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CONTENTS
- PROBLEMS 1.6
- PROBLEMS 2.4
- PROBLEMS 2.7
- PROBLEMS 3.6
- PROBLEMS 4.1
- PROBLEMS 5.8
- PROBLEMS 6.3
- PROBLEMS 7.3
- PROBLEMS 8.8
SECTION 1. 6
- (i)
[
]
R 1 ↔ R 2
[
]
R 1 → 12 R 1
[
]
(ii)
[
]
R 1 ↔ R 2
[
]
R 1 → R 1 − 2 R 2
[
]
(iii)
R^2 →^ R^2 −^ R^1
R 3 → R 3 − R 1
R 1 → R 1 + R 3
R 3 → −R 3
R 2 ↔ R 3
R^2 →^ R^2 +^ R^3
R 3 → −R 3
(iv)
R^3 →^ R^3 + 2R^1
R 1 → 12 R 1
- (a)
R^2 →^ R^2 −^2 R^1
R 3 → R 3 − R 1
R 1 → R 1 − R 2
R 3 → R 3 + 2R 2
R 3 → −^1
8 R^3
R 1 → R 1 − 4 R 3
R 2 → R 2 + 3R 3
The augmented matrix has been converted to reduced row–echelon form and we read off the unique solution x = − 3 , y = 194 , z = 14.
(b)
R^2 →^ R^2 −^3 R^1
R 3 → R 3 + 5R 1
R 3 → R 3 + 2R 2
From the last matrix we see that the original system is inconsistent.
Case 1. t 6 = 2. No solution.
Case 2. t = 2. B =
We read off the unique solution x = 1, y = 0.
- Method 1.
R 1 → R 1 − R 4
R 2 → R 2 − R 4
R 3 → R 3 − R 4
R^4 →^ R^4 −^ R^3 −^ R^2 −^ R^1
Hence the given homogeneous system has complete solution
x 1 = x 4 , x 2 = x 4 , x 3 = x 4 ,
with x 4 arbitrary.
Method 2. Write the system as
x 1 + x 2 + x 3 + x 4 = 4 x 1 x 1 + x 2 + x 3 + x 4 = 4 x 2 x 1 + x 2 + x 3 + x 4 = 4 x 3 x 1 + x 2 + x 3 + x 4 = 4 x 4.
Then it is immediate that any solution must satisfy x 1 = x 2 = x 3 = x 4. Conversely, if x 1 , x 2 , x 3 , x 4 satisfy x 1 = x 2 = x 3 = x 4 , we get a solution.
[ λ − 3 1 1 λ − 3
]
R 1 ↔ R 2
[
1 λ − 3 λ − 3 1
]
R 2 → R 2 − (λ − 3)R 1
[
1 λ − 3 0 −λ^2 + 6λ − 8
]
= B.
Case 1: −λ^2 + 6λ − 8 6 = 0. That is −(λ − 2)(λ − 4) 6 = 0 or λ 6 = 2, 4. Here B is
row equivalent to
[
]
R 2 → (^) −λ (^2) +6^1 λ− 8 R 2
[
1 λ − 3 0 1
]
R 1 → R 1 − (λ − 3)R 2
[
]
Hence we get the trivial solution x = 0, y = 0.
Case 2: λ = 2. Then B =
[
]
and the solution is x = y, with y arbitrary.
Case 3: λ = 4. Then B =
[
]
and the solution is x = −y, with y arbitrary.
[
]
R 1 →
R 1
[
]
R 2 → R 2 − 5 R 1
[
0 −^83 −^23 −^83
]
R 2 →
R 2
[
]
R 1 → R 1 −
R 2
[
]
Hence the solution of the associated homogeneous system is
x 1 = −
x 3 , x 2 = −
x 3 − x 4 ,
with x 3 and x 4 arbitrary.
A =
1 − n 1 · · · 1 1 1 − n · · · 1 .. .
1 1 · · · 1 − n
R 1 → R 1 − Rn R 2 → R 2 − Rn .. . Rn− 1 → Rn− 1 − Rn
−n 0 · · · n 0 −n · · · n .. .
1 1 · · · 1 − n
1 1 · · · 1 − n
Rn → Rn − Rn− 1 · · · − R 1
R 3 → R 3 − R 2
R 2 → − 71 R 2
R 1 → R 1 − 2 R 2
0 0 a^2 − 16 a − 4
R 1 → R 1 − 2 R 2
0 0 a^2 − 16 a − 4
Denote the last matrix by B.
Case 1: a^2 − 16 6 = 0. i.e. a 6 = ±4. Then
R 3 → (^) a (^2) −^116 R 3 R 1 → R 1 − R 3 R 2 → R 2 + 2R 3
(^1 0 0) 7(^8 aa+25+4) (^0 1 0 10) 7(aa+54+4) (^0 0 1) a+4^1
and we get the unique solution
x =
8 a + 25 7(a + 4)
, y =
10 a + 54 7(a + 4)
, z =
a + 4
Case 2: a = −4. Then B =
, so our system is inconsistent.
