Precalculus by Richard Wright

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# 8-03 Multivariable Linear Systems

Summary: In this section, you will:

• Use elementary row operations.
• Solve systems of linear equations by putting them in row-echelon form.
• Write the answer to a three-variable system of equations with many solutions.

SDA NAD Content Standards (2018): PC.6.1

Three friends went to a Mexican fast food restaurant. Joe bought 3 tacos, 2 burritos, and a drink for $6.25. Frank bought 5 tacos and 4 burritos for$7.75. He forgot to buy a drink so Samantha bought an extra drink. She bought 2 tacos, 2 burritos, and 2 drinks for 7.50. How much does Frank owe Samantha for the drink? This problem can be solved by elimination, but there are three variables. Elementary row operations will be used to write the system in row-echelon form. This is called Gaussian Elimination. ###### Elementary Row Operations These operations are allowed in systems of equations and produce equivalent systems. • Interchange two equations. • Multiply one equation by a nonzero constant. • Add a multiple of one equation to another equation and replace one of these two equations. ###### Row-Echelon Form • The first nonzero term in each equation has a coefficient of 1. • All terms under the leading 1 are zeros producing an inverted pyramid shape. • Any equation that is all zeros is at the bottom. \left\{\begin{align} \color{red}{1x} + y + 3z &= 3 \\ \color{red}{1y} + 5z &= 10 \\ \color{red}{1z} &= 7 \end{align}\right. #### Example 1: Perform Row Operations Perform the specified row operations on the system \left\{\begin{align} y + 3z &= 3 \\ x - y + 5z &= 5 \\ 2x + 3y - z &= 7 \end{align}\right. 1. Interchange equations 1 and 2 2. Multiply equation 2 by −2 3. Add −2 times equation 2 with equation 3 and replace equation 3 ###### Solutions 1. Switch the top two equations. \left\{\begin{align} \color{blue}{x - y + 5z} &= \color{blue}{5} \\ \color{purple}{y + 3z} &= \color{purple}{3} \\ 2x + 3y - z &= 7 \end{align}\right. 2. Multiply the second equation by −2 \left\{\begin{align} y + 3z &= 3 \\ \color{blue}{(-2)}(x - y + 5z) &= \color{blue}{(-2)}(5) \\ 2x + 3y - z &= 7 \end{align}\right. \left\{\begin{align} y + 3z &= 3 \\ \color{blue}{-2x + 2y - 10z} &= \color{blue}{-10} \\ 2x + 3y - z &= 7 \end{align}\right. 3. Multiply the second equation by −2 (this was done in part b). \left\{\begin{align} y + 3z &= 3 \\ \color{blue}{-2x + 2y - 10z} &= \color{blue}{-10} \\ 2x + 3y - z &= 7 \end{align}\right. Add the new second equation with the third equation. \begin{align} -2x + 2y - 10z &= -10 \\ \underline{+ \quad 2x + 3y - \quad z} &= \underline{7} \\ \color{purple}{5y - 11z} &= \color{purple}{-3} \end{align} Replace the third equation with this result. \left\{\begin{aligned} y + 3z &= 3 \\ x - y + 5z &= 5 \\ \color{red}{5y - 11z} &= \color{red}{-3} \end{aligned}\right. ##### Try It 1 Add 2 times the 1st equation with −3 times the 3rd equation and replace the 3rd equation in the system \left\{\begin{align} 3x - 3y - z &= 2 \\ y + 2z &= -4 \\ -2x + y + z &= 0 \end{align}\right. ###### Answer \left\{\begin{align} 3x - 3y - z &= 2 \\ y + 2z &= -4 \\ -9y - 5z &= 4 \end{align}\right. #### Example 2: Solve a System Using Gaussian Elimination Use Gaussian Elimination to solve \left\{\begin{align} x + y + z &= 4 \\ 2x + y - z &= 0 \\ x - y + 3z &= 12 \end{align}\right. ###### Solution Want to get rid of 2x in the 2nd equation. Add −2 times 1st equations to 2nd equation and replace 2nd