SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ALGEBRA

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SOLUTIONS TO PROBLEMS
ELEMENTARY
LINEAR ALGEBRA
K. R. MATTHEWS
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF QUEENSLAND
First Printing, 1991
CONTENTS
PROBLEMS 1.6 ............................................ 1
PROBLEMS 2.4 ............................................ 12
PROBLEMS 2.7 ............................................ 18
PROBLEMS 3.6 ............................................ 32
PROBLEMS 4.1 ............................................ 45
PROBLEMS 5.8 ............................................ 58
PROBLEMS 6.3 ............................................ 69
PROBLEMS 7.3 ............................................ 83
PROBLEMS 8.8 ............................................ 91
i
SECTION 1.6
0 0 0 2 4 0 1 2 0
2. (i) R R R 1R ;
2 4 0 1 ↔ 2 0 0 0 1 → 2 1 0 0 0
· ¸ · ¸ · ¸
0 1 3 1 2 4 1 0 2
(ii) 1 2 4 R1 ↔ R2 0 1 3 R1 → R1−2R2 0 1 −3 ;
· ¸ · ¸ · ¸
1 1 1 1 1 0
R R R
(iii) 1 1 0 2 → 2− 1 0 0 1
 1 0 0  R3 → R3−R1  0 1 −1 
− −
   
R R +R 1 0 0 1 0 0
1RR→32→1−RR33 3  00 10 11  R2R→3 →R2−+R3R3  00 10 01 ;
↔ −
   
2 0 0 1 0 0
R R +2R
(iv) 0 0 0 3 → 3 1 0 0 0 .
 4 0 0  R1 → 12R1  0 0 0 
−
   
1 1 1 2 1 1 1 2
R R 2R
3.(a) 2 3 1 8 2 → 2− 1 0 1 3 4
 1 1 −1 8  R3 → R3−R1  0 2 −2 10 
− − − − − −
   
1 0 4 2 1 0 4 2
RR31→→RR31+−2RR22  00 10 −38 −42 R3 → −81R3 00 10 −31 41 
− − 4
   
1 0 0 3
R1 → R1−4R3 0 1 0 −19 .
R2 → R2+3R3  0 0 1 41 
4
 
The augmented matrix has been converted to reduced row–echelon form
and we read off the unique solution x = 3, y = 19, z = 1.
− 4 4
1 1 1 2 10 1 1 1 2 10
(b) 3 1 −7 4 1 R2 → R2−3R1 0 4 −10 2 29
 5 −3 15 6 9  R3 → R3+5R1  0 −8 20 −4 −59 
− − − −
   
1 1 1 2 10
−
R R +2R 0 4 10 2 29 .
3 3 2
→  − − − 
0 0 0 0 1
 
From the last matrix we see that the original system is inconsistent.
1
3 1 7 0 1 1 1 1
− −
2 1 4 1 2 1 4 1
(c)  1 −1 1 12 R1 ↔ R3 3 −1 7 02 
− −
 6 4 10 3   6 4 10 3 
 −   − 
   
1 1 1 1 1 0 3 1
RR23 →→ RR23−−23RR11  00 −21 42 −233  RR14 →→ RR14+−RR23  00 10 20 −−2203 .
R4 → R4−6R1  0 2 4 −3  R3 → R3−2R2  0 0 0 0 
 −   
   
The augmented matrix has been converted to reduced row–echelon form
and we read off the complete solution x = 1 3z, y = 3 2z, with z
−2 − −2 −
arbitrary.
2 1 3 a 2 1 3 a
− −
4. 3 1 5 b R R R 1 2 8 b a
2 2 1
 −  → −  − − 
5 5 21 c 5 5 21 c
− − − −
   
1 2 8 b a 1 2 8 b a
R1 ↔ R2 52 −51 −213 −ac  RR23 →→ RR23−+25RR11  00 −55 −1199 5−b 2b−5+a3+ac 
− − − −
   
1 2 8 b a
R R +R − −
R32→→ −351R22  00 01 −5109 3b 2b−25a3a+c 
−
 
1 0 2 (b+a)
−5 5
R1 → R1−2R2 0 1 −519 2b−53a .
0 0 0 3b 2a+c
−
 
From the last matrix we see that the original system is inconsistent if
3b 2a+c = 0. If 3b 2a+c = 0, the system is consistent and the solution
− 6 −
is
(b+a) 2 (2b 3a) 19
x = + z, y = − + z,
5 5 5 5
where z is arbitrary.
1 1 1 1 1 1
R R tR
5. t 1 t 2 → 2− 1 0 1 t 0
 1+t 2 3  R3 → R3−(1+t)R1  0 1−t 2 t 
− −
   
1 1 1
R R R 0 1 t 0 = B.
3 3 2
→ −  − 
0 0 2 t
−
 
Case 1. t = 2. No solution.
6
2
1 0 1 1 0 1
Case 2. t = 2. B = 0 1 0 0 1 0 .
 −  →  
0 0 0 0 0 0
We read off the unique solutionx =1, y = 0. 
6. Method 1.
3 1 1 1 4 0 0 4
−1 3 1 1 R1 → R1−R4 −0 4 0 4
 1 −1 3 1  R2 → R2−R4  0 −0 4 4 
 1 1 −1 3  R3 → R3−R4  1 1 −1 3 
 −   − 
   
