TEST BANK FOR Applied Linear Algebra By Peter J. Olver and Chehrzad Shakiban

1.1.1.
(a) Reduce the system to x − y = 7, 3y = −4; then use Back Substitution to solve for
x = 17
3 , y = − 4
3 .
(b) Reduce the system to 6u + v = 5, − 5
2 v = 5
2 ; then use Back Substitution to solve for
u = 1, v = −1.
(c) Reduce the system to p + q − r = 0, −3q + 5r = 3, −r = 6; then solve for p = 5, q =
−11, r = −6.
(d) Reduce the system to 2u − v + 2w = 2, − 3
2 v + 4w = 2, −w = 0; then solve for
u = 1
3 , v = − 4
3 ,w = 0.
(e) Reduce the system to 5x1 + 3x2 − x3 = 9, 1
5 x2 − 2
5 x3 = 2
5 , 2x3 = −2; then solve for
x1 = 4, x2 = −4, x3 = −1.
(f ) Reduce the system to x + z − 2w = −3, −y + 3w = 1, −4z − 16w = −4, 6w = 6; then
solve for x = 2, y = 2, z = −3,w = 1.
(g) Reduce the system to 3x1 + x2 = 1, 8
3 x2 + x3 = 2
3 , 21
8 x3 + x4 = 34
, 55
21 x4 = 5
7 ; then
solve for x1 = 3
11 , x2 = 2
11 , x3 = 2
11 , x4 = 3
11 .
1.1.2. Plugging in the given values of x, y and z gives a+2b−c = 3, a−2−c = 1, 1+2b+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
2x = −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.
(b) We will get a diagonal system with the same solution.
(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
assuming 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 ).
Solving the reduced system by Forward Substitution reproduces the same solution (as
it must):
(a) The system reduces to 3
2 x = 17
2 , x + 2y = 3.
(b) The reduced system is 15
2 u = 15
2 , 3u − 2v = 5.
(c) The method doesn’t work since r doesn’t appear in the last equation.
(d) Reduce the system to 3
2 u = 1
2 , 7
2 u − v = 5
2 , 3u − 2w = −1.
(e) Reduce the system to 2
3 x1 = 8
3 , 4x1 + 3x2 = 4, x1 + x2 + x3 = −1.
(f ) Doesn’t work since, after the first reduction, z doesn’t occur in the next to last
equation.
(g) Reduce the system to 55
21 x1 = 5
7 , x2 + 21
8 x3 = 3
4 , x3 + 8
3 x4 = 2
3 , x3 + 3x4 = 1.
1.2.1. (a) 3 × 4, (b) 7, (c) 6, (d) (−2 0 1 2 ), (e)
0
B@
0
2
−6
1
CA
.
1
1.2.2. (a)
0
B@
1 2 3
4 5 6
7 8 9
1
CA
, (b)
1 2 3
1 4 5
!
, (c)
0
B@
1 2 3 4
4 5 6 7
7 8 9 3
1
CA
, (d) ( 1 2 3 4 ),
(e)
0
B@
1
2
3
1
CA
, (f ) ( 1 ).
1.2.3. x = −1
3 , y = 4
3 , z = −1
3 , w = 2
3 .
1.2.4.
(a) A =
1 −1
1 2
!
, x =
x
y
!
, b =
7
3
!
;
(b) A =
6 1
3 −2
!
, x =
u
v
!
, b =
5
5
!
;
(c) A =
0
B@
1 1 −1
2 −1 3
−1 −1 0
1
CA
, x =
0
B@
p
q
r
1
CA
, b =
0
B@
0
3
6
1
CA
;
(d) A =
0
B@
2 1 2
−1 3 3
4 −3 0
1
CA
, x =
0
B@
u
v
w
1
CA
, b =
0
B@
3
−2
7
1
CA
;
(e) A =
0
B@
5 3 −1
3 2 −1
1 1 2
1
CA
, x =
0
B@
x1
x2
x3
1
CA
, b =
0
B@
9
5
−1
1
CA
;
(f ) A =
0
BBB@
1 0 1 −2
2 −1 2 −1
0 −6 −4 2
1 3 2 −1
1
CCCA
, x =
0
BBB@
x
y
z
w
1
CCCA
, b =
0
BBB@
−3
3
2
1
1
CCCA
;
(g) A =
0
BBB@
3 1 0 0
1 3 1 0
0 1 3 1
0 0 1 3
1
CCCA
, x =
0
BBB@
x1
x2
x3
x4
1
CCCA
, b =
0
BBB@
1
1
1
1
1
CCCA
.
1.2.5.
(a) x − y = −1, 2x + 3y = −3. The solution is x = −6
5 , y = −1
5 .
(b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1.
(c) 3x1 − x3 = 1, −2x1 − x2 = 0, x1 + x2 − 3x3 = 1.
The solution is x1 = 1
5 , x2 = −2
5 , x3 = −2
5 .
(d) x + y − z − w = 0, −x + z + 2w = 4, x − y + z = 1, 2y − z + w = 5.
The solution is x = 2, y = 1, z = 0, w = 3.
1.2.6.
(a) I =
0
BBB@
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1
CCCA
, O =
0
BBB@
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
1
CCCA
.
(b) I + O = I , IO = OI = O. No, it does not.
1.2.7. (a) undefined, (b) undefined, (c)
3 6 0
−1 4 2
!
, (d) undefined, (e) undefined,
(f )
0
B@1
11 9
3 −12 −12
7 8 8
1
CA
, (g) undefined, (h)
0
B@
9 −2 14
−8 6 −17
12 −3 28
1
CA
, (i) undefined.