Case 3: a = 4. Then B =
. We read off that the system is
consistent, with complete solution x = 87 − z, y = 107 + 2z, where z is arbitrary.
- We reduce the augmented array of the system to reduced row–echelon form: (^)
R^3 →^ R^3 +^ R^1
R 3 → R 3 + R 2
R 1 → R 1 + R 4
R 3 ↔ R 4
The last matrix is in reduced row–echelon form and we read off the solution of the corresponding homogeneous system:
x 1 = −x 4 − x 5 = x 4 + x 5 x 2 = −x 4 − x 5 = x 4 + x 5 x 3 = −x 4 = x 4 ,
where x 4 and x 5 are arbitrary elements of Z 2. Hence there are four solutions:
x 1 x 2 x 3 x 4 x 5 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 0 1 1 1
- (a) We reduce the augmented matrix to reduced row–echelon form:
R 1 → 3 R 1
R 2 → R 2 + R 1
R 3 → R 3 + 2R 1
R 2 → 4 R 2
R 1 → R 1 + 2R 2
R 3 → R 3 + 3R 2
R^1 →^ R^1 + 2R^3
R 2 → R 2 + 3R 3
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:
R 1 ↔ R 3
R 2 → R 2 + R 1
R 3 → R 3 + 3R 1
R 2 → 3 R 2
or equivalently ∑n
j=
aij (xj − αj ) = 0, 1 ≤ i ≤ m.
So we have (^) n ∑
j=
aij yj = 0, 1 ≤ i ≤ m.
where xj − αj = yj. Hence xj = αj + yj , 1 ≤ j ≤ n, where (y 1 ,... , yn) is a solution of the associated homogeneous system. Conversely if (y 1 ,... , yn) is a solution of the associated homogeneous system and xj = αj + yj , 1 ≤ j ≤ n, then reversing the argument shows that (x 1 ,... , xn) is a solution of the system 1.
- We simplify the augmented matrix using row operations, working towards row–echelon form:
a 1 1 1 b 3 2 0 a 1 + a
R^2 →^ R^2 −^ aR^1 R 3 → R 3 − 3 R 1
0 1 − a 1 + a 1 − a b − a 0 − 1 3 a − 3 a − 2
R 2 ↔ R 3
R 2 → −R 2
0 1 − 3 3 − a 2 − a 0 1 − a 1 + a 1 − a b − a
R 3 → R 3 + (a − 1)R 2
0 1 − 3 3 − a 2 − a 0 0 4 − 2 a (1 − a)(a − 2) −a^2 + 2a + b − 2
= B.
Case 1: a 6 = 2. Then 4 − 2 a 6 = 0 and
B →
0 1 − 3 3 − a 2 − a 0 0 1 a− 21 −a
(^2) +2a+b− 2 4 − 2 a
Hence we can solve for x, y and z in terms of the arbitrary variable w.
Case 2: a = 2. Then
B =
0 0 0 0 b − 2
Hence there is no solution if b 6 = 2. However if b = 2, then
B =
and we get the solution x = 1 − 2 z, y = 3z − w, where w is arbitrary.
- (a) We first prove that 1 + 1 + 1 + 1 = 0. Observe that the elements
1 + 0, 1 + 1, 1 + a, 1 + b
are distinct elements of F by virtue of the cancellation law for addition. For this law states that 1 + x = 1 + y ⇒ x = y and hence x 6 = y ⇒ 1 + x 6 = 1 + y.
Hence the above four elements are just the elements 0, 1 , a, b in some order. Consequently
(1 + 0) + (1 + 1) + (1 + a) + (1 + b) = 0 + 1 + a + b (1 + 1 + 1 + 1) + (0 + 1 + a + b) = 0 + (0 + 1 + a + b),
so 1 + 1 + 1 + 1 = 0 after cancellation. Now 1 + 1 + 1 + 1 = (1 + 1)(1 + 1), so we have x^2 = 0, where x = 1 + 1. Hence x = 0. Then a + a = a(1 + 1) = a · 0 = 0. Next a + b = 1. For a + b must be one of 0, 1 , a, b. Clearly we can’t have a + b = a or b; also if a + b = 0, then a + b = a + a and hence b = a; hence a + b = 1. Then
a + 1 = a + (a + b) = (a + a) + b = 0 + b = b.
Similarly b + 1 = a. Consequently the addition table for F is
- 0 1 a b 0 0 1 a b 1 1 0 b a a a b 0 1 b b a 1 0
We now find the multiplication table. First, ab must be one of 1, a, b; however we can’t have ab = a or b, so this leaves ab = 1.
