Slide 1

GAUSS ELIMINATION

Problem no. 110W + 2X +4Y + 3Z = 385W + 15X + 3Y 8Z = 76 28W + 3X + 20Y + 7Z = 1022W + 2X + 8Y + 20Z = 110

1024338 -5/10 R1+R2R25153876 -8/10 R1+R3R383207102-2/10 R1+R4R422820110

1024338-1.4/14 R2+R3R301416.557 -1.6/14 R2+R4R401.416.84.671.601.67.219.4102.4

12213801416.557 -7.086/16.7R3+R4R40016.84.665.9007.219.495.89

102 4338015257 001165.9000267.92

ANSWER:W=1, X=2, Y=3 & Z= 4

Problem no. 2a + 2b 3c + 4d = 483a + 4b - 3d = 172a - b + 2c + d = 24 4b - 7c + 2d = 24

1 2 -3 4 48 3 4 0 -3 17 r2 - 3r1 2 -1 2 1 24 r3 - 2r1 0 4 -7 2 24

1 2 -3 4 48 0 -2 9 -15 -127 0 -5 8 -7 -72 2r3 5r2 0 4 -7 2 24 r4 + 2r2

1 2 -3 4 48 0 -2 9 -15 -127 0 0 -29 61 491 0 0 11 -28 -230 29r4 + 11r3

1 2 -3 4 48 0 -2 9 -15 -127 0 0 -29 61 491 0 0 0 -141 -1269

In equation 4:-141d = -1269d = 9In equation 3 subst. d:-29c + 61d = 491-29c + 61(9) = 491c = 2In equation 2 subst. d & c:-2b + 9c -15d = -127-2b + 9(2) -15(9) = -127b = 5In equation 1 subst. b ,c & da + 2b -3c + 4d = 48a + 2(5) -3(2) + 4(9) = 48a = 8ANSWER: ( 8, 5, 2, 9 )

Problem no. 3 Solve the following system by using the Gauss- Elimination method.

X + Y + Z = 52X + 3Y + 5Z = 84X + 5Z = 2

Solution: We will perform row operations on the augmented matrix of the system until we obtain amatrix in reduced row echelon form. 1 1 1 5 2 3 5 8 4 0 5 2

R22R1

1 1 1 5 0 1 3 2 4 0 5 2

R34R1

1 1 1 5 0 1 3 2 0 4 1 18

R3+4R2

1 1 1 5 0 1 3 2 0 0 13 26

1/13R3

1 1 1 5 0 1 3 2 0 0 1 2

Z = -2

Y + 3Z = -2Y + 3(-2) = -2Y + (-6) = -2Y = -2 + 6Y = 4

X + Y + Z = 5X + (4) + (-2) = 5X = 5 -4 + 2X = 3

Problem no. 46a + 8b 6c + 2d = 224a 2b + 10d = 5210b 6c +4d = 2614a + 4b 4c = 24

Solution:8 -6 2 4 -2 0 100 10 -6 414 4 -4 0

22522624

6R 4R

8 -6 20 -44 24 520 10 -6 414 4 -6 0 22224 26 24

6R - 14R

8 -6 20 -44 24 52 0 10 -6 40 -88 60 -28 22224 26-164

44R + 10R

8 -6 2 0 -44 24 520 0 -24 6960 -88 60 -28R - 2R222243384-164

8 -6 2 0 -44 24 520 0 -24 6960 0 12 -132 222243384-612

2R + R

8 -6 20 -44 24 520 0 -24 520 0 0 4322222433842160

Well get;

d= 2160/432b= 224 52(5) 24(4) d= 5 -44b= 3c= 3384 696(5) a= 22- 2(5) + 6(4) 8(3)-24 6 c= 4a= 2

Problem no. 5Solve the following system by using the Gauss-Elimination method.

