Solutions for Linear Algebra Midterm I
1. True or False?
(a) If an n x n matrix A is not symmetric, then ATA is not symmetric.
(b) In general, the determinant of the sum of two matrices equals the sum of the
determinants of the matrices.
(c) If A and B are square matrices of order n such that det(AB) = -1, then both A
and B are nonsingular.
(d) We can have AT Ax  AT b if Ax  b , and the reverse condition is also true.
(e) If A is row equivalent to B and B is row equivalent to C, then A is row
equivalent to C.
Sol.
(a) False. ( AT A)T  AT A , so symmetric
(b) False. det( I  I )  det( I )  det( I )
(c) True.
(d) False. If AT Ax  AT b but ( AT )1 is not exist, we cannot promise that Ax  b .
(e) True.
2. (a) Show that if A and B are
1
(b) Show that det( A
n n
matrices, then det(AB) = det(A)det(B).
)  1 det( A) . (Suppose that A1 exists.)
Sol.
(a) (i) Consider the case that B is singular.  det( B)  0
We can know from Theorem 1.4.3 that AB is also singular.
 d e tA( B ) 
0 dA
e t ( )Bd e t ( )
(ii) If B is nonsingular, B can be written as a product of elementary matrices. We
have already seen that the result holds for elementary matrices.(text book
p.110). Thus
det( AB)
 det( AEk Ek 1
E1 )
 det( A) det( Ek ) det( Ek 1 )
 det( A) det( Ek Ek 1
det( E1 )
E1 )
 det( A) det( B)
(b)
A1 A  I
From the result of (a), det( A1 A)  det( A1 ) det( A)  det( I )  1
 det( A1 )  1 det( A)
3. If A is nonsingular, show that adj A is also nonsingular and
(a d j A
)1  d e t (A1 )A
a dj.1 A
Sol.
If A is nonsingular then det( A)  0 , and hence
adj A  det( A) A1
is also nonsingular.
It follows (adj A)1 
1
( A1 ) 1  det( A1 ) A
det( A)
Also adj A1  det( A1 )( A1 )1  det( A1 ) A
4. We let A1 be the inverse of A. Find the inverses of the following matrices.
(a) Interchange the i-th row and the j-th row of A
(b) Multiply the i-th row of A with a constant k ; k  0 .
Sol.
(a) Interchange the i-th column and the j-th column of A1
(b) Multiply the i-th column of A1 with a constant
1
; k  0.
k
2 4 2
5. Compute the LU factorization of A  1 5 2  .


 4 1 9 
Sol.
2 4 2
2 4 2
2 4 2
2 4 2






E3
E1
E2
A  1 5 2  0 3 1   0 3 1  U  0 3 1 
 4 1 9 
 4 1 9 
0 9 5 
0 0 8 
1
1
1
It means that E3 E2 E1 A  U  A  E3 E2 E1 U  LU , where
 1 0 0
1
 1 0 0
1 0 0 
 1

1
E1  
1 0  , E2   0 1 0  , E3  0 1 0 , L  




2

2





2
0
1
0
3
1
 0 0 1
2






0

1 0 ,

3 1 
0
6. Let A and B be symmetric n x n metrics. Prove that AB = BA if and only if AB is
also symmetric.
Sol.
A = A T , B = BT
1. (AB =BA => AB is symmetric)
( AB)T  BT AT  BA  AB . So AB is symmetric.
2. (AB is symmetric => AB = BA)
( AB)T  BT AT  BA . Because AB is symmetric => (AB)T = AB
 AB = BA
7. Two competing companies offer cable television service to a city of 100,000
households. The change in cable subscriptions each year is given by the diagram in
Figure 1. Company A now has 15,000 subscribers an Company B has 20,000
subscribers. How many subscribers will each company have one year from now?
Sol:
The matrix representing the given transition probabilities is
A
From
B
None
 0.7 0.15 0.15 A
P  0.20 0.80 0.15 B To
0.10 0.05 0.70  None
and the state matrix representing the current population in each of the three states is
15, 000  A
X   20, 000  B
 65, 000  None
0.70 0.15 0.15 15, 000  23, 250 
PX  0.20 0.80 0.15  20, 000   28, 750 
0.10 0.05 0.70   65, 000  48, 000 
8. Derive the row echelon form of the following augmented matrix and determine
what kind of solution it will have? (no solution, unique solution, or infinite
solutions)
0 1 2 3 5
1 1 1 1 8


 2 2 3 3 4 


 1 2 2 3 6 
 2 3 1 2 14 
Row echelon form:
1
0

0

0

0

8 
5 
4 

13 
0 0 1 
2
0 0 0
0 
1 1 1
1 2 3
0 1 1
,so these is an unique solution.