solutions manual to accompany with applicationssolutions manual to accompany fundamentals of matrix...

SOLUTIONS MANUAL TO ACCOMPANYFUNDAMENTALS OF MATRIX ANALYSISWITH APPLICATIONS

SOLUTIONS MANUALTO ACCOMPANYFUNDAMENTALS OFMATRIX ANALYSISWITH APPLICATIONS

EDWARD BARRY SAFFVanderbilt University

ARTHUR DAVID SNIDERUniversity of South Florida

Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

Published by John Wiley & Sons, Inc., Hoboken, New Jersey.Published simultaneously in Canada.

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ACKNOWLEDGMENTS

Special thanks to Viktor Maymeskul, Samuel Garrett, Rares Rasdeaconu,and Oleksandr Vlasiuk for contributions to this work.

CONTENTS

1 Systems of Linear Algebraic Equations 1

1.1 Linear Algebraic Equations, 11.2 Matrix Representation of Linear Systems and the

Gauss-Jordan Algorithm, 121.3 The Complete Gauss Elimination Algorithm, 171.4 Echelon Form and Rank, 261.5 Computational Considerations, 34

2 Matrix Algebra 39

2.1 Matrix Multiplication, 392.2 Some Useful Applications of Matrix Operators, 482.3 The Inverse and the Transpose, 542.4 Determinants, 612.5 Three Important Determinant Rules, 68

PART I REVIEW PROBLEMS FOR PART I 79

3 Vector Spaces 89

3.1 General Spaces, Subspaces, and Spans, 89

viii CONTENTS

3.2 Linear Dependence, 933.3 Bases, Dimension, and Rank, 97

4 Orthogonality 105

4.1 Orthogonal Vectors and the Gram-Schmidt Algorithm, 1054.2 Orthogonal Matrices, 1154.3 Least Squares, 1234.4 Function Spaces, 133

PART II REVIEW PROBLEMS FOR PART II 140

5 Eigenvectors and Eigenvalues 144

5.1 Eigenvector Basics, 1445.2 Calculating Eigenvalues and Eigenvectors, 1555.3 Symmetric and Hermitian Matrices, 168

6 Similarity 181

6.1 Similarity Transformations and Diagonalizability, 1816.2 Principal Axes and Normal Modes, 1896.3 Schur Decomposition and Its Implications, 1986.4 The Singular Value Decomposition, 2126.5 The Power Method and the QR Algorithm, 217

7 Linear Systems of Differential Equations 221

7.1 First-Order Linear Systems, 2217.2 The Matrix Exponential Function, 2307.3 The Jordan Normal Form, 2367.4 Matrix Exponentiation via Generalized Eigenvectors, 246

PART III REVIEW PROBLEMS FOR PART III 255

1SYSTEMS OF LINEARALGEBRAIC EQUATIONS

1.1 LINEAR ALGEBRAIC EQUATIONS

1. (a) Solving the last equation for x4 yields x4 = 4/4 = 1. We substitutethis value of x4 into the third equation and solve for x3 to get

x3 + 2(1) = −1 ⇒ x3 = −3.

With these values of x3, the second equation becomes

3x2 + 2(−3) + (1) = 2 ⇒ 3x2 = 7 ⇒ x2 =73

.

For x1, we substitute the values of x2 and x4 into the first equation.

2x1 +

(73

)− (1) = −2 ⇒ 2x1 = −10

3⇒ x1 = −5

3.

Therefore, the answer is: x1 = −5/3, x2 = 7/3, x3 = −3, andx4 = 1.

Solutions Manual to Accompany Fundamentals of Matrix Analysis with Applications,First Edition. Edward Barry Saff and Arthur David Snider.© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.

2 SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS

(b) Solving the first equation for x1, we obtain x1 = 3. We nowsubstitute this value of x1 into the second equation and solve for x2.

−2(3)− 3x2 = −12 ⇒ −3x2 = −6 ⇒ x2 = 2.

The third equation then becomes

(3) + (2) + x3 = 5 ⇒ x3 = 0.

Therefore, the answer is x1 = 3, x2 = 2, and x3 = 0.