Section 2. 4
- Suppose B =
a b c d e f
(^) and that AB = I 2. Then
[
]
a b c d e f
[
]
[
−a + e −b + f c + e d + f
]
Hence −a + e = 1 c + e = 0
−b + f = 0 d + f = 1
e = a + 1 c = −e = −(a + 1)
f = b d = 1 − f = 1 − b
B =
a b −a − 1 1 − b a + 1 b
Next,
(BA)^2 B = (BA)(BA)B = B(AB)(AB) = BI 2 I 2 = BI 2 = B
- Let pn denote the statement
An^ = (
n−1) 2 A^ +^
(3− 3 n) 2 I^2.
Then p 1 asserts that A = (3− 2 1)A + (3− 2 3)I 2 , which is true. So let n ≥ 1 and assume pn. Then from (1),
An+1^ = A · An^ = A
(3n−1) 2 A^ +^
(3− 3 n) 2 I^2
n−1) 2 A
(^2) + (3−^3 n) 2 A = (
n−1) 2 (4A^ −^3 I^2 ) +^
(3− 3 n) 2 A^ =^
(3n−1)4+(3− 3 n) 2 A^ +^
(3n−1)(−3) 2 I^2 = (4·^3
n− 3 n)− 1 2 A^ +^
(3− 3 n+1) 2 I^2 = (
n+1−1) 2 A^ +^
(3− 3 n+1) 2 I^2.
Hence pn+1 is true and the induction proceeds.
- The equation xn+1 = axn + bxn− 1 is seen to be equivalent to
[ xn+ xn
]
[
a b 1 0
] [
xn xn− 1
]
or Xn = AXn− 1 ,
where Xn =
[
xn+ xn
]
and A =
[
a b 1 0
]
. Then
Xn = AnX 0
if n ≥ 1. Hence by Question 3, [ xn+ xn
]
(3n^ − 1) 2
A +
(3 − 3 n) 2
I 2
} [
x 1 x 0
]
(3n^ − 1) 2
[
]
[ (^3) − 3 n 2 0 0 3 −^3 n 2
]} [
x 1 x 0
]
(3n^ − 1)2 + 3 −^3
n 2 (
n (^) − 1)(−3)
3 n− 1 2
3 − 3 n 2
[
x 1 x 0
]
Hence, equating the (2, 1) elements gives
xn =
(3n^ − 1) 2
x 1 +
(3 − 3 n) 2
x 0 if n ≥ 1
- Note: λ 1 + λ 2 = a + d and λ 1 λ 2 = ad − bc. Then
(λ 1 + λ 2 )kn − λ 1 λ 2 kn− 1 = (λ 1 + λ 2 )(λn 1 −^1 + λn 1 −^2 λ 2 + · · · + λ 1 λn 2 −^2 + λn 2 −^1 )
−λ 1 λ 2 (λn 1 − 2 + λn 1 −^3 λ 2 + · · · + λ 1 λn 2 −^3 + λn 2 −^2 )
= (λn 1 + λn 1 −^1 λ 2 + · · · + λ 1 λn 2 −^1 ) +(λn 1 −^1 λ 2 + · · · + λ 1 λn 2 −^1 + λn 2 ) −(λn 1 −^1 λ 2 + · · · + λ 1 λn 2 −^1 ) = λn 1 + λn 1 −^1 λ 2 + · · · + λ 1 λn 2 −^1 + λn 2 = kn+
Hence
An^ =
3 n^ + (−1)n+ 4
A − (−3)
3 n−^1 + (−1)n 4
I 2
3 n^ + (−1)n+ 4
[
]
3 n−^1 + (−1)n 4
} [
]
which is equivalent to the stated result.
- In terms of matrices, we have [ Fn+ Fn
]
[
] [
Fn Fn− 1
]
for n ≥ 1.
