Applied Linear Algebra
by
Peter J. Olver
and Chehrzad Shakiban
Second Edition
Undergraduate Texts in Mathematics
Springer, New York, 2018.
ISBN 978–3–319–91040–6
Current version (v. 2) posted July, 2019
v. 1 posted August, 2018
,To the Student
These solutions are a resource for students studying the second edition of our text Applied
Linear Algebra, published by Springer in 2018. An expanded solutions manual is available
for registered instructors of courses adopting it as the textbook.
Using the Manual
The material taught in this book requires an active engagement with the exercises, and
we urge you not to read the solutions in advance. Rather, you should use the ones in
this manual as a means of verifying that your solution is correct. (It is our hope that
all solutions appearing here are correct; errors should be reported to the authors.) If you
get stuck on an exercise, try skimming the solution to get a hint for how to proceed, but
then work out the exercise yourself. The more you can do on your own, the more you will
learn. Please note: for students taking a course based on Applied Linear Algebra, copying
solutions from this Manual can place you in violation of academic honesty. In particular,
many solutions here just provide the final answer, and for full credit one must also supply
an explanation of how this is found.
Acknowledgements
We thank a number of people, who are named in the text, for corrections to the solutions
manuals that accompanied the first edition. Thanks to Alexander Voronov and his stu-
dents Jacob Boerjan and Cassandra Chanthamontry for further corrections to the current
manual. Of course, as authors, we take full responsibility for all errors that may yet appear.
We encourage readers to inform us of any misprints, errors, and unclear explanations that
they may find, and will accordingly update this manual on a timely basis. Corrections will
be posted on the text’s dedicated web site:
http://www.math.umn.edu/∼olver/ala2.html
Peter J. Olver Cheri Shakiban
University of Minnesota University of St. Thomas
olver@umn.edu cshakiban@stthomas.edu
, Students’ Solutions Manual for
Chapter 1: Linear Algebraic Systems
1.1.1. (b) Reduce the system to 6 u + v = 5, − 25 v = 5
2 ; then use Back Substitution to solve
for u = 1, v = −1.
(d) Reduce the system to 2 u − v + 2 w = 2, − 32 v + 4 w = 2, − w = 0; then solve for
u = 13 , v = − 43 , w = 0.
(f ) Reduce the system to x + z − 2 w = − 3, − y + 3 w = 1, − 4 z − 16 w = − 4, 6 w = 6; then
solve for x = 2, y = 2, z = −3, w = 1.
♥ 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down.
Thus 2 x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and so
y = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7, and
so z = −22.
1.2.5. (b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1.
(c) 3 x1 − x3 = 1, −2 x1 − x2 = 0, x1 + x2 − 3 x3 = 1.
The solution is x1 = 15 , x2 = − 52 , x3 = − 52 .
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
1.2.6. (a) I = 0
0 1 0 0,
O=
0
0 0 0 0.
0 0 0 1 0
0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
(b) I + O = I , I O = O I = O. No, it does not.
1 c 2019 Peter J. Olver and Chehrzad Shakiban
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