(c) From the third equation, we immediately get x4 = 1. This value,when substituted into the first equation, yields

−x3 + 2(1) = 1 ⇒ −x3 = −1 ⇒ x3 = 1.

From the fourth equation, we obtain

x2 + 2(1) + 3(1) = 5 ⇒ x2 = 0.

Finally, from the second equation, we conclude that

4x1 + 2(0) + (1) = −3 ⇒ x1 = −1.

So, the solution is x1 = −1, x2 = 0, x3 = 1, and x4 = 1.

(d) The third equation implies that x1 = 1. Then, the first equationsays

2(1) + x2 = 3 ⇒ x2 = 1.

We now substitute these values of x1 and x2 into the secondequation to get

3(1) + 2(1) + x3 = 6 ⇒ x3 = 1.

The solution to this problem is x1 = x2 = x3 = 1.

3. To eliminate x1 from the first and second equations, we subtract fromthe first equation the third equation multiplied by 3 and subtract from

1.1 LINEAR ALGEBRAIC EQUATIONS 3

the second equation the third equation multiplied by 2, resp. Thus weget an equivalent system

−x2 = −2−9x2 −x3 = −16

x1 +3x2 +x3 = 3.

From the first equation, we get x2 = 2. Making the back substitutionsinto the second and third equations yields

−9(2)− x3 = −16 ⇒ −x3 = 2 ⇒ x3 = −2

x1 + 3(2) + (−2) = 3 ⇒ x1 = −1.

The solution is x1 = −1, x2 = 2, and x3 = −2.

5. We eliminate x1 and x4 from the first and the the last equations bysubtracting from them the second equation. We also eliminate x1 fromthe third equation by subtracting from it twice the second equation.The new system is

x2 +x3 = 1x1 +x4 = 0

2x2 −x3 −x4 = 62x2 −x3 = 0

Next, we eliminate x2 from the third and fourth by subtracting fromthem the first equation multiplied by 2. This gives

x2 +x3 = 1x1 +x4 = 0

−3x3 −x4 = 4−3x3 = −2

We now can go with the back substitution. The last equation givesx3 = 2/3. With this value, we find x1 and x4 from the first and thirdequations, resp.


x2 +

(23

)= 1 ⇒ x2 =

13

−3

(23

)− x4 = 4 ⇒ −x4 = 6 ⇒ x4 = −6.

Finally, the second equation says x1 = −x4 = 6. so, the answer isx1 = 6, x2 = 1/3, x3 = 2/3, and x4 = −6.

7. To eliminate x1 from the second equation, we multiply the firstequation by 0.987/0.123 and subtract the result from the second one.Thus, we get

0.123x1 +0.456x2 = 0.789−3.005x2 = −6.010

From the second equation, we get x2 = (−6.010)/(−3.005) = 2.000.Substituting this value into the first equation, we get

0.123x1 + 0.456(2) = 0.789 ⇒ 0.123x1 = −0.123

⇒ x1 = −1.000

The solution, rounded to three decimal places, is x1 = −1.000, x2 =

2.000. The number of arithmetic operations required is 3 divisions, 3multiplications, and 3 additions.

9. Following the notations in Problem 8, the coefficients in Problem 7are

a = 0.123, b = 0.456, c = 0.789,

d = 0.987, e = 0.654, f = 0.321.

Applying the formulas given in Problem 8, we obtain

x =(0.789)(0.654)− (0.456)(0.321)(0.123)(0.654)− (0.456)(0.987)

= −1.000;

y =(0.123)(0.321)− (0.789)(0.987)(0.123)(0.654)− (0.456)(0.987)

= 2.000.

The number of arithmetic operations required is 2 divisions, 6 multi-plications, and 6 additions.


11. To eliminate x1 from the second equation, we subtract from it the firstequation multiplied by 0.987/0.123. Similarly, we eliminate x1 fromthe third equation using the factor of 0.333/0.123, and obtain

0.123x1 +0.456x2 +0.789x3 = 0.111−3.005x2 −6.010x3 = −0.446−1.790x2 −2.913x3 = 0.587

We subtract from the third equation the second one multiplied by(1.790/3.005):

0.123x1 +0.456x2 +0.789x3 = 0.111−3.005x2 −6.010x3 = −0.446

0.667x3 = 0.853

Going from the third equation up, using the back substitution we findthat

x3 =0.8530.667

= 1.279;

x2 =−0.446 + 6.010(1.279)

−3.005= −2.410;

x1 =0.111 − 0.456(−2.410)− 0.789(1.279)

0.123= 1.633.

