Chapter 3 System of Linear Equations

3.1 Linear Equations in Two Variables Forms of Linear Equation Forms of Linear Equation  ax + by = c (a, b, c are constants)  y = mx + b (b = y-intercept, m = slope)

Solution of System of Linear Equations Given: Given: a)x + y = 7 b)x - y = -1 What are the coordinate values such that both equations are satisfied? What are the coordinate values such that both equations are satisfied? Answer: (3, 4) Answer: (3, 4) x + y = 7 x – y = -1

Solution: Substitution Method Given: y = -2x + 4 7x – 2y = 3 Given: y = -2x + 4 7x – 2y = 3 Solve the system of equations 7x – 2(-2x + 4) = 3 7x + 4x – 8 = 3 11 x = 11 x = 1 y = -2(1) + 4 = 2 Solve the system of equations 7x – 2(-2x + 4) = 3 7x + 4x – 8 = 3 11 x = 11 x = 1 y = -2(1) + 4 = 2 Solution (1, 2) Solution (1, 2) Check solution Check solution

Solution: Substitution Method Given: 5x + 2y = 1 x – 3y = 7 Given: 5x + 2y = 1 x – 3y = 7 Solve the system of equations x = 3y + 7 5(3y + 7) + 2y = 1 15y y = 1 17y = -34 y = -2 x = 3(-2) + 7 = 1 Solve the system of equations x = 3y + 7 5(3y + 7) + 2y = 1 15y y = 1 17y = -34 y = -2 x = 3(-2) + 7 = 1 Solution (1, -2) Solution (1, -2) Check solution Check solution

Solution: Eliminating One Variable Given: 3x + 4y = -10 5x – 2y = 18 Given: 3x + 4y = -10 5x – 2y = 18 Solve the system of equations 3x + 4y = -10 2(5x – 2y) = 2(18) 3x + 4y = x – 4y = 36 13x = 26 x = 2 3(2) + 4y = -10 4y = -16 y = -4 Solve the system of equations 3x + 4y = -10 2(5x – 2y) = 2(18) 3x + 4y = x – 4y = 36 13x = 26 x = 2 3(2) + 4y = -10 4y = -16 y = -4 Solution (2, -4) Solution (2, -4)

Lines with No Solutions Given 3x – 2y = 6 6x – 4y = 18 -2(3x – 2y) = -2(6) -6x + 4y = 12 6x – 4y = 18 0 = 6 Given 3x – 2y = 6 6x – 4y = 18 -2(3x – 2y) = -2(6) -6x + 4y = 12 6x – 4y = 18 0 = 6 Are there values of (x, y) for which 0 = 6? Are there values of (x, y) for which 0 = 6?

Lines with No Solutions Given 3x – 2y = 6 y = (2/3)x - 3 6x – 4y = 18 y = (2/3)x – (9/2) Given 3x – 2y = 6 y = (2/3)x - 3 6x – 4y = 18 y = (2/3)x – (9/2)

Your Turn Solve by substitution method Solve by substitution method a)x – 3y = 8 y = 2x - 9 Solve by elimination method Solve by elimination method b)2x + 3y = 6 2x – 3y = 6 a)x - 3(2x - 9) = 8 x – 6x + 27 = 8 -5x = -19 x = 19/5 y = 2(19/5) – 9 = 38/5 – 9 = (38 – 45)/5 = -7/5 Solution: (19/5, -7/5) b)2x + 3y = 6 2x – 3y = 6 4x = 12 x = 3 2(3) + 3y = 6 3y = 0 y = 0 Solution: (3, 0)

Your Turn Solve by elimination method Solve by elimination method 2x – 5y = 13 5x + 3y = 17 3(2x – 5y) = 3(13) 5(5x + 3y) = 5(17) 6x – 15y = 39 25x + 15y = x = 124 x = 4 6(4) – 15y = – 15y = y = 15 y = -1 Solution: (4, -1)

3.3 Systems of Linear Equations in Three Variables {(x, y, z) | ax + by + cz = d} {(x, y, z) | ax + by + cz = d} (x, y, z) y-axis z-axis x-axis

Graph of Linear Equation in 3 Variables {(x, y, z) | 2x + 3y + z = 1} Set of all points (x, y, z) in a particular plane in 3-D. {(x, y, z) | 2x + 3y + z = 1} Set of all points (x, y, z) in a particular plane in 3-D. z-axis A(1/2, 0, 0) y-axis x-axis C(0, 0, 1) B(0, 1/3, 0)

X-Y Plane x-axis y-axis z-axis {(x, y, z) | z = 0}

X-Z Plane x-axis y-axis z-axis {(x, y, z) | y = 0}

Y-Z Plane x-axis y-axis z-axis {(x, y, z) | x = 0}

Systems of Linear Equations in 3 Variables Given: Given:  air fare + hotel + car for $210  hotel + car for $112  air fare + hotel for $180 What is the cost of air fare only, hotel only, or car only? What is the cost of air fare only, hotel only, or car only?

