

x−2⋅7−10(−12)=−6x−14+5=−6x−9=−6x=3.
x−2⋅7−10(−12)x−14+5x−9x====−6−6−63.

The triple (3,7,−12)(3,7,−12) checks in the original system of equations, so it is the solution.

Now Try Exercise 27.

The procedure followed in Example 1 is called Gaussian elimination with matrices. The last matrix in Example 1 is in row-echelon form. To be in this form, a matrix must have the following properties.

Example 2

Which of the following matrices are in row-echelon form? Which, if any, are in reduced row-echelon form?

  1. [1−350  1−40  01|−2   310]⎡⎣⎢100−3  1  05−41∣∣∣∣−2   310⎤⎦⎥

  2. [0−101|25][00−11∣∣∣25]

  3. [1−2−64035−80019|7   −12]⎡⎣⎢100−230−6514−89∣∣∣∣7   −12⎤⎦⎥

  4. [100010001|−2.40.85.6]⎡⎣⎢100010001∣∣∣∣−2.40.85.6⎤⎦⎥

  5. [1000010000100000|23−14670]⎡⎣⎢⎢⎢⎢⎢1000010000100000∣∣∣∣∣∣∣23−14670⎤⎦⎥⎥⎥⎥⎥

  6. [1−4200001−3|50−8]⎡⎣⎢100−40120−3∣∣∣∣50−8⎤⎦⎥

Solution

The matrices in (a), (d), and (e) satisfy the row-echelon criteria and, thus, are in row-echelon form. In (b) and (c), the first nonzero elements of the first and second rows, respectively, are not 1. In (f), the row consisting entirely of 0’s is not at the bottom of the matrix. Thus the matrices in (b), (c), and (f) are not in row-echelon form. In (d) and (e), not only are the row-echelon criteria met but each column that contains a leading 1 also has 0’s elsewhere, so these matrices are in reduced row-echelon form.

Gauss–Jordan Elimination

We have seen that with Gaussian elimination we perform row-equivalent operations on a matrix to obtain a row-equivalent matrix in row-echelon form. When we continue to apply these operations until we have a matrix in reduced row-echelon form, we are using Gauss–Jordan elimination. This method is named for Karl Friedrich Gauss and Wilhelm Jordan (1842–1899).

Example 3

Use Gauss–Jordan elimination to solve the system of equations in Example 1.

Solution

Using Gaussian elimination in Example 1, we obtained the matrix

[1−2−10018001|−63−12].
⎡⎣⎢100−210−1081∣∣∣∣∣−63−12⎤⎦⎥.

We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form. We multiply the third row by 10 and add it to the first row. We also multiply the third row by −8−8 and add it to the second row.

[1−20010001|−117−12]   New row 1 = 10(row 3) + row 1   New row 2 = −8(row 3) + row 2
⎡⎣⎢100−210001∣∣∣∣∣−117−12⎤⎦⎥   New row 1 = 10(row 3) + row 1   New row 2 = −8(row 3) + row 2

Next, we multiply the second row by 2 and add it to the first row.

[100010001|37−12]   New row 1 = 2(row 2) + row 1
⎡⎣⎢100010001∣∣∣∣∣37−12⎤⎦⎥   New row 1 = 2(row 2) + row 1

Writing the system of equations that corresponds to this matrix, we have

x=3,y=7,z=−12.
xyz===3,7,−12.

We can actually read the solution, (3,7,−12)(3,7,−12), directly from the last column of the reduced row-echelon matrix.

Now Try Exercise 27.

Example 4

Solve the following system:

3x−4y−z=6,2x−y+z=−1,4x−7y−3z=13.
3x2x4x−−−4yy7y−+−zz3z===6,−1,13.

Solution

First, we write the augmented matrix and use Gauss–Jordan elimination.

[3−4−12−114−7−3|6−113]
⎡⎣⎢324−4−1−7−11−3∣∣∣∣6−113⎤⎦⎥

We then multiply the second and third rows by 3 so that each number in the first column below the first number, 3, is a multiple of that number.

[3−4−16−3312−21−9|6−339]    New row 2 = 3(row 2)    New row 3 = 3(row 3)
⎡⎣⎢3612−4−3−21−13−9∣∣∣∣6−339⎤⎦⎥    New row 2 = 3(row 2)    New row 3 = 3(row 3)

Next, we multiply the first row by −2−2 and add it to the second row. We also multiply the first row by −4−4 and add it to the third row.