A + B + 2C = 12A B + D = 2A B C 2D = 42A B + 2C D = 0

Solution: We will perform row operations on the augmented matrix of the system until we obtain amatrix in reduced row echelon form. 1 1 2 0 1 2 1 0 1 2 1 1 1 2 4 2 1 2 1 0

R22R1

1 1 2 0 1 0 3 4 1 4 1 1 1 2 4 2 1 2 1 0

R3R1

1 1 2 0 1 0 3 4 1 4 0 2 3 2 3 2 1 2 1 0

R42R1

1 1 2 0 1 0 3 4 1 4 0 2 3 2 3 0 3 2 1 2

R4R2

1 1 2 0 1 0 3 4 1 4 0 2 3 2 3 0 0 2 2 2

R2R3

1 1 2 0 1 0 2 3 2 3 0 3 4 1 4 0 0 2 2 2

R3+3R2

1 1 2 0 1 0 1 3/2 1 3/2 0 0 1/2 4 17/2 0 0 2 2 2

2R3

1 1 2 0 1 0 1 3/2 1 3/2 0 0 1 8 17 0 0 2 2 2

R42R3

1 1 2 0 1 0 1 3/2 1 3/2 0 0 1 8 17 0 0 0 18 36

-1/18R4

1 1 2 0 1 0 1 3/2 1 3/2 0 0 1 8 17 0 0 0 1 2

D = -2

C + 8D = -17C + 8(-2) = -17C 16 = -17C = -17 + 16C = -1

B + 3/2C + D = -3/2B + 3/2(-1) + (-2) = -3/2B + (-3/2) + (-2) = -3/2B = -3/2 +3/2 + 2B = 2

A + B + 2C = 1A + (2) + 2(-1) = 1A + (2) + (-2) = 1A = 1 -2 + 2A = 1

Problem no. 6

23-105-110205r1(-3)+r3

23-105-1100-4-98r2(4/5)+r3

23-105-1100-4-98r2(4/5)+r3

23-105-110r2(1/5)00-49/516r3(-5/49)

23-101-1/52001-80/49

Gauss Elimination

Solve the following system using Gaussian Elimination:

The augmented matrix which represent this system is

First is to produce zeros below the first entry in the first column

The third row of the final matrix translates into

The solution of this system is therefore

Problem no. 8Solve the following system using Gaussian Elimination:

The augmented matrix which represent this system is

First is to produce zeros below the first entry in the first column

The third row of the final matrix translates into

The solution of this system is therefore

Problem no. 9Using gaussian elimination method:

2x + 3y + z = 1x + y + z = 3 3x + 4y + 2z = 4

SOLUTION:2 3 1 1 1 1 1 3 r1 r2 3 4 2 4 1 1 1 32 3 1 1 r2 2r1 3 4 2 4 r3 3r1

1 1 1 3 0 1 -1 -5 0 1 -1 -5 r3 r1

1 1 1 3 0 1 -1 -5 0 0 -2 -8

-2z = -8 z = 4 y z = -5 y 4 = -5 y = -1 x + y + z = 3 x + (-1) + 4 = 3 x = 0 ( x, y, z ) = ( 0, -1, 4 )

Problem no. 10

x + -3y + z = 42x + -8y +8z = -2-6x + 3y + -15z = 9

Augmented matrix: Now use -2R1+R2=R2: 1 -3 1 4 1 -3 1 4 2 -8 8 -2 0 -2 6 -10-6 3 -15 9 -6 3 -15 9

Then use 6R1+R3=R3: Then use (1/2)R2=R2, and (1/3)R3=R3:1 -3 1 4 1 -3 1 40 -2 6 -10 0 -1 3 -50 -15 -9 33 0 -5 -3 11

Then use -5R2+R3=R3: Now turn the matrix into an equation:1 -3 1 4 x + -3y +z = 40 -1 3 -5 -y + 3z = -50 0 -18 36 -18z = 36

From this point, we can easily solve for (x, y, z) by substitution:From the last equation solve for z by dividing (36/-18), and we can get z=-2.Then plug the -2 into the z of the 2nd equation and the new equation is -y -6 = -5.Then move the -6 to the other side to get -y = 1, so at the end we get y=-1.Once we have the values for the y and the z, just plug them into the first equation and solve for x.So the equation looks like this, x +3 -2 = 4, which is the same as x +1 = 4.Then move the 1 to the other side and we get x=3.The results for the Gaussian elimination example are: x=3y=-1z=-2

