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(math111)[2010](f)midterm~1406^_10436.pdf
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MATH 111 Linear Algebra Midterm Test, Fall 2010-11
L2: 18-Oct-2010, 6:00pm C 7:30pm, LTC
Answer all questions. Total marks: 100

Qn. 1 (15 marks) Find an equation involving g, h, and k that makes the following augmented

matrix correspond to a consistent system:
. 1 .4 7 1 g .
. 0 3 .5 2 h . .
.2 5 .9 .4 k
Solution: By row operations:

. 1 .4 7 1 g . . 1 .4 7 1 g . . 1 .4 7 1 g .
. 0 3 .5 2 h . . 0 3 .5 2 h . . 0 3 .5 2 h . .
.2 5 .9 .4 k 0 .3 5 .2 2g + k 0 0 0 0 2g + h + k
For the system to be consistent, we need 2g + h + k = 0.

Qn. 2 (15 marks) Determine if the columns of the following matrix A span R4:
..
131 01
..
11 .11 3
..
A = .
..
04 4 .21
20 .43 .1
Solution: By row operations:
. .. ...
13101 13101 13101
. .. ...
0 .2 .212 0 .2 .212 0 .2 .212
. .. ...
A .
. 04 4 .21 . . 00 005 . . 00 005 . 0 .6 .63 .3 00 00 .9 00 000
Since the 4th row contains no pivot position, the columns of A cannot span R4 .
Qn. 3 (10 marks) Find the value(s) of h for which the following vectors are linearly dependent.
.. .. ..
1 24 a1 = . 3 . , a2 = . 2 . , a3 = . h . . .217
Solution: Consider the coe.cient matrix of x1a1 + x2a2 + x3a3 = 0:
. .. ...
124 124 124
.. .. ..
32 h 0 .4 h . 12 00 h. .217 05 15 0515
For the vector equation to have non-trivial solutions, we need h = 0.
1

Qn. 4 (10 marks) Suppose vectors v1,..., vp span Rn, and let T : Rn Rm be a linear transfor-mation such that T (vi)= 0 for i =1,...,p. Show that for any vector x in Rn, we always have T (x)= 0.
Solution: Let x be any vector in Rn. Since v1,..., vp span Rn, there exist suitable scalars c1,...,cp such that:
x = c1v1 + ... + cpvp.
As T is linear, we have:
T (x)= T (c1v1 + ... + cpvp)= c1T (v1)+ ... + cpT (vp)= c1 0 + ... + cp 0 = 0.
Qn. 5 (10 marks) Let A be an m n matrix and suppose that there exists an n m matrix C such that CA = In. Show that the homogeneous system Ax = 0 has only the trivial solution. Explain why in this case we cannot have n>m.
Solution: Let x0 be a solution of Ax = 0, i.e. Ax0 = 0. Then:
C Ax0 = C 0 . (CA)x0 = 0 . In x0 = 0 . x0 = 0,
so Ax = 0 can only have the trivial solution x = 0. When n>m, we will have more variables than equations in Ax = 0, and hence non-trivial solutions exist. Thus in our case, we cannot have n>m.
Qn. 6 (10 marks) Let v1,..., vp be vectors in Rn and set S = {v1,..., vp}. Write down the original de.nition that says S is linearly independent. De.nition: The vector equation: x1v1 + x2v2 + ... + xpvp = 0 has only the trivial solution / has unique zero solution x1 = x2 = ... = xp = 0.
Qn. 7 (15 marks) Let T : R2 R3 be a linear transformation satisfying:
.. ..
.. 2 .. 1
2 .1
.. ..
T = 3 and T = .2 .
.22
2 .1
10
Find T (e1) and T (e2), where e1 =, e2 =.
01
2 .11
Solution: First we note that + = = e1. As T is linear, we h