equation. \left\{\begin{align} \color{blue}{\left(-2\right)}\left(x + y + z\right) &= \color{blue}{\left(-2\right)}4 \\ 2x + y - z &= 0 \\ x - y + 3z &= 12 \end{align}\right. \left\{\begin{align} x + y + z &= 4 \\ \color{blue}{-y - 3z} &= \color{blue}{-8} \\ x - y + 3z &= 12 \end{align}\right. Now get rid of the x in the 3rd equation. Add −1 times 1st equation to 3rd equation and replace 3rd equation. \left\{\begin{align} \color{blue}{\left(-1\right)}\left(x + y + z\right) &= \color{blue}{\left(-1\right)}\left(4\right) \\ -y - 3z &= -8 \\ x - y + 3z &= 12 \end{align}\right. \left\{\begin{align} x + y + z &= 4 \\ -y - 3z &= -8 \\ \color{blue}{-2y + z} &= \color{blue}{8} \end{align}\right. Next get rid of the −2y in the 3rd equation. Add −2 times the 2nd equation with the 3rd equation and replace the 3rd equation. \left\{\begin{align} x + y + z &= 4 \\ \color{blue}{\left(-2\right)}\left(-y - 3z\right) &= \color{blue}{\left(-2\right)}\left(-8\right) \\ -2y + z &= 8 \end{align}\right. \left\{\begin{align} x + y + z &= 4 \\ -y - 3z &= -8 \\ \color{blue}{8z} &= \color{blue}{24} \end{align}\right. Finish getting Row-Echelon Form by turning the leading coefficients into 1's. Multiply the 2nd equation by −1 and the 3rd equation by $$\frac{1}{8}$$. \left\{\begin{align} x + y + z &= 4 \\ \color{red}{\left(-1\right)}\left(-y - 3z\right) &= \color{red}{\left(-1\right)}\left(-8\right) \\ \color{blue}{\left(\tfrac{1}{8}\right)}\left(8z\right) &= \color{blue}{\left(\tfrac{1}{8}\right)}\left(24\right) \end{align}\right. \left\{\begin{align} x + y + z &= 4 \\ \color{red}{y + 3z} &= \color{red}{8} \\ \color{blue}{z} &= \color{blue}{3} \end{align}\right. The 3rd equation now says z = 3. Substitute this back into the 2nd equation and solve for y. $$y + 3z = 8$$ $$y + 3(3) = 8$$ $$y = -1$$ Finally, substitute both y and z into the 1st equation and solve for x. $$x + y + z = 4$$ $$x + (-1) + 3 = 4$$ $$x = 2$$ The solution is (2, −1, 3). ##### Try It 2 Using Gaussian Elimination to solve \left\{\begin{align} x - y + 2z &= -2 \\ x + 2y - 3z &= 11 \\ 3x - 2y + z &= 3 \end{align}\right. ###### Answer (2, 0, −3) #### Example 3: Solve a System Using Gaussian Elimination Use Gaussian Elimination to solve \left\{\begin{align} 2x - 4y + z &= 1 \\ 3x + 2y + 2z &= 8 \\ -5x - 2y - 2z &= -6 \end{align}\right. ###### Solution Want to get rid of 3x in the 2nd equation. Add −3 times 1st equations to 2 times 2nd equation and replace 2nd equation. \left\{\begin{align} \color{blue}{\left(-3\right)}\left(2x - 4y + z\right) &= \color{blue}{\left(-3\right)}1 \\ \color{blue}{\left(2\right)}\left(3x + 2y + 2z\right) &= \color{blue}{\left(2\right)}\left(8\right) \\ -5x - 2y - 2z &= -6 \end{align}\right. \left\{\begin{align} 2x - 4y + z &= 1 \\ \color{blue}{16y + z} &= \color{blue}{13} \\ -5x - 2y - 2z &= -6 \end{align}\right. Now get rid of the −5x in the 3rd equation. Add 5 times 1st equation to 2 times 3rd equation and replace 3rd equation. \left\{\begin{align} \color{blue}{\left(5\right)}\left(2x - 4y + z\right) &= \color{blue}{\left(5\right)}\left(1\right) \\ 16y + z &= 13 \\ \color{blue}{\left(2\right)}\left(-5x - 2y - 2z\right) &= \color{blue}{\left(2\right)}\left(-6\right) \end{align}\right. \left\{\begin{align} 2x - 4y + z &= 1 \\ 16y + z &= 13 \\ \color{blue}{-24y + z} &= \color{blue}{-7} \end{align}\right. Next get rid of the −24y in the 3rd equation. Add 3 times the 2nd equation with 2 times the 3rd equation and replace the 3rd equation. \left\{\begin{align} 2x - 4y + z &= 1 \\ \color{blue}{\left(3\right)}\left(16y + z\right) &= \color{blue}{\left(3\right)}\left(13\right) \\ \color{blue}{\left(2\right)}\left(-24y + z\right) &= \color{blue}{\left(2\right)}\left(-7\right) \end{align}\right. \left\{\begin{align} 2x - 4y + z &= 1 \\ 16y + z &= 13 \\ \color{blue}{5z} &= \color{blue}{25} \end{align}\right. Finish getting Row-Echelon Form by turning the leading coefficients into 1's. Multiply the 1st equation by $$\frac{1}{2}$$, the 2nd equation by $$\frac{1}{16}$$, and the 3rd equation by \frac{1}{5}\). \left\{\begin{align} \color{purple}{\left(\tfrac{1}{2}\right)}\left(2x - 4y + z\right) &= \color{purple}{\left(\tfrac{1}{2}\right)}\left(1\right) \\ \color{red}{\left(\tfrac{1}{16}\right)}\left(16y + z\right) &= \color{red}{\left(\tfrac{1}{16}\right)}\left(13\right) \\ \color{blue}{\left(\tfrac{1}{5}\right)}\left(5z\right) &= \color{blue}{\left(\tfrac{1}{5}\right)}\left(25\right) \end{align}\right. \left\{\begin{align} \color{purple}{x - 2y + \tfrac{1}{2}z} &= \color{purple}{\tfrac{1}{2}} \\ \color{red}{y + \tfrac{1}{16}z} &= \color{red}{\tfrac{13}{16}} \\ \color{blue}{z} &= \color{blue}{5} \end{align}\right. The 3rd equation now says z = 5. Substitute this back into the 2nd equation and solve for y. $$y + \frac{1}{16}z = \frac{13}{16}$$ $$y + \frac{1}{16}(5) = \frac{13}{16}$$ $$y = \frac{1}{2}$$ Finally, substitute both y and z into the 1st equation and solve for x. $$2x - 4y + z = 1$$ $$2x - 4\left(\frac{1}{2}\right) + 5 = 1$$ $$x = -1$$ The solution is $$\left(-1, \tfrac{1}{2}, 5\right)$$. ##### Try It 3 Using Gaussian Elimination to solve \left\{\begin{align} -2x + y - 2z &= -1 \\ 3x + 2y + z &= 7 \\ -2x - y + 4z &= -9 \end{align}\right. ###### Answer (2, 1, −1) ## Many Solutions and No Solutions Linear equations in three dimensions graph planes. One solution to a system occurs when the three planes intersect in one point. No solution occurs with the plans never intersect in any point. Many solutions occur when the planes all intersect in a single line. A line in three dimensions is not a simple equation. The line can be described by variable expressions that give points on the line, one expression per variable. ###### Classifying Types of Solutions for Linear Systems If all the variables disappear while solving the linear system and • a false statement remains such as 0 = 5, then no solution • a true statement remains such as 0 = 0, then many solutions ###### Solution to Many Solutions for Three Dimensions To write the solution to many solutions, an expression for each dimension describes points on the line of intersection. When solving the linear system and all the variables disappear leaving a true statement such as 0 = 0. 1. Solve the system using Gaussian Elimination. 2. The 3rd equation will become 0 = 0. 3. Let z be equal to some constant times a. Choose the constant to be the common denominator of the coefficients of y and z in the first two equation. 4. Substitute this expression into the 2nd equation to obtain the expression for y. 5. Substitute the expressions for y and z into the 1st equation to obtain the expression for x. 6. The solution is the expressions for (x, y, z). #### Example 4: Solve a System Using Gaussian Elimination with Many or No Solutions Use Gaussian Elimination to solve \left\{\begin{align} x + 2y - z &= 7 \\ 3x + 7y + z &= 23 \\ -2x - 3y + 6z &= -12 \end{align}\right. ###### Solution Want to get rid of 3x in the 2nd equation. Add −3 times 1st equations to the 2nd equation and replace 2nd equation. \left\{\begin{align} \color{blue}{\left(-3\right)}\left(x + 2y - z\right) &= \color{blue}{\left(-3\right)}7 \\ 3x + 7y + z &= 23 \\ -2x - 3y + 6z &= -12 \end{align}\right. \left\{\begin{align} x + 2y - z &= 7 \\ \color{blue}{y + 4z} &= \color{blue}{2} \\ -2x - 3y + 6z &= -12 \end{align}\right. Now get rid of the −2x in the 3rd equation. Add 2 times 1st equation to the 3rd equation and replace 3rd equation. \left\{\begin{align} \color{blue}{\left(2\right)}\left(x + 2y - z\right) &= \color{blue}{\left(2\right)}\left(7\right) \\ y + 4z &= 2 \\ -2x - 3y + 6z &= -12 \end{align}\right. \left\{\begin{align} x + 2y - z &= 7 \\ y + 4z &= 2 \\ \color{blue}{y + 4z} &= \color{blue}{2} \end{align}\right. Next get rid of the y in the 3rd equation. Add −1 times the 2nd equation with the 3rd equation and replace the 3rd equation. \left\{\begin{align} x + 2y - z &= 7 \\ \color{blue}{\left(-1\right)}\left(y + 4z\right) &= \color{blue}{\left(-1\right)}\left(2\right) \\ y + 4z &= 2 \end{align}\right. \left\{\begin{align} x + 2y - z &= 7 \\ y + 4z &= 2 \\ \color{blue}{0} &= \color{blue}{0} \end{align}\right. Check that the leading coefficients are 1. The third equation is now 0 = 0 which is true, so this is many solutions. Because the denominators are all 1 in the coefficients of y and z, let $$z = a$$ Substitute this into the 2nd equation and solve for y. $$y + 4z = 2$$ $$y + 4\color{blue}{a} = 2$$ $$y = -4a + 2$$ Substitute y and z into the first equation and solve for x. $$x + 2y - z = 7$$ $$x + 2\color{blue}{(-4a + 2)} - \color{blue}{a} = 7$$ $$x - 8a + 4 - a = 7$$ $$x = 9a + 3$$ The solution is (x, y, z). $$\left(9a + 3, -4a + 2, a\right)$$ ##### Try It 4 Use Gaussian Elimination to solve \left\{\begin{align} x - y + 2z &= 3 \\ -2x + 4y - 5z &= -2 \\ 2x + 3z &= 9 \end{align}\right. ###### Answer No solution #### Example 5: Solve a System Using Gaussian Elimination Missing One Equation Use Gaussian Elimination to solve \left\{\begin{align} x + y + z &= 2 \\ -5x + y - 2z &= 2 \end{align}\right. ###### Solution Want to get rid of −5x in the 2nd equation. Add 5 times 1st equations to the 2nd equation and replace 2nd equation. \left\{\begin{align} \color{blue}{\left(5\right)}\left(x + y + z\right) &= \color{blue}{\left(5\right)}2 \\ -5x + y - 2z &= 2 \end{align}\right. \left\{\begin{align} x + y + z &= 2 \\ \color{blue}{6y + 3z} &= \color{blue}{12} \end{align}\right. Check that the leading coefficients are 1. Multiply the 2nd equation by $$\frac{1}{6}$$. \left\{\begin{align} x + y + z &= 2 \\ \color{blue}{\left(\tfrac{1}{6}\right)}\left(6y + 3z\right) &= \color{blue}{\left(\tfrac{1}{6}\right)}\left(12\right) \end{align}\right. \left\{\begin{align} x + y + z &= 2 \\ \color{blue}{y + \tfrac{1}{2}z} &= \color{blue}{2} \end{align}\right. There is no third equation. Two planes will always intersect in a line, so this is many solutions. Or imagine that the third equation is now 0 = 0, so this is many solutions. Because the 2nd equation has a denominator of 2 in the coefficient of 2z, let $$z = 2a$$ Substitute this into the 2nd equation and solve for y. $$y + \frac{1}{2}z = 2$$ $$y + \frac{1}{2}\color{blue}{\left(2a\right)} = 2$$ $$y = -a + 2$$ Substitute y and z into the first equation and solve for x. $$x + y + z = 2$$ $$x + \color{blue}{(-a + 2)} + \color{blue}{2a} = 2$$ $$x = -a$$ The solution is (x, y, z). $$\left(-a, -a + 2, 2a\right)$$ ##### Try It 5 Use Gaussian Elimination to solve \left\{\begin{align} x + 2y - z &= -3 \\ 2x + 3y - 5z &= 1 \end{align}\right. ###### Answer $$\left(7a + 17, -3a - 7, a\right)$$ ##### Lesson Summary ###### Elementary Row Operations These operations are allowed in systems