Solution Let x = air fare y = hotel z = car Let x = air fare y = hotel z = car x + y + z = 210 y + z = 112 x + y = 180 x + y + z = 210 y + z = 112 x + y = 180 Solution: (x, y, z) = (98, 82, 30) Solution: (x, y, z) = (98, 82, 30)

Solving Equations in 3 Variables (1) Given: 5x – 2y – 4z = 3 3x + 3y + 2z = -3 -2x + 5y + 3z = 3 5x–3y–4z=3 5x – 2y – 4z = 3 3x+3y+2z=-3 2(3x + 3y + 2z) = 2(-3) 5x – 2y – 4z = 3 6x + 6y + 4z = -6 11x + 4y = -3 5x – 2y – 4z = 3 6x + 6y + 4z = -6 11x + 4y = -3 3x+3y+2z=-3 3(3x + 3y + 2z) = 3(-3) -2x+5y+3z=3 -2(-2x + 5y + 3z) = -2(3)

Solving Equations in 3 Variables (2) 11x + 4y = -3 11x + 4y = -3 3x+3y+2z=-3 3(3x + 3y + 2z) = 3(-3) -2x+5y+3z=3 -2(-2x + 5y + 3z) = -2(3) 9x + 9y + 6z = -9 4x - 10y - 6z = -6 13x - y = -15 9x + 9y + 6z = -9 4x - 10y - 6z = -6 13x - y = x+4y=-3 11x + 4y = -3 13x- y=-15 4(13x – y) = 4(-15) 11x + 4y = -3 52x – 4y = x = x + 4y = -3 52x – 4y = x = -63

Solving Equations in 3 Variables (3) 63x = -63 x = -1 63x = -63 x = -1 We had: 11x + 4y = -3 11(-1) + 4y = -3 4y = = 8 y = 2 We had: 3x + 3y + 2z = -3 3(-1) + 3(2) + 2z = z = -3 2z = -6 z = -3 Solution: (-1, 2, -3) Check the solution

Solving Equations in 3 Variables (4) Given: x + z = 8 x + y + 2z = 17 x + 2y + z = 16 x+y+2z = 17 -2(x + y + 2z) = -2(17) x+2y+z = 16 x + 2y + z) = 16 -2x - 2y - 4z = -34 x + 2y + z = 16 -x - 3z = x - 2y - 4z = -34 x + 2y + z = 16 -x - 3z = -18 x + z = 8 -x - 3z = 18 2z = 10 z = 5 x + z = 8 -x - 3z = 18 2z = 10 z = 5

Solving Equations in 3 Variables (5) z = 5 We had: x + z = 8 x + (5) = 8 x = 3 We had: x + y + 2z = 17 (3) + y + 2(5) = y + 10 = 17 y = 4 Solution: (3, 4, 5) Check your solution

Incosistent Systems Given: 2x + 5y + z = 12 x – 2y + 4z = x + 6y – 12z = 20 x-2y+4z=10 3(x – 2y + 4z) = 3(-10) -3x+6y-12z=20 -3x + 6y – 12z = 20 x-2y+4z=10 3(x – 2y + 4z) = 3(-10) -3x+6y-12z=20 -3x + 6y – 12z = 20 3x – 6y + 12z = x + 6y – 12z = 20 0 = -10 ??? 3x – 6y + 12z = x + 6y – 12z = 20 0 = -10 ??? No (x, y, z) to satisfy this condition.No (x, y, z) to satisfy this condition. No common point of intersectionNo common point of intersection System of equations is inconsistentSystem of equations is inconsistent

Solution to air-hotel-car problems Let x = air fare y = hotel z = car Let x = air fare y = hotel z = car a) x + y + z = 210 b) y + z = 112 c) x + y = 180 a) x + y + z = 210 b) y + z = 112 c) x + y = 180

a) x + y + z = 210 b) y + z = 112 c) x + y = 180 Use a) and b) to eliminate z x + y + z = ( y + z) = -(112) x + y + z = y - z = d) x = 98

a) x + y + z = 210 b) y + z = 112 c) x + y = 180 Use b) and c) to eliminate y y + z = 112 -(x + y) = -(180) y + z = 112 -x – y = e) -x + z = -68

a) x + y + z = 210 b) y + z = 112 c) x + y = 180 Use d) and e) to find z d) x = 98 e) –x + z = -68 -(98) + z = -68 z = z = 30

a) x + y + z = 210 b) y + z = 112 c) x + y = 180 Use c) and d) to find y c) x + y = 180 d) x = y = 180 y = 180 – 98 = 82 Solution: (98, 82, 30) … (x: air, y: hotel, z: car)

