Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 10, Problem 3P

(a) Solve the following system of equations by LU decomposition without pivoting

8 x 1 + 4 x 2 x 3 = 11 2 x 1 + 5 x 2 + x 3 = 4 2 x 1 x 2 + 6 x 3 = 7

(b) Determine the matrix inverse. Check your results by verifying that [ A ] [ A ] 1 = [ l ] .

(a)

Expert Solution
Check Mark
To determine

To calculate: The solution of the system of equations given below by LU decomposition without pivoting.

8x1+4x2x3=112x1+5x2+x3=42x1x2+6x3=7

Answer to Problem 3P

Solution:

The solution of the system of equations is x1=1, x2=1 and x3=1.

Explanation of Solution

Given:

The system of equations,

8x1+4x2x3=112x1+5x2+x3=42x1x2+6x3=7

Formula used:

(1) The forward substitution equations for L can be expressed as,

[L][D]=[B]

(2) The backward substitution equation for U can be expressed as,

[U][X]=[D]

Calculation:

Consider the system of equations,

8x1+4x2x3=11…… (1)

2x1+5x2+x3=4…… (2)

2x1x2+6x3=7 …… (3)

The coefficient a21 is eliminated by multiplying equation (1) by f21= a21a11,

f21=28

And subtracting the result from equation (2).

Thus, multiply equation (1) by (28),

8(28)x1+4(28)x2(28)x3=11(28)2x1x2+0.25x3=228

Now subtract this equation from equation (2),

2x1+5x2+x3+2x1+x20.25x3=42286x2+0.75x3=1.25 …… (4)

The coefficient a31 is eliminated by multiplying equation (1) by f31= a31a11,

f31=28

And subtracting the result from equation (3).

Thus, multiply equation (1) by (28),

8(28)x1+4(28)x2(28)x3=11(28)2x1+x20.25x3=228

Now subtract this equation from equation (3),

2x1x2+6x32x1x2+0.25x3=72282x2+6.25x3=4.25 …… (5)

Now the set of equations is,

8x1+4x2x3=116x2+0.75x3=1.252x2+6.25x3=4.25

The factors f21 and f31 can be stored in a21 and a31.

[8410.2560.750.2526.25]

The coefficient a32 is eliminated by multiplying equation (4) by f32= a32a22,

f32=26=0.33333

And subtracting the result from equation (5). Thus, multiply equation (4) by (26),

6(26)x2+0.75(26)x3=1.25(26)2x20.25x3=0.41666

Now, subtract this equation from equation (5),

2x2+6.25x3+2x2+0.25x3=4.25+0.416666.5x3=4.6666 …… (6)

The factor f32 can be stored in a32. Thus, the matrix obtained is:

[8410.2560.750.250.333336.5]

Therefore, the LU decomposition is

[L]=[1000.25100.250.333331] [U]=[841060.75006.5]

Now, to find the solution of the given system:

The forward substitution equations for L can be expressed as,

[L][D]=[B]

[1000.25100.250.333331]{d1d2d3}={1147}

Solve for d1,

d1=11

Solve for d2,

0.25d1+d2=4d2=4+0.25d1d2=4+0.25(11)d2=6.75

Solve for d3,

0.25d10.33333d2+d3=7d3=70.25d1+0.33333d2d3=70.25(11)+0.33333(6.75)d3=6.5

Thus, d1=11, d2=6.75 and d3=6.5. That is, [D]T=[116.756.5].

Now, perform backward substitution:

[U][X]=[D][841060.75006.5][x1x2x3]=[116.756.5]

Solve for x3,

6.5x3=6.5x3=6.56.5x3=1

Solve for x2,

6x2+0.75x3=6.756x2=6.750.75(1)x2=66x2=1

Solve for x1,

8x1+4x2x3=118x1=114(1)+(1)x1=88x1=1

Thus, x1=1, x2=1 and x3=1.

(b)

Expert Solution
Check Mark
To determine

To calculate: The matrix inverse for given system of equations and check the result by verifying that [A][A]1=[I].