of equations and produce equivalent systems. • Interchange two equations. • Multiply one equation by a nonzero constant. • Add a multiple of one equation to another equation and replace the one of these two equations. ###### Row-Echelon Form • The first nonzero term in each equation has a coefficient of 1. • All terms under the leading 1 are zeros producing an inverted pyramid shape. • Any equation that is all zeros is at the bottom. \left\{\begin{align} \color{red}{1x} + y + 3z &= 3 \\ \color{red}{1y} + 5z &= 10 \\ \color{red}{1z} &= 7 \end{align}\right. ###### Classifying Types of Solutions for Linear Systems If all the variables disappear while solving the linear system and • a false statement remains such as 0 = 5, then no solution • a true statement remains such as 0 = 0, then many solutions ###### Solution to Many Solutions for Three Dimensions To write the solution to many solutions, an expression for each dimension describes points on the line of intersection. When solving the linear system and all the variables disappear leaving a true statement such as 0 = 0. 1. Solve the system using Gaussian Elimination. 2. The 3rd equation will become 0 = 0. 3. Let z be equal to some constant times a. Choose the constant to be the common denominator of the coefficients of y and z in the first two equation. 4. Substitute this expression into the 2nd equation to obtain the expression for y. 5. Substitute the expressions for y and z into the 1st equation to obtain the expression for x. 6. The solution is the expressions for (x, y, z). Helpful videos about this lesson. ## Practice Exercises Perform the indicated row operations. What did it accomplish? 1. \left\{\begin{align} 2x - y - 3z &= 0 \\ x + 2y - z &= 3 \\ x + y + z &= 1 \end{align}\right. Interchange the first two equations. 2. \left\{\begin{align} 2x - 3y + 2z &= -1 \\ -2x + y + z &= 5 \\ 3x + 2y + 2z &= 1 \end{align}\right. Add equation 1 to equation 2 and replace equation 2. 3. \left\{\begin{align} x - 3y + 2z &= -2 \\ x + 2y + 2z &= -1 \\ x - y - z &= 4 \end{align}\right. Add −1 times equation 1 to equation 3 and replace equation 3. 4. Solve using Gaussian Elimination. 5. \left\{\begin{align} x + 3y - 2z &= -3 \\ y + z &= 5 \\ z &= 1 \end{align}\right. 6. \left\{\begin{align} 2x + y + 2z &= 4 \\ y + 3z &= 4 \\ z &= -2 \end{align}\right. 7. \left\{\begin{align} 2x + y - z &= -1 \\ 2y + 3z &= -1 \\ x + y \qquad &= -1 \end{align}\right. 8. \left\{\begin{align} x - 5y + 2z &= -11 \\ x + 4y + z &= 7 \\ -x + 2y + z &= 5 \end{align}\right. 9. \left\{\begin{align} x - y + z &= 8 \\ 2x + y + z &= 8 \\ x + y + z &= 6 \end{align}\right. 10. \left\{\begin{align} x + y + 2z &= 1 \\ -2x - y - 3z &= 2 \\ 4x + 5y + 9z &= 8 \end{align}\right. 11. \left\{\begin{align} x \qquad + z &= -2 \\ x + y - 2z &= -1 \\ 3x + y \qquad &= 8 \end{align}\right. 12. \left\{\begin{align} 3x + y - 3z &= -18 \\ 2x - 2y + z &= 17 \\ -2x + y - 2z &= -21 \end{align}\right. 13. \left\{\begin{align} 2x - y + z &= 2 \\ -2x + 3y + 2z &= -1 \\ 4x \qquad + 5z &= 2 \end{align}\right. 14. \left\{\begin{align} x + 2y - 7z &= 5 \\ y + z &= 3 \end{align}\right. 15. \left\{\begin{align} x + 2y + z &= 1 \\ 2x + 2y + z &= 4 \end{align}\right. 16. Problem Solving 17. Three friends went to a Mexican fast food restaurant. Joe bought 3 tacos, 2 burritos, and a drink for6.25. Frank bought 5 tacos and 4 burritos for $7.75. He forgot to buy a drink so Samantha bought an extra drink. She bought 2 tacos, 2 burritos, and 2 drinks for$7.50. How much does Frank owe Samantha for the drink?