Your Turn Solve the following system x + y + 2z = 11 x + y + 3z = 14 x + 2y – z = 5 Solve the following system x + y + 2z = 11 x + y + 3z = 14 x + 2y – z = 5 Is it consistent or inconsistent Is it consistent or inconsistent

Your Turn a) x + y + 2z = 11 b) x + y + 3z = 14 c) x + 2y – z = 5 To rid of x a) –(x + y + 2z) = -(11) -x – y – 2z = -11 b) x + y + 3z = 14 x + y + 3z = z = 3 a) x + y + 2(3) = 11 x + y = 5 b) x + y + 3(3) = 14 x + y = 9 5 = 9 No values of x, y will make this true. Thus, no solution. (Inconsistent system)

3.4 Matrix Solution to Linear Systems Matrix Matrix  Arrangement of number in rows and columns  Population (in mil.) men women

Augmented Matrix 4x + 4y = 19 2y + 3z = 8 4x - 5z = 7 4x + 4y = 19 2y + 3z = 8 4x - 5z = a b c d e f a b c d e f

Matrix Row Operations 4x + 3y = -15 x + 2y = a b 0 1 c Interchange i th and j th rows: R i R j Multiply each element in the ith row by k: kR i Add k times the elements in row i to corresponding elements in Row j: kR i + R j Row-echelon form

Matrix Row Operations (4) 3(3) 3(-15) Interchange R 1 and R 2 : R 1 R 2 Multiply each element in the 1 st row by 3: 3R 1 Given a 2 x 3 matrix: =

Matrix Row Operations (4)+1 3(3)+2 3(-15)-1 Add 3 times the elements in row 1 to corresponding elements in Row 2: 3R 1 + R 2 =

Your Turn Perform the following operation: R 1 R Given:

Your Turn Perform the following operation: 2R 2 Given: (4) 2(-3) 2(1) =

Your Turn Perform the following operation: 3R 1 + R (3)+4 3(-2)-3 3(5)+1 = Given:

Matrix Row Operations 4x + 4y = 19 2y + 3z = 8 4x - 5z = R1 R kR2 (k = -2) kR1+R3 (k = -1)

Solving Linear System in 2 Variables 4x – 3y = -15 x + 2y = a b 0 1 c Recall the strategy. Then, the last row says: y = c The first row says: x + ay = b

Solving thy Linear System in 2 Variables 4x – 3y = -15 x + 2y = R 1 R 2 (to make e 11 = 1) R 1 + R 2 (to make e 21 = 0) (-1/11)R 2 (to make e 12 = 1)

Solving System in 2 Variables y = 1 x + 2(1) = -1 x = -3 Solution: (-3, 1) Check: 4x – 3y = -15 4(-3) – 3(1) = -12 – 3 = -15 x + 2y = -1 (-3) + 2(1) = = -1

Your Turn Solve the following using matrix method. Solve the following using matrix method. 3x – 6y = 1 2x – 4y = 2/ /3 1 a b 0 1 c (1/3)R 1 (to make e 11 = 1) (1/3)3 (1/3(-6) (1/3) / / /3 =

Your Turn (1/3)3 (1/3(-6) (1/3) / / /3 = 2R 1 -R 2 (to make e 21 = 0) /3 2(1)-2 2(-2)-4 2(1/3) – 2/ / = (-1/8)R 2 (to make e 22 = 1) /3 (-1/8)0 (-1/8)(-8) (-1/8) / =

Your Turn / y = 0 x – 2y = 1/3 x – 0 = 1/3 x = 1/3 Solution: (1/3, 0) Check: 3x – 6y = 1 3(1/3) – 6(0) = 1 1 = 1 2x – 4y = 2/3 2(1/3) – 4(0) = 2/3 2/3 – 0 = 2/3 2/3 = 2/3

Your Turn Solve the following using matrix method. Solve the following using matrix method. -3x + 4y = 12 2x + y = 3

Solving System in 3 Variables x + y + z = 210 y + z = 112 x + y = (-1)R1+R (-1)R3 x + y + z = 210 y + z = 112 z = z = 30 y + 30 = 112 → y = 82 x = 210 → z = 98

Your Turn Solve the following system of equations Solve the following system of equations  3x + y + 2z = 31 x + y + 2z = 19 x + 3y + 2z = 25

3.5: 3.5: Determinants 2 x 2 Matrix 2 x 2 Matrix a 1 b 1 a 2 b 2 Determinant Determinant a 1 b 1 a 2 b 2 = a 1 b 2 – a 2 b 1

Determinants = 2(-5) – (-3)4 = = = ? = ?