The number of arithmetic operations required is 6 divisions, 11 multi-plications, and 11 additions.

13. (a) From the third equation we find that x3 = 5.16/1.42 ≈ 3.6338(rounded to four decimal places). Substituting this value into thesecond equation, we get

x2 =1.11 − 1.34(3.6338)

2.73≈ −1.3770

Finally, we use the values of x2 and x3 to find x1 from the firstequation.

x1 =−4.22 − 7.29(−1.3770) + 3.21(3.6338)

1.23≈ 14.2137.


Rounding the results to two decimal places, the answer is

x1 = 14.21, x2 = −1.38, x3 = 3.63.

The total number of arithmetic operations required is 3 divisions,3 multiplications, and 3 additions.

(b) We start from the fourth equation to find x4. Substituting itsvalue into the third equation, we evaluate x3, and so on. Thesecomputations give

x4 =−1

0.250= −4.0000;

x3 =1 − 0.888(−4)

0.999= 4.5565;

x2 =−1 − 0.222(4.5565)− 0.333(−4)

0.111= −6.1220;

x1 =1 − 0.333(−6.1220)− 0.250(4.5565)− 0.200(−4)

0.500= 5.3990.

Rounding the results to three decimal places yields

x1 = 5.399, x2 = −6.122, x3 = 4.557, x4 = −4.000.

The number of arithmetic operations required is 4 divisions, 6multiplications, and 6 additions.

15. No, the suggested procedure is inconsistent with the Gauss eliminationrules. First of all, these rules require, on each step, the eliminationof one of the variables from all but one equations. One can do somepreliminary steps adding to an equation a multiple of another equation,but on each such step the new system must be considered.

In the problem, let’s follow steps. Subtracting the second equationfrom the first, we get a new system:

−x1 +x2 +x3 = 02x1 +x2 +x3 = 6

x1 +x2 +3x3 = 6


Subtracting the third equation from the second, yields a new system:

−x1 +x2 +x3 = 0x1 −2x3 = 0x1 +x2 +3x3 = 6

Performing the last step suggested in the problem, we obtain

−x1 +x2 +x3 = 0x1 −2x3 = 0

2x1 +2x3 = 6

After these preliminary steps (that were not really necessary), we cannow go with Gauss elimination procedure. Adding the second equationto the third yields

−x1 +x2 +x3 = 0x1 −2x3 = 0

3x1 = 6

We can now use the back substitution method to solve the system.

x1 =63= 2; x3 =

x1

2= 1; x2 = x1 − x3 = 1.

17. In the “derivation”, a standard mistake was made: the equation x2 =

(y − 1)2 does not conclude that x = y − 1. The correct conclusion isx = ±(y − 1). With the sign “+”, we get the answer given; i.e. x = 1,y = 2. Choosing the sign “−”, we get

−(y − 1) + 2y = 5 ⇒ y = 4 ⇒ x = −(4 − 1) = −3.

The figure below indicates these two solutions – the points of intersec-tion of the graphs of the equations.


5

4

3

2

y

x

–1

–4 –3 –2 –1 1 2 3 40

1

19. The coordinates of both points must satisfy the equation ax + by = c.Thus, we have a system of two linear equations with three unknowns– a, b, and c:

ax1 + by1 = c

ax2 + by2 = c.

Since there is always a line passing through two given points, this sys-tem is consistent. If the triple (a0, b0, c0) is a solution, multiplying bothequations by k �= 0, we get an equivalent system

(ka0) x1 + (kb0) y1 = (kc0)

(ka0) x2 + (kb0) y2 = (kc0) .

Thus, the triple (ka0, kb0, kc0) also satisfies the conditions and, actu-ally, determines the same line because

a0x + b0y = c0 ⇔ (ka0) x + (kb0) y = (kc0) .