18. In electrical circuit analysis, it is important to know the currents through each part of a circuit. Kirchhoff's Laws are used to generate a system of equations to find the currents. For this diagram, the Junction Rule says the total current into a junction equals the total current leaving a junction. This generates the 1st equation in the system below. The Loop Rule says that for any complete loop, the voltage rises equals the voltage drops. This generates the 2nd and 3rd equations below. Solve the system of equations to find the size of the currents (I1, I2, and I3).
\left\{\begin{align} I_1 + I_2 - I_3 &= 0 \\ 10I_1 \quad + 5I_3 &= 4.5 \\ 3I_2 + 5I_3 &= 3.4 \end{align}\right.
19. Mixed Review

20. (8-02) Solve the system of equations \left\{\begin{align} \tfrac{1}{2}x + 3y = -1 \\ y = \tfrac{1}{4}x - \tfrac{5}{4} \end{align}\right.
21. (8-01) Solve by substitution \left\{\begin{align} x = y^2 \\ x + 3y^2 = 1 \end{align}\right.
22. (7-05) Classify the conic $$2x^2 - 3xy + 2y^2 - 5x + 7y + 10 = 0$$.
23. (6-03) If $$\overset{\rightharpoonup}{m} = \langle 2, 5 \rangle$$ and $$\overset{\rightharpoonup}{n} = \langle -3, -1 \rangle$$, find $$\overset{\rightharpoonup}{m} + 2\overset{\rightharpoonup}{n}$$.

1. \left\{\begin{align} x + 2y - z &= 3 \\ 2x - y - 3z &= 0 \\ x + y + z &= 1 \end{align}\right.; Created a leading coefficient of 1 in the 1st equation
2. \left\{\begin{align} 2x - 3y + 2z &= -1 \\ -2y + 3z &= 4 \\ 3x + 2y + 2z &= 1 \end{align}\right.; Eliminated the first term in the 2nd equation
3. \left\{\begin{align} x - 3y + 2z &= -2 \\ x + 2y + 2z &= -1 \\ 2y - 3z &= 6 \end{align}\right.; Eliminated the first term in the 3rd equation
4. (−13, 4, 1)
5. (−1, 10, −2)
6. (1, −2, 1)
7. (−1, 2, 0)
8. (2, −1, 5)
9. $$\left(-a - 3, -a + 4, a\right)$$
10. No solution
11. (2, −3, 7)
12. No solution
13. $$\left(9a - 1, -a + 3, a\right)$$
14. $$\left(3, -a - 1, 2a\right)$$
15. \$2
16. $$I_1 = 0.2 A, I_2 = 0.3 A, I_3 = 0.5 A$$
17. (2.2, −0.7)
18. $$\left(\frac{1}{4}, \frac{1}{2}\right), \left(\frac{1}{4}, -\frac{1}{2}\right)$$
19. ellipse
20. $$\langle -4, 3 \rangle$$