8x1+4x2x3=112x1+5x2+x3=42x1x2+6x3=7

Answer to Problem 3P

Solution:

The matrix inverse is A1=[0.0993590.0737180.0288460.0448720.1602560.019230.0256410.0512820.153846].

Explanation of Solution

Given:

The system of equations,

8x1+4x2x3=112x1+5x2+x3=42x1x2+6x3=7

And the LU decomposition is

[L]=[1000.25100.250.333331] [U]=[841060.75006.5]

Formula used:

(1) The forward substitution equations for L can be expressed as,

[L][D]=[B]

(2) The backward substitution equation for U can be expressed as,

[U][X]=[D]

Calculation:

Consider the given system of equations:

8x1+4x2x3=112x1+5x2+x3=42x1x2+6x3=7

The matrix [A] is:

[A]=[841251216]

The lower and upper triangular matrix after decomposition are given as:

[A]=[L][U]=[1000.25100.250.333331][841060.75006.5]

The first column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the first row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

[L][D]=[B]

Where,

[B]=[100]

Determine D by substituting L and B as shown below,

[1000.25100.250.333331][d1d2d3]=[100]

Solve for d1,

d1=1

Solve for d2,

0.25d1+d2=0d2=0.25d1d2=0.25(1)d2=0.25

Solve for d3,

0.25d10.33333d2+d3=0d3=0.25d1+0.33333d2d3=0.25(1)+0.33333(0.25)d3=0.16667

Hence, the values obtained are d1=1, d2=0.25, and d3=0.16667.

Solve with forward substitution of DT=[10.250.16667],

This vector can be used as right-hand side vector of equation,

[U][X]=[D][841060.75006.5][x1x2x3]=[10.250.16667]

Solve the above matrix by back substitution, which gives the first column of the inverse matrix as:

[X]=[0.0993590.0448720.025641]

Similarly, the second column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the second row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

[L][D]=[B]

Where,

[B]=[010]

Determine D by substituting L and B as shown below,

[1000.25100.250.333331][d1d2d3]=[010]

Solve for d1,

d1=0

Solve for d2,

0.25d1+d2=1d2=1+0.25d1d2=1+0.25(0)d2=1

Solve for d3,

0.25d10.33333d2+d3=0d3=0.25d1+0.33333d2d3=0.25(0)+0.33333(1)d3=0.33333

Hence, the values obtained are d1=0, d2=1, and d3=0.33333.

Solve with forward substitution of DT=[010.33333],

This vector can be used as right-hand side vector of equation,

[U][X]=[D][841060.75006.5][x1x2x3]=[010.33333]

Solve the above matrix by back substitution, which gives the second column of the inverse matrix as:

[X]=[0.0737180.1602560.051282]

Similarly, the third column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the third row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

[L][D]=[B]

Where,

[B]=[001]

Determine D by substituting L and B as shown below,

[1000.25100.250.333331][d1d2d3]=[001]

Solve for d1,

d1=0

Solve for d2,

0.25d1+d2=0d2=0+0.25d1d2=0+0.25(0)d2=0

Solve for d3,

0.25d10.33333d2+d3=1d3=10.25d1+0.33333d2d3=10.25(0)+0.33333(0)d3=1

Hence, the values obtained are d1=0, d2=0 and d3=1.

Solve with forward substitution of DT=[001],

This vector can be used as right-hand side vector of equation,

[U][X]=[D][841060.75006.5][x1x2x3]=[001]

Solve the above matrix by back substitution, which gives the third column of the inverse matrix as:

[X]=[0.0288460.019230.153846]

Thus, the inverse matrix is:

A1=[0.0993590.0737180.0288460.0448720.1602560.019230.0256410.0512820.153846]

Now, check the result obtained.

[A][A]1=[841251216][0.0993590.0737180.0288460.0448720.1602560.019230.0256410.0512820.153846]=[1.00000.00000.00000.00001.00000.0000001.0000]=[100010001]=[I]

Hence, verified.

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