[3−4−10550−5−5|6−1515]    New row 2 = −2(row 1) + row 2    New row 3 = −4(row 1) + row 3
⎡⎣⎢300−45−5−15−5∣∣∣∣6−1515⎤⎦⎥    New row 2 = −2(row 1) + row 2    New row 3 = −4(row 1) + row 3

Now we add the second row to the third row.

[3−4−1055000|6−150]        New row 3 = row 2 + row 3
⎡⎣⎢300−450−150∣∣∣∣6−150⎤⎦⎥        New row 3 = row 2 + row 3

We can stop at this stage because we have a row consisting entirely of 0’s. The last row of the matrix corresponds to the equation 0=00=0, which is true for all values of x, y, and z. Therefore, the equations are dependent and the system is equivalent to

3x−4y−z=6,5y+5z=−15.
3x−4y5y−+z5z==6,−15.

This particular system has infinitely many solutions. (A system containing dependent equations could be inconsistent.)

Solving the second equation for y gives us

y=−z-3.
y=−z−3.

Substituting −z−3−z−3 for y in the first equation and solving for x, we get

3x−4(−z−3)−z=63x+4z+12−z=63x+3z+12=63x=−3z−6x=−z−2.
3x−4(−z−3)−z3x+4z+12−z3x+3z+123xx=====666−3z−6−z−2.

Then the solutions of this system are of the form

(−z-2,−z-3,z),
(−z−2,−z−3,z),

where z can be any real number.

Now Try Exercise 33.

Similarly, if we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix

[103015000|−246],
⎡⎣⎢100010350∣∣∣∣−246⎤⎦⎥,

the last row corresponds to the false equation 0=60=6, so we know the original system of equations has no solution.

9.3 Exercise Set

Determine the order of the matrix.

  1. 1. [1−6−3205]⎡⎣⎢1−30−625⎤⎦⎥

  2. 2. [7−5−13]⎡⎣⎢⎢⎢⎢7−5−13⎤⎦⎥⎥⎥⎥

  3. 3. [2−409][2−409]

  4. 4. [−8][−8]

  5. 5. [1−5−864−2−307]⎡⎣⎢16−3−540−8−27⎤⎦⎥

  6. 9. [132−64−1185−12][13−1218−654−12]

Write the augmented matrix for the system of equations.

  1. 7. 2x−y=7,x+4y=−52xx−+y4y==7,−5

  2. 8. 3x+2y=8,2x−3y=153x2x+−2y3y==8,15

  3. 9. x−2y+3z=12,2x−4z=8,3y+z=7x2x−2y3y+−+3z4zz===12,8,7

  4. 10. x+y−z=7,3y+2z=1,−2x−5y=6x−2x+−y3y5y−+z2z===7,1,6

Write the system of equations that corresponds to the augmented matrix.

  1. 11. [3−514|1−2][31−54∣∣∣1−2]

  2. 12. [1241|−6−3][1421∣∣∣−6−3]

  3. 13. [21−43051−11|12−12]⎡⎣⎢23110−1−451∣∣∣∣12−12⎤⎦⎥

  4. 14. [−1−230412−10|629]⎡⎣⎢−102−24−1310∣∣∣∣629⎤⎦⎥

Solve the system of equations using Gaussian elimination or Gauss–Jordan elimination.