Thus, the equation of a line is determined up to a non-zero constantmultiple.

21. Since the points (x0, y0) , . . . , (xn, yn) are on the graph y = a0 + · · · +anxn, they must satisfy the equation. Substituting the coordinates of


these points into the equation, yield the system of (n + 1) equationswith (n + 1) unknowns – a0, . . . , an.

a0 +a1x0 + · · ·+ anxn0 = y0

a0 +a1x1 + · · ·+ anxn1 = y1

...a0 +a1xn + · · ·+ anxn

n = yn

23. Multiplying the second equation by (1 + i) we get

(1 + i)x1 + (1 + i)ix2 + (1 + i)(2 − i)x3 = (1 + i)2

⇔ (1 + i)x1 + (i − 1)x2 + (3 + i)x3 = 2i.

We eliminate now x1 from the first equation by subtracting the aboveresult from it (keeping the original second equation unchanged). Thuswe get

(3 − i)x2 −(3 + i)x3 = 0x1 +ix2 +(2 − i)x3 = 1 + i

(−1 + 2i)x2 +(−2 − 3i)x3 = 0

To eliminate x2 from the first equation, we multiply it by (−1+2i) andthen subtract from it the third equation multiplied by (3 − i). Thus,we get

[−(3 + i)(−1 + 2i)− (−2 − 3i)(3 − i)] x3 = 0

⇒ (9 + 7i)x3 = 0,

so that we have a new system

−(9 + 7i)x3 = 0x1 +ix2 +(2 − i)x3 = 1 + i

(−1 + 2i)x2 +(−2 − 3i)x3 = 0

From the first equation, we find that x3 = 0. Substituting this valueinto the third equation yields x2 = 0. With zero values of x2 and x3,the second equation says that x1 = 1 + i. Therefore, the solution is

x1 = 1 + i, x2 = 0, x3 = 0.


25. If x and y are integers, then 6x and 4y are even numbers. Thus, theirdifference, 6x−4y, is even, and so cannot be equal to an odd number 9.

27. The first equation is equivalent to

3(2x + 4y) = 3(3) ⇒ 6x + 12y = 9.

In the integers modulo 6, we have the sides as

(6x + 12y) mod 6 = 6(x + 2y) mod 6 = 0, 9 mod 6 = 3.

Thus, the equation is inconsistent in integers modulo 6, and so is thesystem.

Concerning a solution in integers modulo 7, there are different waysto go. One of them is the following. First, we get a system that is equiv-alent to the given system by multiplying the first equation by 3 and thesecond equation by 2.

6x +12y = 96x +4y = 6

Subtracting the second equation from the first one yields 8y = 3.Rewriting this equation in integers modulo 7 (since 8 mod 7 = 1)as y mod 7 = 3, the back substitution into the second equation in theoriginal system yields

3x = 3 − 2y ⇒ (3x) mod 7 = (3 − 2y)

mod 7 = 4 mod 7.

Solving for x, we get x mod 7 = 6. Therefore, in integers modulo 7,the solution to the given system is

x mod 7 = 6, y mod 7 = 3.


29. To simplify computations, we note that, for large n, the total numberof each arithmetic operation can be approximated with a good relativeaccuracy by

Additions :(n − 1)n(2n + 5)

6≈ n · n · (2n)

6=

n3

3;

Multiplications :(n − 1)n(2n + 5)

6≈ n · n · (2n)

6=

n3

3;

Divisions :n(n + 1)

2≈ n2

2.

Thus, the total number N(n) of arithmetic operations can be approxi-mated as

N(n) ≈ n3

3+

n3

3+

n2

2≈ 2n3

3.

For the “Thermal stress”, n= 10, 000= 104. Therefore, using the per-formance of computers, we get that computers require approximately

Typical PC:2(104

)3/3

5 × 109=

2 × 1012

15 × 109≈ 133.33 (sec.)

Tianhe − 2:2(104

)3/3

3.38 × 1016≈ 1012

5.07 × 1016≈ 1.97 × 10−5 (sec.)

to solve the system. Performing similar computations for other sys-tems, we fill in the following table.

