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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
• PROBLEMS 2.4
• PROBLEMS 2.7
• PROBLEMS 3.6
• PROBLEMS 4.1
• PROBLEMS 5.8
• PROBLEMS 6.3
• PROBLEMS 7.3
• PROBLEMS 8.8

(c)

 R^1 ↔^ R^3

R 2 → R 2 − 2 R 1

R 3 → R 3 − 3 R 1

R 4 → R 4 − 6 R 1

R 1 → R 1 + R 2

R 4 → R 4 − R 3

R 3 → R 3 − 2 R 2

The augmented matrix has been converted to reduced row–echelon form and we read off the complete solution x = − 12 − 3 z, y = − 32 − 2 z, with z arbitrary.

2 − 1 3 a 3 1 − 5 b − 5 − 5 21 c

 R 2 → R 2 − R 1

2 − 1 3 a 1 2 − 8 b − a − 5 − 5 21 c

R 1 ↔ R 2

1 2 − 8 b − a 2 − 1 3 a − 5 − 5 21 c

 R^2 →^ R^2 −^2 R^1

R 3 → R 3 + 5R 1

1 2 − 8 b − a 0 − 5 19 − 2 b + 3a 0 5 − 19 5 b − 5 a + c

R 3 → R 3 + R 2

R 2 → − 51 R 2

1 2 − 8 b − a 0 1 − 519 2 b− 53 a 0 0 0 3 b − 2 a + c

R 1 → R 1 − 2 R 2

1 0 − 52 (b+ 5 a) 0 1 − 519 2 b− 53 a 0 0 0 3 b − 2 a + c

From the last matrix we see that the original system is inconsistent if 3 b − 2 a + c 6 = 0. If 3b − 2 a + c = 0, the system is consistent and the solution is

x = (b + a) 5

z, y = (2b − 3 a) 5

z,

where z is arbitrary.

t 1 t 1 + t 2 3

 R^2 →^ R^2 −^ tR^1 R 3 → R 3 − (1 + t)R 1

0 1 − t 0 0 1 − t 2 − t

R 3 → R 3 − R 2

0 1 − t 0 0 0 2 − t

 = B.

Case 1. t 6 = 2. No solution.

Case 2. t = 2. B =

We read off the unique solution x = 1, y = 0.

6. 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.

with xn arbitrary. Alternatively, writing the system in the form

x 1 + · · · + xn = nx 1 x 1 + · · · + xn = nx 2 .. . x 1 + · · · + xn = nxn

shows that any solution must satisfy nx 1 = nx 2 = · · · = nxn, so x 1 = x 2 = · · · = xn. Conversely if x 1 = xn,... , xn− 1 = xn, we see that x 1 ,... , xn is a solution.

10. Let A =

[

a b c d

]

and assume that ad − bc 6 = 0.

Case 1: a 6 = 0. [ a b c d

]

R 1 → (^1) a R 1

[

1 b a c d

]

R 2 → R 2 − cR 1

[

(^1) ab 0 ad− abc

]

R 2 → (^) ada−bc R 2

[

(^1) ab 0 1

]

R 1 → R 1 − (^) ab R 2

[

]

Case 2: a = 0. Then bc 6 = 0 and hence c 6 = 0.

A =

[

0 b c d

]

R 1 ↔ R 2

[

c d 0 b

]

[

1 d c 0 1

]

[

]

So in both cases, A has reduced row–echelon form equal to

[

]

11. We simplify the augmented matrix of the system using row operations:  

4 1 a^2 − 14 a + 2

 R^2 →^ R^2 −^3 R^1

R 3 → R 3 − 4 R 1

0 − 7 a^2 − 2 a − 14

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.

12. 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 ,

14. Suppose that (α 1 ,... , αn) and (β 1 ,... , βn) are solutions of the system of linear equations ∑n

j=

aij xj = bi, 1 ≤ i ≤ m.

Then (^) n ∑

j=

aij αj = bi and

∑^ n

j=

aij βj = bi

for 1 ≤ i ≤ m. Let γi = (1 − t)αi + tβi for 1 ≤ i ≤ m. Then (γ 1 ,... , γn) is a solution of the given system. For

∑^ n

j=

aij γj =

∑^ n

j=

aij {(1 − t)αj + tβj }

∑^ n

j=

aij (1 − t)αj +

∑^ n

j=

aij tβj

= (1 − t)bi + tbi = bi.

15. Suppose that (α 1 ,... , αn) is a solution of the system of linear equations

∑^ n

j=

aij xj = bi, 1 ≤ i ≤ m. (1)

Then the system can be rewritten as

∑^ n

j=

aij xj =

∑^ n

j=

aij αj , 1 ≤ i ≤ m,

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.

16. We simplify the augmented matrix using row operations, working to- wards 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.

17. (a) We first prove that 1 + 1 + 1 + 1 = 0. Observe that the elements

1 + 0, 1 + 1, 1 + a, 1 + b

R 2 ↔ R 3

1 a b a 0 b a 0 0 0 a a

 R^2 →^ aR^2 R 3 → bR 3

1 a b a 0 1 b 0 0 0 1 1

R 1 ↔ R 1 + aR 2

1 0 a a 0 1 b 0 0 0 1 1

 R^1 →^ R^1 +^ aR^3 R 2 → R 2 + bR 3

0 1 0 b 0 0 1 1

The last matrix is in reduced row–echelon form.

Section 2. 4

2. 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

4. 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.

5. The equation xn+1 = axn + bxn− 1 is seen to be equivalent to [ xn+ xn

]

[

a b 1 0

] [

xn xn− 1

]

If λ 1 6 = λ 2 , we see that

(λ 1 − λ 2 )kn = (λ 1 − λ 2 )(λn 1 −^1 + λn 1 −^2 λ 2 + · · · + λ 1 λn 2 −^2 + λn 2 − 1 ) = 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 − λn 2.

Hence kn = λ

n 1 −λn 2 λ 1 −λ 2.

We have to prove An^ = knA − λ 1 λ 2 kn− 1 I 2. ∗ n=1:

A^1 = A; also k 1 A − λ 1 λ 2 k 0 I 2 = k 1 A − λ 1 λ 20 I 2 = A.

Let n ≥ 1 and assume equation ∗ holds. Then

An+1^ = An^ · A = (knA − λ 1 λ 2 kn− 1 I 2 )A = knA^2 − λ 1 λ 2 kn− 1 A.

Now A^2 = (a + d)A − (ad − bc)I 2 = (λ 1 + λ 2 )A − λ 1 λ 2 I 2. Hence

An+1^ = kn(λ 1 + λ 2 )A − λ 1 λ 2 I 2 − λ 1 λ 2 kn− 1 A = {kn(λ 1 + λ 2 ) − λ 1 λ 2 kn− 1 }A − λ 1 λ 2 knI 2 = kn+1A − λ 1 λ 2 knI 2 ,

and the induction goes through.

8. Here λ 1 , λ 2 are the roots of the polynomial x^2 − 2 x − 3 = (x − 3)(x + 1). So we can take λ 1 = 3, λ 2 = − 1. Then

kn = 3 n^ − (−1)n 3 − (−1)

3 n^ + (−1)n+ 4

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.

9. 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

10. From Question 5, we know that [ xn yn

]

[

1 r 1 1

]n [ a b

]

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

[

∣∣ 1 /^13 −^4 /^13

]

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).

2. 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 results 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.

If k = −3, then B =

 (^) = B and consequently A is singu-

lar, as it is row–equivalent to a matrix containing a zero row.