Solutions to Math 163 Online Problems Supplemental Lessons on Gaussian
Elimination
1. Change to the augmented matrix and solve by back substitution.
2x + 3y – 4z = 19
2y – 4z = -2
3z = 9
2 3 4 19
0 2 4 2 3z = 9, z = 3
0 0 3 9
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2y –4z = -2, 2y –4(3) = -2, 2y –12 = -2, 2y = 10, y = 5
2x +3y –4z = 19, 2x +3(5) –4(3) = 19, 2x +15 –12 = 19, 2x +3 = 19, 2x = 16,
x=8
x = 8, y = 5, and z = 3
2 1 1
2. Solve the matrix by back substitution:
0 5 15
5y = -15, y = -3, 2x + y = -1, 2x –3 = -1, 2x = 2, x = 1
x = 1, and y = -3
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3.
1
0
0
0
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Use back substitution to solve the augmented matrix in x,y,z,w:
1 1 1 19
2 3 4 0
0 1 0 8
0 0 2 4
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2w = 4, w = 2
z + 0w = 8, z = 8
2y + 3z – 4w =0, 2y +3(8) –4(2) = 0, 2y +24 –12 = 0, 2y +12 =0, 2y = -12, y=-6
x + y + z + w = 19, x +(-6) + 8 + 2 = 19, x + 4 = 19, x = 15
x = 15, y = -6, z = 8, w = 2
4. Solve the system of equations by forming the augmented matrix and
simplifying by using gaussian elimination method and back substitution.
x  3 y  7
3x  10 y  z  23
4 x  10 y  2 z  24
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3
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3 0 7
10 1 23
10 2 24
3 9 0 21
3R1+R2 = 3 10 1
0 1 1
1 3
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0 1
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0 2
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23
2
4 12 0 28
-4R1+R3 = 4 10 2 24
0
2 2 4
0 7
1 2
2 4
0 2 2 4
-2R2+R3 = 0
0
2
0
2
0
4
0
Since we lost our variables but the equation 0 = 0 is true, our solution is all
numbers such that z = any number,
Y + z = 2, y = 2-z
X –3y = -7, x –3(2-z) = -7, x –6 + 3z = -7, x = -1 + 3z
Z = any number, x = 3z – 1, y = 2 - z
5. Solve the system of equations by forming the augmented matrix and
simplifying by using gaussian elimination method and back substitution.
3x  3 y  6z  12
1x  4 z  0
2 x  4 y  2 z  17
First I would multiply equation 2 by –1 and make it equation 1
1 0 4
0
3 3 6 12
2 4 2 17
3 0 12 0
-3R1 + R2 = 3 3 6 12 which could be simplified by dividing by 3
0 3 6 12
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0 1 -2 -4
2 0 8
0
-2R1 + R3 = 2 4 2 17
0 4 10 17
1 0
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0 1
The updated matrix is now: M
M
0 4
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0
2 4
10 17
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0 4
Now –4R2+R3 = 0
0
4
0
8
16
10 17
2
1
1 0
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0 1
The final matrix form is M
M
0 0
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4
0
2 4
2 1
Back Substitution: -2z = -1, z = ½
Y –2z = -4, y –2(1/2) = -4, y –1 = -4, y = -3
X +4z = 0, x + 4(1/2) = 0 , x +2 = 0 , x = -2
X = -2, y = -3, z = 1/2
6. Solve the system of equations by forming the augmented matrix and
simplifying by using gaussian elimination method and back substitution.
2x  2 y  z  8
x  y  z  7
12 x  12 y  6z  49
Multiply equation 1 by –1 and move to equation 1 before
1
 1 1 7
2
2 1
8
12 12 6 49
2 2 2 14
12 12 12
-2R1 + R2 = 2 2 1 8 12R1 + R3 = 12 12 6
0
0 1 22
0
0
6
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forming the matrix.
84
49
133
Which gives us the matrix form of
1 1 1
7
0 0 1 22 Back substitution gives us two incompatible statements:
0 0 6 133
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6z = -133, z = -22
1
6
-z=22, z=-22
Since z cannot be two different numbers we must conclude there are no
valid solutions for this matrix.