Model n Typical PC Tianhe-2Thermal Stress 104 133.333 1.97 × 10−5

American Sign Language 105 1.33 × 105 0.020Chemical Plant Modeling 3 × 105 3.59 × 106 0.532Mechanics of Composite

Materials 106 1.33 × 108 19.723Electromagnetic Modeling 108 1.33 × 1014 1.97 × 107

Computation Fluid Dynamics 109 1.33 × 1017 1.97 × 1010


1.2 MATRIX REPRESENTATION OF LINEARSYSTEMS AND THE GAUSS-JORDANALGORITHM

1. We subtract the first row the second row multiplied by 2.

⎡⎣ 2 1

... 8

1 −3... −3

⎤⎦ ⇒

⎡⎣ 2 1

... 8

0 −7... −14

⎤⎦

Therefore, x2 = (−14)/(−7) = 2, and the back substitution yields

2x1 + (2) = 8 ⇒ x1 = 3.

The solution is x1 = 3, x2 = 2.

3. We perform elementary row operations to reduce the given augmentedcoefficient matrix to upper triangular form.

⎡⎢⎢⎢⎣

4 2 2... 8 0

1 −1 1... 4 2

3 2 1... 2 0

⎤⎥⎥⎥⎦ ρ2 ↔ ρ1

⎡⎢⎢⎢⎣

1 −1 1... 4 2

4 2 2... 8 0

3 2 1... 2 0

⎤⎥⎥⎥⎦

ρ2 − 4ρ1 → ρ2

ρ3 − 3ρ1 → ρ3

⎡⎢⎢⎢⎣

1 −1 1... 4 2

0 6 −2... −8 −8

0 5 −2... −10 −6

⎤⎥⎥⎥⎦

6ρ3 − 5ρ2 → ρ3

⎡⎢⎢⎢⎣

1 −1 1... 4 2

0 6 −2... −8 −8

0 0 −2... −20 4

⎤⎥⎥⎥⎦

Evaluating x3 from the third equation and making back substitutions weobtain the following solutions:

For the first system,

x3 =−20−2

= 10, x2 =−8 + 2(10)

6= 2, x1 = 4 + (2)− (10) = −4.

1.2 MATRIX REPRESENTATION OF LINEAR SYSTEMS 13

For the second system,

x3 =4−2

= −2, x2 =−8 + 2(−2)

6= −2,

x1 = 2 + (−2)− (−2) = 2.

5. For an m×n matrix A, let’s find a formula for the address of the elementaij, 1 ≤ i ≤ m, 1 ≤ j ≤ n. This element is in the j’s column. Therefore,the number of preceding columns is (j − 1). Each of them contain melements, and so the number of preceding elements is m(j − 1). Theelement aij is the i’s element in the j’s column, and so it has the number

# of aij = m(j − 1) + i, 1 ≤ i ≤ m, 1 ≤ j ≤ n.

This is the answer to the part (c).(a) We have m = 7 and n = 8. Using the formula derived for the part

(c) yields

# of a1,7 = 7(7 − 1) + 1 = 43; # of a7,1 = 7(1 − 1) + 7 = 7;

# of a5,5 = 7(5 − 1) + 5 = 33; # of a7,8 = 7(8 − 1) + 7 = 56.

(d) Let’s first find the answer to this part, and then apply it in thepart (b). From the formula in the part (c), we have

# of aij − im

= j − 1 ⇔ j =# of aij − i + m

m.

Since j is a natural number and 1 ≤ i ≤ m, we can eliminate the(unknown) i from this equation by writing

j =

⌈# of aij

m

⌉,

where · means the “ceiling function”, i.e., x is the smallest inte-ger that is not less than x. Once j is found, we can compute the indexi using

i = # of aij − m(j − 1).


(b) Here, m = 5, n = 12. With these values, the formulas in the part(d) give

#4: j =

⌈45

⌉= 1, i = 4 − 5(1 − 1) = 4 ⇒ a4,1;

#20: j =

⌈205

⌉= 4, i = 20 − 5(4 − 1) = 5 ⇒ a5,4;

#50: j =

⌈505

⌉= 10, i = 50 − 5(10 − 1) = 5 ⇒ a5,10.