1 0 0 1 1 0 0 1
− −
0 1 0 1 0 1 0 1
→  0 0 1 −1 R4 → R4−R3−R2−R1 0 0 1 −1 .
− −
 1 1 1 3   0 0 0 0 
 −   
   
Hence the given homogeneous system has complete solution
x = x , x = x , x = x ,
1 4 2 4 3 4
with x arbitrary.
4
Method 2. Write the system as
x +x +x +x = 4x
1 2 3 4 1
x +x +x +x = 4x
1 2 3 4 2
x +x +x +x = 4x
1 2 3 4 3
x +x +x +x = 4x .
1 2 3 4 4
Then it is immediate that any solution must satisfy x = x = x = x .
1 2 3 4
Conversely, if x , x , x , x satisfy x = x = x = x , we get a solution.
1 2 3 4 1 2 3 4
7.
λ 3 1 1 λ 3
−1 λ 3 R1 ↔ R2 λ 3 −1
· − ¸ · − ¸
1 λ 3
R2 → R2−(λ−3)R1 0 λ2+−6λ 8 = B.
· − − ¸
Case 1: λ2+6λ 8 = 0. That is (λ 2)(λ 4) = 0 or λ = 2, 4. Here B is
− − 6 − − − 6 6
1 0
row equivalent to :
0 1
· ¸
1 λ 3 1 0
R2 → λ2+16λ 8R2 0 −1 R1 → R1−(λ−3)R2 0 1 .
− − · ¸ · ¸
Hence we get the trivial solution x = 0, y = 0.
3
1 1
Case 2: λ = 2. Then B = − and the solution is x = y, with y
0 0
· ¸
arbitrary.
1 1
Case 3: λ = 4. Then B = and the solution is x = y, with y
0 0 −
· ¸
arbitrary.
8.
3 1 1 1 1 1 1 1 1
R R 3 3 3
5 1 1 1 1 → 3 1 5 1 1 1
· − − ¸ · − − ¸
1 1 1 1
R R 5R 3 3 3
2 → 2− 1 0 8 2 8
· −3 −3 −3 ¸
3 1 1 1 1
R − R 3 3 3
2 → 8 2 0 1 1 1
· 4 ¸
1 1 0 1 0
R R R 4 .
1 → 1− 3 2 0 1 1 1
· 4 ¸
Hence the solution of the associated homogeneous system is
1 1
x = x , x = x x ,
1 3 2 3 4
−4 −4 −
with x and x arbitrary.
3 4
9.
1 n 1 1 R R R n 0 n
1 1 n
− ··· → − − ···
1 1 n 1 R R R 0 n n
2 2 n
A =  . −. ··· .  → . −  . −. ··· . 
. . . . . . .
. . . . . . .
 ···   ··· 
 1 1 1 n  R R R  1 1 1 n 
n 1 n 1 n
 ··· −  − → − −  ··· − 
   
1 0 1 1 0 1
··· − ··· −
0 1 1 0 1 1
→  ... ... ··· −... Rn → Rn−Rn−1···−R1 ... ... ··· −... .
 ···   ··· 
 1 1 1 n   0 0 0 
 ··· −   ··· 
   
The last matrix is in reduced row–echelon form.
Consequently the homogeneous system with coefficient matrix A has the
solution
x = x , x = x ,...,x = x ,
1 n 2 n n 1 n
−
4
with x arbitrary.
n
Alternatively, writing the system in the form
x + +x = nx
1 n 1
···
x + +x = nx
1 n 2
···
.
.
.
x + +x = nx
1 n n
···
shows that any solution must satisfy nx = nx = = nx , so x = x =
1 2 n 1 2
···
= x . Conversely if x = x ,...,x = x , we see that x ,...,x is a
n 1 n n 1 n 1 n
··· −
solution.
a b
10. Let A = and assume that ad bc = 0.
c d − 6
· ¸
Case 1: a = 0.
6
a b 1 b 1 b
R 1R a R R cR a
· c d ¸ 1 → a 1· c d ¸ 2 → 2− 1· 0 ad−abc ¸
1 b 1 0
R a R a R R bR .
2 → ad bc 2 0 1 1 → 1− a 2 0 1
− · ¸ · ¸
Case 2: a = 0. Then bc = 0 and hence c = 0.
6 6
0 b c d 1 d 1 0
A = R R c .
c d 1 ↔ 2 0 b → 0 1 → 0 1
· ¸ · ¸ · ¸ · ¸
1 0
So in both cases, A has reduced row–echelon form equal to .
0 1
· ¸
11. We simplify the augmented matrix of the system using row operations:
1 2 3 4 1 2 3 4
3 1 −5 2 R2 → R2−3R1 0 7 −14 10
 4 −1 a2 14 a+2  R3 → R3−4R1  0 −7 a2 2 a− 14 
− − − −
   