  1. 15. 4x+2y=11,3x-y=24x3x+−2yy==11,2

  2. 16. 2x+y=1,3x+2y=−22x3x++y2y==1,−2

  3. 17. 5x-2y=−3,2x+5y=−245x2x−+2y5y==−3,−24

  4. 18. 2x+y=1,3x-6y=42x3x+−y6y==1,4

  5. 19. 3x+4y=7,−5x+2y=103x−5x++4y2y==7,10

  6. 20. 5x-3y=−2,4x+2y=55x4x−+3y2y==−2,5

  7. 21. 3x+2y=6,2x-3y=−93x2x+−2y3y==6,−9

  8. 22. x-4y=9,2x+5y=5x2x−+4y5y==9,5

  9. 23. x-3y=8,−2x+6y=3x−2x−+3y6y==8,3

  10. 24. 4x-8y=12,−x+2y=−34x−x−+8y2y==12,−3

  11. 25. −2x+6y=4,3x-9y=−6−2x3x+−6y9y==4,−6

  12. 26. 6x+2y=−10,−3x-y=66x−3x+−2yy==−10,6

  13. 27. x+2y-3z=9,2x-y+2z=−8,3x-y-4z=3x2x3x+−−2yyy−+−3z2z4z===9,−8,3

  14. 28. x-y+2z=0,x-2y+3z=−1,2x-2y+z=−3xx2x−−−y2y2y+++2z3zz===0,−1,−3

  15. 29. 4x-y-3z=1,8x+y-z=5,2x+y+2z=54x8x2x−++yyy−−+3zz2z===1,5,5

  16. 30. 3x+2y+2z=3,x+2y-z=5,2x-4y+z=0

  17. 31. x-2y+3z=−4,3x+y-z=0,2x+3y-5z=1

  18. 32. 2x-3y+2z=2,x+4y-z=9,−3x+y-5z=5

  19. 33. 2x-4y-3z=3,x+3y+z=−1,5x+y-2z=2

  20. 34. x+y-3z=4,4x+5y+z=1,2x+3y+7z=−7

  21. 35. p+q+r=1,p+2q+3r=4,4p+5q+6r=7

  22. 36. m+n+t=9,m-n-t=−15,3m+n+t=2

  23. 37. a+b-c=7,a-b+c=5,3a+b-c=−1

  24. 38. a-b+c=3,2a+b-3c=5,4a+b-c=11

  25. 39. −2w+2x+2y-2z=−10,w+x+y+z=−5,3w+x-y+4z=−2,w+3x-2y+2z=−6

  26. 40. −w+2x-3y+z=−8,−w+x+y-z=−4,w+x+y+z=22,−w+x-y-z=−14

Use Gaussian elimination or Gauss–Jordan elimination in Exercises 41–44.

  1. 41. Borrowing. Greenfield Manufacturing borrowed $30,000 to buy a new piece of equipment. Part of the money was borrowed at 8%, part at 10%, and part at 12%. The annual interest was $3040, and the total amount borrowed at 8% and at 10% was twice the amount borrowed at 12%. How much was borrowed at each rate?

  2. 42. Time of Return. The Patels pay their babysitter $11 per hour before 11 p.m. and $14.50 after 11 p.m. One evening, they went out for 6 hr and paid the sitter $73. What time did they return?

  3. 43. Stamp Purchase. For her business, Olivia spent $86.80 on both 49¢ and 21¢ stamps. She bought a total of 200 stamps. How many of each type did she buy?

  4. 44. Advertising Expense.  eAuction.com spent a total of $11 million on advertising in fiscal years 2010, 2011, and 2012. The amount spent in 2012 was three times the amount spent in 2010. The amount spent in 2011 was $3 million less than the amount spent in 2012. How much was spent on advertising each year?

Skill Maintenance

In Exercises 45–52, classify the function as linear, quadratic, cubic, quartic, rational, exponential, or logarithmic.

  1. 45. f(x)=3x−1 [5.2]

  2. 46. f(x)=3x−1 [1.3]

  3. 47. f(x)=3x-1x2+4 [4.5]

  4. 48. f(x)=−34x4+92x3+2x2-4 [4.1]

  5. 49. f(x)=ln(3x-1) [5.3]

  6. 50. f(x)=34x3-x [4.1]

  7. 51. f(x)=3 [1.3]

  8. 52. f(x)=2-x-x2 [3.2]

Synthesis

In Exercises 53 and 54, three solutions of the equation y=ax2+bx+c are given. Use a system of three equations in three variables and Gaussian elimination or Gauss–Jordan elimination to find the constants a, b, and c and write the equation.

  1. 53. (−3,12), (−1,−7), and (1,−2)

  2. 54. (−1,0), (1,−3), and (3,−22)

  3. 55. Find two different row-echelon forms of

    [1532].

  4. 56. Consider the system of equations

    x−y+3z=−8,2x+3y−z=5,3x+2y+2kz=−3k.

    For what value(s) of k, if any, will the system have:

    1. no solution?

    2. exactly one solution?

    3. infinitely many solutions?

Solve using matrices.

  1. 57. y=x+z,3y+5z=4,x+4=y+3z

  2. 58. x+y=2z,2x-5z=4,x-z=y+8

  3. 59. x-4y+2z=7,3x+y+3z=−5

  4. 60. x-y-3z=3,−x+3y+z=−7

  5. 61. 4x+5y=3,−2x+y=9,3x-2y=−15

  6. 62. 2x−3y=−1,−x+2y=−2,3x-5y=1

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