Cramer’s Rule Given: a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 c 1 b 1 c 2 b 2 a 1 b 1 a 2 b 2 a 1 c 1 a 2 c 2 a 1 b 1 a 2 b 2 x = y = Dx x = D Dy y = D

Cramer’s Rule Given: 5x - 4y = 2 6x - 5y = x = y = -10 – (-4) -6 = = ---- = – (-24) -1 5 – = = ---- = (6, 7)

Your Turn Use Cramer’s rule to solve Use Cramer’s rule to solve 1.12x + 3y = 15 2x – 3y = 13 2.x – 3y = 4 3x – 4y = 12

Solution for 1 12x + 3y = 15 2x – 3y = D = 2 -3 = 12(-3) – 2(3) = D = 2 -3 = 12(-3) – 2(3) = D x = = 15(-3) – 3(12) = D x = = 15(-3) – 3(12) = -84 D x -84 x = ---- = = 2 D -42 D x -84 x = ---- = = 2 D -42

Solution for 1 12x + 3y = 15 2x – 3y = D = 2 -3 = 12(-3) – 2(3) = D = 2 -3 = 12(-3) – 2(3) = D y = 2 13 = 12(13) – 2(15) = D y = 2 13 = 12(13) – 2(15) = 126 D y 126 y = ---- = = -3 D -42 D y 126 y = ---- = = -3 D -42 Solution: (2, -3)

Solution for 2 x - 36 = 4 3x – 3y = 12 x - 36 = 4 3x – 3y = D = 3 -3 = 1(-3) – 3(-3) = D = 3 -3 = 1(-3) – 3(-3) = D x = = 4(-3) – 12(-3) = D x = = 4(-3) – 12(-3) = 24 D x 24 y = ---- = ---- = 4 D 6 D x 24 y = ---- = ---- = 4 D 6

Solution for 2 x - 36 = 4 3x – 3y = 12 x - 36 = 4 3x – 3y = D = 3 -3 = 1(-3) – 3(-3) = D = 3 -3 = 1(-3) – 3(-3) = D y = 3 12 = 1(12) – 3(4) = D y = 3 12 = 1(12) – 3(4) = 0 D y 0 y = ---- = ---- = 0 D 6 D y 0 y = ---- = ---- = 0 D 6 Solution: (4, 0)

Determinant of a 3 x 3 Matrix a b c b c b c b c a b c = a b c – a b c + a b c a b c a 1 b 1 c 1 b 2 c 2 b 1 c 1 b 1 c 1 a 2 b 2 c 2 = a 1 b 3 c 3 – a 2 b 3 c 3 + a 3 b 2 c 2 a 3 b 3 c 3E.g = ? = ? 4 3 6

Determinant of a 3 x 3 Matrix Given: a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 a 1 b 1 c 1 D = a 2 b 2 c 2 a 3 b 3 c 3 a 1 b 1 c 1 D = a 2 b 2 c 2 a 3 b 3 c 3 d 1 b 2 c 3 D x = d 2 b 2 c 3 d 3 b 3 c 3 d 1 b 2 c 3 D x = d 2 b 2 c 3 d 3 b 3 c 3 a 1 d 1 c 1 a 1 b 1 d 1 D y = a 2 d 2 c 2 D z = a 2 b 2 d 2 a 3 d 3 c 3 a 3 b 3 d 3 a 1 d 1 c 1 a 1 b 1 d 1 D y = a 2 d 2 c 2 D z = a 2 b 2 d 2 a 3 d 3 c 3 a 3 b 3 d 3 D x x = ---- D D x x = ---- D D y y = ---- D D z z = ---- D D y y = ---- D D z z = ---- D

Example Given: x + 2y - z = -4 x + 4y - 2z = -6 2x + 3y + z = D = D = D x = D x = D y = D z = D y = D z =

Example D = = = (4 – (-6)) – (2 – (-3)) + 2(-4 – (-4)) = = D = = = (4 – (-6)) – (2 – (-3)) + 2(-4 – (-4)) = = D x = = (-6) = (-4)(4 + 6) – (-6)(2 + 3) + (3)(-4 + 4) = = D x = = (-6) = (-4)(4 + 6) – (-6)(2 + 3) + (3)(-4 + 4) = = x = = x = = -2 5

Your Turn Calculate Calculate 1.y = D y / D 2.Z = D z / D