Problem no. 112x - 2y + z = 3 3x + y - z = 7 x - 3y + 2z = 0

2 -2 1 3 1 -3 2 0 3 1 -1 7 R3 R1 3 1 -1 7 -3R1 + R2 R2 1 -3 2 0 2 -2 1 3 -2R1 + R3 R3

1 -3 2 0 1 -3 2 0 0 10 -7 7 (1/10)R2R2 0 1 -7/10 7/10 3R2 + R1 R1 0 4 -3 3 0 4 -3 3 -4R2 + R3R3

1 0 - 1/10 21/10 1 0 -1/10 21/100 1 -7/10 7/10 3R2 + R1 R1 0 1 -7/10 7/10 -5R3 R30 4 -3 3 -4R2 + R3R3 0 0 -1/5 1/5

1 0 -1/10 21/10 1 0 0 20 1 -7/10 7/10 (1/10)R3 + R1 R1 0 1 0 0 Reduced row-echelon form0 0 1 -1 (7/10)R3 + R2 R2 0 0 1 -1

Answerx1 = 2, x2 = 0, x3 = -1

Problem no. 13Problem:

Solve the following system using Gauss Elimination:x + 2y + z = 82x + y z = 1x + y 2z = -3

1218 -2r1 + r221-1111-2-3

1218 -r1 + r30-3 -3-1511-2-3

1218 -1/3 r2 0-3-3-150-1-3-11

1218 r 2+ r301150-1-3-11

1218 - r3011500-2-6

121801150013

z = 3 x + 2y + z = 8y + z = 5 x + 2(2) + 3 = 8y + 3 =5 x = 1y = 2

The solution of this system is therefore (x, y, z) = (1, 2, 3)

Problem no. 15

Solution:

2 -3 5 16 2 -3 5 16 2 -3 5 16 1 -2 3 10 0 - 2 0 - 2 4 2 12 0 7 -3 -4 0 0 4 24

Problem no. 16 2x+4y+7z=15 8x+2y+5z=45 2y+z=25

2 4 7 15 8 2 5 45 0 2 1 25

Solution

2 4 7 15 r2-4r 2 4 7 15 8 2 5 45 0 -14 -23 -15 0 2 1 25 0 2 1 25

67

2 4 7 15 7r3+r2 2 4 7 15 0 -14 -23 -15 0 -14 -23 -15 0 2 1 25 0 0 -16 160

Problem no. 17Solve the following system using Gaussian elimination:4x+y-z=37x+3y+z=51/2x+y+2z=7

71

Solution:

Add (-7 * row1) to row2Divide row1 by 411/4-1/43/405/411/4-1/41/2127

11/4-1/43/405/411/4-1/407/817/853/8

11/4-1/43/40111/5-1/507/817/853/8

Divide row2 by 5/4Add (-1/2 * row1) to row311/4-1/43/473151/2127

Add (-7/8 * row2) to row3Equation1:1/5z = 34/5z = 34Equation2:y+11/5z = -1/5Substitute the value of z;y+(11/5)(34) = -1/5y = -75Equation3:x+1/4y-1/4z = Substitute the value of y and z;x = 28

11/4-1/43/40111/5-1/5001/534/5

Problem no. 18x + y + z = 62x y +z =3x + z = 4

(-2)r1+r2(-1)r1+r3(-1)r2(-1)r3(-3)r2+r3

z = 3y = 2x + 2 + 3 = 6x = 1(x, y, z) = (1, 2, 3)

Problem no. 19

Problem no. 20Solve for the values of x, y, and z.2x + y 3z = -33x 4y + 2z = -7X + 3y 4z = 1

2 1 -3 -33 -4 2 -71 3 -4 12 1 -3 -30 -11 13 -51 3 -4 1-3R1 + 2R2 2 1 -3 -3 0 -11 13 -5 0 -5 5 -5-2R3 + R1 2 1 -3 -3 0 -11 13 -5 0 0 10 30-11R3 + 5R2 2 1 -3 -3 0 -11 13 -5 0 0 1 3