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Instructors’ Solutions Manual for Applied Linear Algebra 2nd Edition By Peter J.Olver and Chehrzad Shakiban (All Chapters) Complete Guide A+$12.99
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, Instructors’ Solutions Manual for
Chapter 1: Linear Algebraic Systems
Note: Solutions marked with a ⋆ do not appear in the Students’ Solutions Manual.
1.1.1. (b) Reduce the system to 6 u + v = 5, − 52 v = 52 ; then use Back Substitution to solve
for u = 1, v = −1.
⋆ (c) Reduce the system to p + q − r = 0, −3 q + 5 r = 3, − r = 6; then solve for
p = 5, q = −11, r = −6.
(d) Reduce the system to 2 u − v + 2 w = 2, − 32 v + 4 w = 2, − w = 0; then solve for
u = 31 , v = − 43 , w = 0.
1 2 2
⋆ (e) Reduce the system to 5 x1 + 3 x2 − x3 = 9, 5 x2 − 5 x3 = 5, 2 x3 = −2; then solve for
x1 = 4, x2 = −4, x3 = −1.
(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.
⋆ 1.1.2. Plugging in the values of x, y and z gives a + 2 b − c = 3, a − 2 − c = 1, 1 + 2 b + c = 2.
Solving this system yields a = 4, b = 0, and c = 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.
⋆ (c) Start with the last equation and, assuming the coefficient of the last variable is 6= 0, use
the operation to eliminate the last variable in all the preceding equations. Then, again as-
suming the coefficient of the next-to-last variable is non-zero, eliminate it from all but the
last two equations, and so on.
⋆ (d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ). Solv-
ing the reduced system by Forward Substitution reproduces the same solution (as it must):
(a) The system reduces to 23 x = 17 15
2 , x + 2 y = 3. (b) The reduced system is 2 u = 2 ,
15
3 u − 2 v = 5. (d) Reduce the system to 32 u = 21 , 72 u − v = 52 , 3 u − 2 w = −1. (f ) Doesn’t
work since, after the first reduction, z doesn’t occur in the next to last equation.
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 = − 25 , x3 = − 25 .
⋆ (d) x + y − z − w = 0, −x + z + 2 w = 4, x − y + z = 1, 2 y − z + w = 5.
The solution is x = 2, y = 1, z = 0, w = 3.
1.2.11. (a) True, ⋆ (b) true.
! ! !
x y ax by ax ay
⋆ ♥ 1.2.12. (a) Let A =
z w
. Then A D =
az bw
=
bz bw
= D A, so if a 6= b these
!
a 0
are equal if and only if y = z = 0. (b) Every 2 × 2 matrix commutes with = a I.
0 a
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