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(MATH113)[2010](f)midterm~nykwok^_10451.pdf
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.. .. ....
12 1 b1
1. Let a1 = . 2 ., a2 = . 0 ., a3 = . 6 ., b = . b2 ., and A =[ a1 a2 a3 ]. 03 .3 b3
(a) Does the homogeneous equation Ax = 0 have a nontrivial solution? If yes, .nd the solution.
Solution: Yes. Row operations on augmented matrix [A | 0] yields: . 1 2 1 0 . . 1 2 1 0 . . 1 2 1 0 . . 1 0 3 0 .
. 2 0 6 0 . . 0 .4 4 0 . . 0 1 .1 0 . . 0 1 .1 0 . .
0 3 .3 0 0 3 .3 0 0 0 0 0 0 0 0 0

Set x3 = s to be a free parameter. Then the general solution of Ax = 0 is:
.. ..
x1 .3
. x2 . = s . 1 . ,s R. x3 1
Any nonzero s leads to a nontrivial solution.
(b) If a vector u is in Span{a1, a2, a3}, then is it also in Span{a1, a2}? Solution: Yes. [From (a) we have .3a1 + a2 + a3 = 0, i.e. a3 =3a1 . a2. So for any u Span {a1, a2, a3}, we
can write:
u = c1a1 + c2a2 + c3(3a1 . a2)=(c1 +3c3)a1 +(c2 . c3)a2

which shows that u is in Span{a1, a2}.]
(c) Determine the condition for b to be in Span{a1, a2, a3}.
Solution: b Span {a1, a2, a3} means x1a1 + x2a2 + x3a3 = b has solution in (x1,x2,x3). So we perform row operations on the corresponding augmented matrix [A | b]:
.
12 1
.
20 6 03 .3
. .. .
b1 12 1
b1 b2 . 2b1 .
b2 . . 0 .44 b3 03 ..3
b3
. .
12 1
b1 12 1
b1
.
01 .1
.b2/4+ b1/2 . . 01 .1
.b2/4+ b1/2 .
01 .1
b3/3 000
b3/3+ b2/4 . b1/2
The system is consistent i. b3/3+ b2/4 . b1/2=0 (or 6b1 . 3b2 . 4b3 = 0).
(d) Consider the matrix transformation T de.ned by x T (x)= Ax. Is T one-to-one and why?
Is T onto the codomain and why?
Solution: It is not one-to-one because of the existence of nontrivial solutions in (a). It is not
onto the codomain because {a1, a2, a3} dont span R3 according to (c).

[30 points]
1

2. (a) Find the inverse (if it exists) of the following matrix:

.
.
1 .10 0
.11 .10

A =

...

...

.

0 .11 .1
00 .11

Solution: Perform row operations to the combined matrix [ A | I4 ]: 1 .10 0
1000
.
.
.
.

1 .100
1000
...

.11 .10
0100
0 .11 .1
0010
...

.r4+r3

.

.r1+r2
...


00 .10
1100
0 .100
0011
...

00 .11

0001 00 .11
0001
.
.
.
.
10 00

10 .1 .1 1000
10 .1 .1
.r2+r4
.


.r3+r1
...

00 .10

1100
0 .100

0011
...

r2.r3
.


scalings

...


0100
00 .1 .1
0010
.1 .10 0
...


.

00 01

.1 .10 1
0001
.1 .10 1
Hence A is invertible, and A.1 is given by:
.
.
...


10 .1 .1
00 .1 .1

.1 .10 0
.1 .10 1

...

A.1

=
.

(b) Consider the following linear system:

. ... .
.

x1 . x3 . x4 = b1 x3 + x4 = b2 x1 + x2 = b3
x1 + x2 . x4 = b4

Is the above system always consistent for any numbers b1,b2,b3,b4? Why?
When the solution of the above system exists, is it unique? Why?
Solution: The coe.cient matrix B of the system is row-equivalent to A.1 in (a):

.
.
.
.
10 .1 .1 10 .1 .1

B =


...

001 1

110 0

...

scalings