7. (a) Following the Gauss-Jordan elimination procedure, we proceed asfollows.

⎡⎢⎢⎢⎣

3 8 3... 7

2 −3 1... −10

1 3 1... 3

⎤⎥⎥⎥⎦

(1/3)ρ1 → ρ1

⎡⎢⎢⎢⎣

1 8/3 1... 7/3

2 −3 1... −10

1 3 1... 3

⎤⎥⎥⎥⎦

ρ2 − 2ρ1 → ρ2

ρ3 − ρ1 → ρ3

⎡⎢⎢⎢⎣

1 8/3 1... 7/3

0 −25/3 −1... −44/3

0 1/3 0... 2/3

⎤⎥⎥⎥⎦

(−3/25)ρ2 → ρ2

⎡⎢⎢⎢⎣

1 8/3 1... 7/3

0 1 3/25... 44/25

0 1/3 0... 2/3

⎤⎥⎥⎥⎦

ρ1 − (8/3)ρ2 → ρ1

ρ3 − (1/3)ρ2 → ρ3

⎡⎢⎢⎢⎣

1 0 17/25... −177/75

0 1 3/25... 44/25

0 0 −1/25... 2/25

⎤⎥⎥⎥⎦

1.2 MATRIX REPRESENTATION OF LINEAR SYSTEMS 15

(−25)ρ3 → ρ3

⎡⎢⎢⎢⎣

1 0 17/25... −177/75

0 1 3/25... 44/25

0 0 1... −2

⎤⎥⎥⎥⎦

ρ1 − (17/25)ρ3 → ρ1

ρ2 − (3/25)ρ3 → ρ2

⎡⎢⎢⎢⎣

1 0 0... −1

0 1 0... 2

0 0 1... −2

⎤⎥⎥⎥⎦

Therefore, the answer is x1 = −1, x2 = 2, and x3 = −2.

(b) This problem, actually, contains three systems of linear equationswith the same coefficient matrix but different right-hand sidecolumns. They are

⎡⎣1

00

⎤⎦ ,

⎡⎣0

10

⎤⎦ ,

⎡⎣0

01

⎤⎦ .

So, can proceed with the Gauss-Jordan elimination process for thecoefficient matrix keeping a track of the right-hand sides on eachstep. For convenience, we first rearrange the equations.

⎡⎢⎢⎢⎣

1 2 1... 1 0 0

1 3 2... 0 1 0

1 0 1... 0 0 1

⎤⎥⎥⎥⎦

ρ3 → ρ1

ρ1 → ρ2

ρ2 → ρ3

⎡⎢⎢⎢⎣

1 0 1... 0 0 1

1 2 1... 1 0 0

1 3 2... 0 1 0

⎤⎥⎥⎥⎦

Then we proceed as follows:

ρ2 − ρ1 → ρ2

ρ3 − ρ1 → ρ3

⎡⎢⎢⎢⎣

1 0 1... 0 0 1

0 2 0... 1 0 −1

0 3 1... 0 1 −1

⎤⎥⎥⎥⎦

2ρ3 − 3ρ2 → ρ3

⎡⎢⎢⎢⎣

1 0 1... 0 0 1

0 2 0... 1 0 −1

0 0 2... −3 2 1

⎤⎥⎥⎥⎦


(1/2)ρ2 → ρ2

(1/2)ρ3 → ρ3

⎡⎢⎢⎢⎣

1 0 1... 0 0 1

0 1 0... 1/2 0 −1/2

0 0 1... −3/2 1 1/2

⎤⎥⎥⎥⎦

ρ1 − ρ3 → ρ1

⎡⎢⎢⎢⎣

1 0 0... 3/2 −1 1/2

0 1 0... 1/2 0 −1/2

0 0 1... −3/2 1 1/2

⎤⎥⎥⎥⎦

Therefore, the solutions to the given systems, respectively, are

x1 =32

, x2 =12

, x3 = −32

;

x1 = −1, x2 = 0, x3 = 1;

x1 =12

, x2 = −12

, x3 =12

.