R R R 1 2 3 4 1 0 1 8
3 → 3− 2 − 7
R 1R 0 1 2 10 R R 2R 0 1 2 10 .
2 → −7 2  − 7  1 → 1− 2  − 7 
R R 2R 0 0 a2 16 a 4 0 0 a2 16 a 4
1 1 2
→ − − − − −
   
Denote the last matrix by B.
5
Case 1: a2 16 = 0. i.e. a = 4. Then
− 6 6 ±
R 1 R 1 0 0 8a+25
3 → a2 16 3 7(a+4)
R R− R 0 1 0 10a+54
1 → 1− 3  7(a+4) 
R2 R2+2R3 0 0 1 1
→  a+4 
 
and we get the unique solution
8a+25 10a+54 1
x = , y = , z = .
7(a+4) 7(a+4) a+4
1 0 1 8
7
Case 2: a = 4. Then B = 0 1 2 10 , so our system is inconsistent.
−  − 7 
0 0 0 8
−
 
1 0 1 8
7
Case 3: a = 4. Then B = 0 1 2 10 . We read off that the system is
 − 7 
0 0 0 0
consistent, with complete solution x= 8 z, y = 10 +2z, where z is
7 − 7
arbitrary.
12. We reduce the augmented array of the system to reduced row–echelon
form:
1 0 1 0 1 1 0 1 0 1
0 1 0 1 1 0 1 0 1 1
 1 1 1 1 0  R3 → R3+R1  0 1 0 1 1 
 0 0 1 1 0   0 0 1 1 0 
   
   
1 0 1 0 1 1 0 0 1 1
0 1 0 1 1 R R +R 0 1 0 1 1
R3 → R3+R2  0 0 0 0 0  1R→3 1R4 4  0 0 1 1 0 .
↔
 0 0 1 1 0   0 0 0 0 0 
   
   
The last matrix is in reduced row–echelon form and we read off the solution
of the corresponding homogeneous system:
x = x x = x +x
1 4 5 4 5
− −
x = x x = x +x
2 4 5 4 5
− −
x = x = x ,
3 4 4
−
6
wherex andx arearbitraryelementsofZ . Hencetherearefoursolutions:
4 5 2
x x x x x
1 2 3 4 5
0 0 0 0 0
1 1 0 0 1 .
1 1 1 1 0
0 0 1 1 1
13. (a) We reduce the augmented matrix to reduced row–echelon form:
2 1 3 4 1 3 4 2
4 1 4 1 R 3R 4 1 4 1
1 1
  →  
3 1 2 0 3 1 2 0
   
1 3 4 2 1 3 4 2
R R +R
R32→→R32+2R11  00 42 30 34  R2 → 4R2  00 12 20 24 
   
1 0 3 1 1 0 0 1
R R +2R R R +2R
1 → 1 2 0 1 2 2 1 → 1 3 0 1 0 2 .
R3 → R3+3R2  0 0 1 0  R2 → R2+3R3  0 0 1 0 
   
Consequently the system has the unique solution x = 1, y = 2, z = 0.
(b) Again we reduce the augmented matrix to reduced row–echelon form:
2 1 3 4 1 1 0 3
4 1 4 1 R R 4 1 4 1
1 3
  ↔  
1 1 0 3 2 1 3 4
   
1 1 0 3 1 1 0 3
R R +R
R32→→R32+3R11  00 24 43 43  R2 → 3R2  00 14 23 23 
   
1 0 3 1
R R +4R
1 → 1 2 0 1 2 2 .
R3 → R3+R2  0 0 0 0 
 
We read off the complete solution
x = 1 3z = 1+2z
−
y = 2 2z = 2+3z,
−
where z is an arbitrary element of Z .
5
7
14. Suppose that (α ,...,α ) and (β ,...,β ) are solutions of the system
1 n 1 n
of linear equations
n
a x = b , 1 i m.
ij j i
≤ ≤
j=1
X
Then
n n
a α = b and a β = b
ij j i ij j i
j=1 j=1
X X
for 1 i m.
≤ ≤
Let γ = (1 t)α +tβ for 1 i m. Then (γ ,...,γ ) is a solution of
i i i 1 n
− ≤ ≤
the given system. For
n n
a γ = a (1 t)α +tβ
ij j ij j j
{ − }
j=1 j=1
X X
n n
= a (1 t)α + a tβ
ij j ij j
−
j=1 j=1
X X
= (1 t)b +tb
i i
−
= b .
i
15. Suppose that (α ,...,α ) is a solution of the system of linear equations
1 n
n
a x = b , 1 i m. (1)
ij j i
≤ ≤
j=1
X
Then the system can be rewritten as
n n
a x = a α , 1 i m,
ij j ij j
≤ ≤
j=1 j=1
X X
or equivalently
n
a (x α ) = 0, 1 i m.
ij j j
− ≤ ≤
j=1
X
So we have
n
a y = 0, 1 i m.
ij j
≤ ≤
j=1
X
where x α = y . Hence x = α +y , 1 j n, where (y ,...,y ) is
j j j j j j 1 n
− ≤ ≤
a solution of the associated homogeneous system. Conversely if (y ,...,y )
1 n
8