1
MATH 221 - October 9, 2015 Sample Test II.
“No one can make you feel inferior without
your consent.” Eleanor Roosevelt
1 1 0
−5 −4



1. Let A =  0 1 3  is B =  6
4
2 3 4
−2 −1



x1


the unique solution to A 
 x2  = ~b = 
x3




3

−3 . Find AB and write down A−1 . Find
1

1

0 . What is B −1 ? Find B −1~x.
1
2 −1 3


2. Find the inverse of A =  4 0 2  by row reductions [ A | I ] ∼ [ I | A−1 ]. In the
0 −1 3
1
1
answer (page 256) to problem 3.6(3) (page 96) replace by .
4
2


"
3. The homogeneous solution to the system A~x =
1 2 3
2 4 7
#


"
#
x1
1


is given by
 x2  =
3
x3
−2


~xh = x2  1  where the x2 is arbitrary.
0


(a) Find a basis for CA .
(b) What is the dimRA ?
(c) What is rank(A)?
(d) Find N (A). (see ~xh )
(e) What is dimN (A)?
(f) What is dimN (AT )?
(g) Are there any vectors ~b ∈ R2 so that the system A~x = ~b has no solution?
"
4. Change the coefficient matrix A in #3 to A =
#
1 2 3
. The homogeneous solution
2 4 6
−2
−3



reads ~xh = x2  1  + x3  0 
, x2 and x3 are arbitrary. Answer (a)-(g) from #3.
0
1





1


5. Suppose that ~v1 =  2 
1

−2

 1
sume that Aw
~1 = A 
 3

0
(i) (a) The size of A is
of CA



2


and ~v2 =  1  and that CA = span {~v1 , ~v2 }. Further, as−1





1
0




 −5 


 = Aw

~2 = A 
~ 1, w
~ 2 }.

 0  =  0  and N (A) = span {w



0
3
. (b) The dimension of N (A) is
and the dimension
. (c) A harder arithmetic problem: # of columns of A =
2
. True or False: A~x = ~b is always solvable.
+
= dimN (A)+ dimCA




1

(ii) Is it possible to add ~v3 =  1 
 ∈ CA and leave everything else as stated?
0
0


(iii) Same as part (b) for ~v3 =  1  ∈ CA .
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For (ii) and (iii), it is relevant to know whether the set V = {~v1 , ~v2 , ~v3 } is a linearly
dependent or independent set of vectors.
Answers
1. A−1 = B, B −1 = A and B −1~x = ~b. see Theorem (3.12) pages 77. 2. see page 256.
("
3. (a) CA = span
1
2
# "
,
3
7




span 

#)
("
= span
1
0
# "
,
0
1
#)
, (b), (c) dimRA = 2 = rankA,







−2
(e) dimN (A) = 1 and (f) dimN (AT ) = 0 (g) No
1
0
2
since dim CA = 2 = dimR the problem is always solvable.
(d) N (A) = ~xh =
("
4. (a) CA = span
1
2
#)
(b), (c) dimRA = 1 = rankA, (d) N (A) =



span 



("
T
T
(e) so dimN (A) = 2 and (f) dimN (A ) = 1, note that N (A ) = span
"
(g) If A~x = ~b has no solution unless ~b is a multiple of
of CA
 
−2
1
#)
#
1
.
2
5. (i) The size of A is 3 × 4 . (b) The dimension of N (A) is
2 and the dimension
2 . (c) # of columns of A = 4 = dimN (A)+ dimCA 2 + 2 . A~x = ~b
is not always solvable since dimN (AT ) = 2. (ii) Since ~v3 = (1/3)~v1 + (1/3)~v2 , the
answers in (i) are unchanged. (iii) The set {~v1 , ~v2 , ~v3 } is linearly independent so there
is one of two possibilities:
First: the scenario is impossible. The nullspace of A is “too large” due to the Dimension
Theorem (never underestimate a counting theorem) or
Second: one could remove one vector from the nullspace of A, say N (A) = span {w
~ 1}
then all is well. The dimension of N (A) is 1 and the dimension of CA 3 .
The # of columns of A = 4 = dimN (A)+ dimCA 1 + 3 . A~x = ~b is solvable
since dim CA = 3 so that dimN (AT ) = 0.

−2
−3 

 

1  ,  0 

0
1