9. Let us find out which approach, the standard one or the suggested mod-ified one, is more efficient (meaning less time consuming). In otherwords, we have to compare the number of arithmetic operations, Ns

and Nm, resp.Intuitively, the modified approach should be more efficient:In the standard way, given 1≤ j≤ n, to eliminate xij, 1≤ i≤ n, i �= j,

from the ith column, we have to perform arithmetics with entries onall rows ρi. In the modified procedure, in obtaining an upper triangularmatrix, we deal only with the rows ρi, i > j. It is, approximately, twiceless time consuming. Going then from the upper triangular coefficientmatrix to a diagonal one, as described, essentially requires arithmeticsperformed only with the last column of the augmented matrix for therows ρi, 1 ≤ i < j. These arguments suggest that Nm ≤ Ns.

As an example, let us compute the number of arithmetic operationsused in Problem 9(a), where we followed the standard Gauss-Jordanelimination process: Ns = 27. (Computations with obvious zero resultsare not counted.)

Using the modified way, we get Nm = 20. (Check it!)Rigorous computations show that, for n large, the suggested

modification of the Gauss-Jordan elimination procedure can saveapproximately 30% of the time spent for solving an n× n linear systemvs the standard procedure.

1.3 THE COMPLETE GAUSS ELIMINATION ALGORITHM 17

1.3 THE COMPLETE GAUSS ELIMINATIONALGORITHM

1. To simplify computations, we rearrange the equations first.

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

2 2 −1 1... 0

1 0 0 1... 0

1 2 −1 1... 0

⎤⎥⎥⎥⎥⎥⎦

ρ1 → ρ2

ρ2 → ρ4

ρ3 → ρ1

ρ4 → ρ3

⎡⎢⎢⎢⎢⎢⎣

1 0 0 1... 0

1 1 1 1... 1

1 2 −1 1... 0

2 2 −1 1... 0

⎤⎥⎥⎥⎥⎥⎦

ρ2 − ρ1 → ρ2

ρ3 − ρ1 → ρ3

ρ4 − 2ρ1 → ρ4

⎡⎢⎢⎢⎢⎢⎣

1 0 0 1... 0

0 1 1 0... 1

0 2 −1 0... 0

0 2 −1 −1... 0

⎤⎥⎥⎥⎥⎥⎦

ρ4 − ρ3 → ρ4

ρ3 − 2ρ2 → ρ3

⎡⎢⎢⎢⎢⎢⎣

1 0 0 1... 0

0 1 1 0... 1

0 0 −3 0... −2

0 0 0 −1... 0

⎤⎥⎥⎥⎥⎥⎦

From the third and fourth equations we find that

x3 =−2−3

=23

, x4 =0−1

= 0.

Substituting these values in the first two equations, we obtain

x1 + (0) = 0 ⇒ x1 = 0, x2 +

(23

)= 1 ⇒ x2 =

13

.


Therefore, the solution is

x1 = 0, x2 =13

, x3 =23

, x4 = 0.

3. We follow the Gauss elimination procedure with back substitution.

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

2 2 −1 1... 0

−1 −1 0 1... 0

0 0 3 −4... 11

⎤⎥⎥⎥⎥⎥⎦

ρ2 − 2ρ1 → ρ2

ρ3 + ρ1 → ρ3

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

0 0 −3 −1... −2

0 0 1 2... 1

0 0 3 −4... 11

⎤⎥⎥⎥⎥⎥⎦

ρ4 + ρ2 → ρ4

−ρ2 → ρ2

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

0 0 3 1... 2

0 0 1 2... 1

0 0 0 −5... 9

⎤⎥⎥⎥⎥⎥⎦

ρ2 − 3ρ3 → ρ2

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

0 0 0 −5... −1

0 0 1 2... 1

0 0 0 −5... 9

⎤⎥⎥⎥⎥⎥⎦

ρ4 − ρ2 → ρ4

⎡⎢⎢⎢⎢⎢⎣

1 1 1 1... 1

0 0 0 −5... −1

0 0 1 2... 1

0 0 0 0... 10

⎤⎥⎥⎥⎥⎥⎦

The last equation is inconsistent, so is the original system.

solutions manual to accompany with applicationssolutions manual to accompany fundamentals of matrix...

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