[ Fn+ Fn
]
[
]n [ F 1 F 0
]
[
]n [ 1 0
]
Now λ 1 , λ 2 are the roots of the polynomial x^2 − x + 1 here. Hence λ 1 = 1+
√ 5 2 and^ λ^2 =^
1 −√ 5 2 and
kn =
1+√ 5 2
)n− 1 −
1 −√ 5 2
)n− 1
1+ √ 5 2 −
1 − √ 5 2
1+ √ 5 2
)n− 1 −
1 − √ 5 2
)n− 1
√ 5
Hence
An^ = knA − λ 1 λ 2 kn− 1 I 2 = knA − kn− 1 I 2
So [ Fn+ Fn
]
= (knA − kn− 1 I 2 )
[
]
= kn
[
]
− kn− 1
[
]
[
kn − kn− 1 kn
]
Hence
Fn = kn =
1+ √ 5 2
)n− 1 −
1 − √ 5 2
)n− 1
√ 5
- From Question 5, we know that [ xn yn
]
[
1 r 1 1
]n [ a b
]
Now by Question 7, with A =
[
1 r 1 1
]
An^ = knA − λ 1 λ 2 kn− 1 I 2 = knA − (1 − r)kn− 1 I 2 ,
where λ 1 = 1 +
r and λ 2 = 1 −
r are the roots of the polynomial x^2 − 2 x + (1 − r) and
kn =
λn 1 − λn 2 2
r
Hence [ xn yn
]
= (knA − (1 − r)kn− 1 I 2 )
[
a b
]
([
kn knr kn kn
]
[
(1 − r)kn− 1 0 0 (1 − r)kn− 1
]) [
a b
]
[
kn − (1 − r)kn− 1 knr kn kn − (1 − r)kn− 1
] [
a b
]
[
a(kn − (1 − r)kn− 1 ) + bknr akn + b(kn − (1 − r)kn− 1 )
]
Hence, in view of the fact that
kn kn− 1
λn 1 − λn 2 λn 1 −^1 − λn 2 −^1
λn 1 (1 − { λ λ^21 }n) λn 1 −^1 (1 − { λ λ^21 }n−^1 )
→ λ 1 , as n → ∞,
we have [ xn yn
]
a(kn − (1 − r)kn− 1 ) + bknr akn + b(kn − (1 − r)kn− 1 )
=
a( (^) kknn− 1 − (1 − r)) + b (^) kknn− 1 r a (^) kknn− 1 + b( (^) kknn− 1 − (1 − r))
→
a(λ 1 − (1 − r)) + bλ 1 r aλ 1 + b(λ 1 − (1 − r))
16
Section 2. 7
1. [A|I 2 ] =
[
]
R 2 → R 2 + 3R 1
[
]
R 2 → 131 R 2
[
]
R 1 → R 1 − 4 R 2
[
]
Hence A is non–singular and A−^1 =
[
]
Moreover E 12 (−4)E 2 (1/13)E 21 (3)A = I 2 ,
so A−^1 = E 12 (−4)E 2 (1/13)E 21 (3).
Hence
A = {E 21 (3)}−^1 {E 2 (1/13)}−^1 {E 12 (−4)}−^1 = E 21 (−3)E 2 (13)E 12 (4).
- Let D = [dij ] be an m × m diagonal matrix and let A = [ajk] be an m × n matrix. Then
(DA)ik =
∑^ n
j=
dij ajk = diiaik,
as dij = 0 if i 6 = j. It follows that the ith row of DA is obtained by multiplying the ith row of A by dii. Similarly, post–multiplication of a matrix by a diagonal matrix D re- sults in a matrix whose columns are those of A, multiplied by the respective diagonal elements of D. In particular,
diag (a 1 ,... , an)diag (b 1 ,... , bn) = diag (a 1 b 1 ,... , anbn),
as the left–hand side can be regarded as pre–multiplication of the matrix diag (b 1 ,... , bn) by the diagonal matrix diag (a 1 ,... , an). Finally, suppose that each of a 1 ,... , an is non–zero. Then a− 1 1 ,... , a− n^1 all exist and we have
diag (a 1 ,... , an)diag (a− 1 1 ,... , a− n 1 ) = diag (a 1 a− 1 1 ,... , ana− n 1 ) = diag (1,... , 1) = In.
Hence diag (a 1 ,... , an) is non–singular and its inverse is diag (a− 1 1 ,... , a− n 1 ). Next suppose that ai = 0. Then diag (a 1 ,... , an) is row–equivalent to a matix containing a zero row and is hence singular.
3. [A|I 3 ] =
R 1 ↔ R 2
R 3 → R 3 − 3 R 1
R 2 ↔ R 3
R 3 → 12 R 3
R 1 → R 1 − 2 R 2
R 1 → R 1 − 24 R 3
R 2 → R 2 + 9R 3
Hence A is non–singular and A−^1 =
Also
E 23 (9)E 13 (−24)E 12 (−2)E 3 (1/2)E 23 E 31 (−3)E 12 A = I 3.
Hence A−^1 = E 23 (9)E 13 (−24)E 12 (−2)E 3 (1/2)E 23 E 31 (−3)E 12 ,
so A = E 12 E 31 (3)E 23 E 3 (2)E 12 (2)E 13 (24)E 23 (−9).
A =
1 2 k 3 − 1 1 5 3 − 5
1 2 k 0 − 7 1 − 3 k 0 − 7 − 5 − 5 k
1 2 k 0 − 7 1 − 3 k 0 0 − 6 − 2 k
= B.
Hence if − 6 − 2 k 6 = 0, i.e. if k 6 = −3, we see that B can be reduced to I 3 and hence A is non–singular.
