1
4.1 Solve the following system of equations using the Gauss elimination method:
Solution
Step 1: Write the system of equations in matrix form:
Using the first element of the matrix as a pivot, and .
2x1x2x3
–+1=
x12x2x3
++ 8=
x–1x2x3
–+5–=
21 1–
121
1–11–
x1
x2
x3
1
8
5–
=
a11 2=
m21
a21
a11
——–1
2
—–0.5===
1
4.2 Given the system of equations , where , , and ,
determine the solution using the Gauss elimination method.
Solution
Following the same procedure as in Problem 4.1, the row elimination operations proceed as follows:
Step 1: Multiply the first row by , subtract the result from the second row, and
replace the second row with the final result:
a[]x[] b[]=
a
22–1
32 5–
1–23
=
x
x1
x2
x3
=
b
10
16–
8
=
m21
a21
a11
——–3
2
—–1.5===
22–1
0 5 6.5–
1–23
x1
x2
x3
10
31–
8
=
m31
1
4.3 Consider the following system of two linear equations: .
(a) Solve the system with the Gauss elimination method using rounding with four significant figures.
(b) Switch the order of the equations, and solve the system with the Gauss elimination method using
rounding with four significant figures.
Check the answers by substituting the solution back in the equations.
Solution
(a) First, write the system in matrix form: . Next, apply Gaussian elimina-
tion with rounding. With , we have . Note that
(b) Switching the order of the equations, we have: . Applying Gaussian
elimination with rounding to 4 significant figures yields: and
0.0003x11.566x2
+1.569=
0.3454x12.436x2
–1.018=
0.0003 1.566
0.3454 2.436–
x1
x2
1.569
1.018
=
m21
a21
a11
——–0.3454
0.0003
—————– 1 1 5 1== =
0.0003 1.566
0.0001 1804–
x1
x2
1.569
1805–
=
0.3454 2.436–
0.0003 1.566
x1
x2
1.018
1.569
=
m21
a21
a11
——–0.0003
0.3454
—————– 0 . 0008686== =
1
4.4 Solve the following system of equations using the Gauss elimination method.
Solution
The system of equations in matrix form is:
2x1x2x3
–2x4
++0=
x12x2
–x34x4
–+3=
3x1x2
–2x3
–x4
–3–=
x–12x2x32x4
–++ 13=
21 1–2
12–14–
31–2–1–
1–212–
x1
x2
x3
x4
0
3
3–
13
=
1
4.5 Solve the following system of equations with the Gauss elimination method.
Solution
The system of equations in matrix form is:
With , , and , the first pass with
Gaussian elimination yields:
2x1x2x3
–4x4
++19=
x–12x2
–x32x4
++ 3–=
2x14x22x3x4
+++ 25=
x1
–x2x3
–2x4
–+5–=
21 1–4
1–2–12
2421
1–11–2–
x1
x2
x3
x4
19
3–
25
5–
=
m21
a21
a11
——–1–
2
—––0.5–===
m31
a31
a11
——–2
2
—–1===
m41
a41
a11
——–1–
2
—––0.5–===
m43
1
4.6 Solve the following system of equations using the Gauss–Jordan method.
Solution
First, form the augmented matrix, including the right hand side column vector: .
Step 1: The pivot element is . Normalize the second row by dividing it by 4:
The pivot element is now 1.
Use the first (pivot) row to eliminate the entries below the pivot element:
=
4x1x22x3
++ 21=
2x12x2
–2x3
+8=
x12x2
–4x3
+16=
41221
22–28
12–416
a11 4=
10.250.55.25
22–28
12–416
a11
10.250.55.25
22–28
12–416
←
←
21 0.25 0.5 5.25
–
11 0.25 0.5 5.25
–
10.250.55.25
02.5– 1 2.5–
02.25– 3.5 10.75
a33
1
4.7 Solve the system of equations given in Problem 4.2 using the Gauss–Jordan method.
Solution
First, form the augmented matrix, including the right hand side column vector: .
Step 1: The pivot element is . Normalize the second row by dividing it by 4:
The pivot element is now 1.
Use the first (pivot) row to eliminate the entries below the pivot element:
=
22–110
32 5–16–
1–238
a11 2=
11–0.55
32 5–16–
1–238
a11
11–0.55
32 5–16–
1–238
←
←
311–0.55
–
1–()
11–0.55
–
11–0.55
0 5 6.5–31–
01 3.5 13
2
1
4.8 Given the system of equations , where , , and , deter-
mine the solution using the Gauss–Jordan method.
Solution
First, form the augmented matrix, including the right hand side column vector: .
Step 1: The pivot element is . Normalize the first row by dividing it by 4:
Use the first (pivot) row to eliminate the entries below the pivot element:
=
a[]x[] b[]=
a
45 2–
25–2
62 4
=
x
x1
x2
x3
=
b
6–
24
30
=
45 2–6–
25–224
62 430
a11 4=
11.25 0.5–1.5–
25–224
62 4 30
11.25 0.5–1.5–
25–224
62 4 30
←
←
211.25 0.5–1.5–
–
611.25 0.5–1.5–
–
1 1.25 0.5–1.5–
07.5–327
05.5–739
2
1
4.9 Solve the following system of equations with the Gauss–Jordan elimination method.
Solution
First, form the augmented matrix, including the right hand side column vector:
Step 1: The pivot element is . Normalize the first row by dividing it by 4:
4x13x22x3x4
+++ 17=
2x1x2
–2x34x4
–+11=
x12x22x–3x4
–+8=
2x1
–4x25x3x4
–++ 15=
432117
21–24–11
12 2–1–8
2–451–15
a11 4=
1 0.75 0.5 0.25 4.25
21–24–11
12 2–1–8
2–45 1–15
2
=
Step 3: Normalize the third row by dividing it by -2:
Step 4: Normalize the fourth row by dividing it by -24.75:
The pivot element is now 1. Use the third (pivot) row to eliminate the entries above the pivot element:
1 0.75 0.5 0.25 4.25
0 1 0.4–1.81–
01.25 2.5–1.25–3.75
05.5 6 0.5– 23.5
←
←
←
←
0.75 0 1 0.4–1.81–
–
1.25 0 1 0.4–1.81–
–
5.5 01 0.4–1.81–
–
10 0.8 1.1–5
01 0.4–1.81–
00 2–3.5–5
0 0 8.2 10.4–29
10 0.8 1.1–5
01 0.4–1.81–
100 2.5–7
010 2.5 2–
0 0 1 1.75 2.5–
000 1 2–
a44
100 2.5–7
←
2.5–()
0001 2–
–
1000 2
1
4.10 Determine the LU decomposition of the matrix using the Gauss elimination procedure.
Solution
LU decomposition using Gausian elimination transforms the above matrix into a lower triangular matrix
multiplied by an upper triangular matrix . is the upper triangular matrix that would normally
result after applying Gaussian elimination to the given matrix. consists of the multipliers that are used
a
246
351
62–2
=
L[]
U[]
U[]
L[]
1
4.11 Determine the LU decomposition of the matrix using Crout’s method.
Solution
The LU decomposition is done by following the procedure described in Section 4.5.2.
a
61224
21129
41024
=
1
4.12 Solve the following system with LU decomposition using Crout’s method.
Solution
First the LU decomposition of the matrix of coefficients is done by following the procedure described in
Section 4.5.2.
26–6
37–13
2–211–
x
y
z
2
13–
21
=
2
Next is substituted in Eq. (4.22):
x‘
y‘
z‘
13–3
012
001
x
y
z
1
8–
3–
=
1
4.13 Find the inverse of the matrix using the Gauss–Jordan method.
Solution
First form the augmented matrix with the identity matrix:
Step 1: The pivot element is . Normalize the first row by dividing it by :
Step 2: The pivot element is . Normalize the second row by dividing it by :
10 12 0
028
248
10120100
0 2 8010
2 4 8001
a11 10=
10
a22 2=
2
1 1.2 0 0.1 0 0
01400.50
01.66 0.2–01
2
Step 3: The pivot element is . Normalize the third row by dividing it by :
a33 0.4–=
0.4–
1 0 4.8– 0.1 0.6–0
01 4 0 0.5 0
0 0 1 0.5 2 2.5–
1
4.14 Given the matrix , determine the inverse of using the Gauss–Jordan method.
Solution
First form the augmented matrix with the identity matrix:
Step 1: The pivot element is . Normalize the first row by dividing it by :
a
1–22
0 2 0.5–
0.5 1 2–
=
a[]
1–22100
0 2 0.5–010
0.5 1 2–001
a11 1–=
1–
12–2–1–00
0 2 0.5–010
0.5 1 2–001
2
Step 3: The pivot element is . Normalize the third row by dividing it by :
a33 0.5–=
0.5–
1 0 2.5–1–10
01 0.25–00.50
00 1 1–22–