1

1 In both cases, the system of equations can be written as

A Bxyz

= , so that we have xyz

= −A B1 .

a For:

The solution is:

x y z= − = =1 4 0, , .

b For:

The solution is:

x y z= = =1 1 2, , .

2 Matrix A is singular if det(A) = 0. So, in each case, we have:

a

det( )A a aa a

a a a

a

=−

= + =

=

2 2 2

2

2

2 0

Therefore, the matrixx is singular for a

Aa

a aa a

a

=

= −

=

−−

0

12

121

2

.

112

12

12

0a

a a

a

≠for

b

det( )Ba

a aa a a a

a

= = − ⋅ =

=

3

23 2

0

2

2 2

2

Therefore, the mattrix is singular for a

Ba

a aa

=

= −−

−

0

1

2 31

2

2

.

==−

−

≠1

1

2 30

2

a

a a

afor

c

det( )C e e

e ee e e e e

e

a a

a a

a a a a a

a

= = ⋅ − ⋅ = −

− =

−−

3 2

2

3 2 2 4

4

1

1 00 14 4 0⇒ = ⇒ =e e ea a

Therefore, the matrix is singularr for

for

a

Ce

e e

e eaa

a a

a a

=

=−

−−

−−

0

1

11

4

2

2 3

.

≠≠ 0

d det( ) sin coscos sin

sin cosD a aa a

a a= − = + =2 2 1

For matrix D, det(D) ≠ 0 for any real number a, so D is always regular (not singular).

D a a

a a− =

−

1 sin coscos sin

3 det( ) ( )

det( )

A

A

=−

= − − = −

= ⇒ −

32 1

3 1 2 3 5

2 3 5

2

2

x xx

x x x x x

x x == ⇒ − − =

⇒ = −

2 3 5 2 0

13

2

2

1 2

x x

x , ,

The possible values are: x = −1

3 or x = 2.

4 A k=−

33 1

a A k k k kk

223

3 13

3 19 3 3

3 3 10=

−

−

= + −−

b We are given that A2 13 33 10

=

. By comparing

the corresponding elements, we obtain the following equations:

k

k

2 9 13

3 3 3

+ =− =

Therefore, the unique solution is k = 2.

Practice questions

Chapter 5

2

Chapter 5

c The system can be written as:

A Axy

xy

=

⇒

=

−133

133

1 .

A− =

−

− −−

=− −

− −−

= −

1 1

2 33 1

1 33 2

12 9

1 33 2

111

−− −−

⇒

= − − −

−

1 33 2

111

1 33 2

133

xy

= − −−

=

111

2233

23

The solution is x = 2, y = 3.

5 MN =

3 38 8

and M =

2 13 5

N M=

= −−

=

−1 3 38 8

1

2 13 5

5 13 2

3 38 8

1

10 −−

=

=

3

7 77 7

1

77 77 7

1 11 1

6

A kI k kk

− = −

−

= − −−

4 13 5

1 00 1

4 13 5

det(( ) ( )( )A kI k k k k− = − − + = − +4 5 3 9 232

A – kl is singular if det(A – kl ) = 0. So, k should be the solution of the quadratic equation k k2 9 23 0− + = .However, for this equation Δ = –11, so there is no real solution.

7 a

ABm

n

m n

=−−

− −

− −

=−

1 12 3 21 2 2

2 0 12 01 1 1

−− −− −

+

=

m mn

n

10 3 2 00 2 2 1

1 0 00 1 00 0 1

By comparing the corresponding elements, we obtain the following system of equations:

m

n m

m

n

n

=− − =− =

− − =+ =

1

0

1 0

3 2 1

2 2 0 The solution is m n= = −1 1, .

b The system of equations can be written as

Axyz

=−

497

. So:

xyz

=−

=−

=

−A B1497

497

2 0 122 1 01 1 1

497

11

2−− −

−

= −

The solution is x = –1, y = –1, z = 2.

8 The system can be written as: Axy

=

712

, where

A mm

= −− +

4 32 1

. It is inconsistent if A is singular,

i.e. det(A) = 0.

Therefore:

det( ) ( )( )A mm

m m m m

m m

= −− +

= − + + = − + ⇒

− + =

4 32 1

4 1 6 3 2

3 2 0

2

2 ⇒⇒ = =m m1 21 2,

So, m = 1 or m = 2.

9

AB BA m n n m= ⇒

=

34 2

28 4

28 4

34 2

⇒ + ++

= +2 24 1224 4 8

2 4 2m mnn

m n nnm

++

68 16 32

By comparing the corresponding elements, we obtain the following system of equations:

2 24 2 4

12 2 6

8 16 24

4 8 32

m m n

mn n

m

n

+ = ++ = ++ =+ =

The solution m = 1, n = 6 satisfies all the equations.

10 A is singular if det(A) = 0. So:

det( )

( ) ( )

A =−

+ −−

= − − + + −[ ] +

1 2 21 2 2 0

1 2 1

1 2 0 2 1 2 0 2

xx

x x 22 1 2 2

8 42

( )+ −[ ]= +

x

x x

8 4 0 2 1 0 0

12

21 2x x x x x x+ = ⇒ + = ⇒ = = −( ) ,

Therefore, x = −1

2 or x = 0.

3

11

AB B C

m np q

m np q

+ =

+

=

−

2 17 4

2 93 8

+ ++ +

+

=

−

2 2

7 4 7 42 93 8

m p n q

m p n q

m np q

+ ++ +

=

−

3 3

7 5 7 52 93 8

m p n q

m p n q

By comparing the corresponding elements, we obtain two systems of linear equations (each with two variables):

3 2 3 9

7 5 3 7 5 8

138

238

378

m p n q

m p n q

m p n q

+ = + =+ = − + =

= = − = =, , −− 398

So, m = 13

8, n =

37

8, p = −

23

8, and q = −

39

8.

12 M aa

= +

6 1 23 1

a

det( ) ( )

det( )

M aa

a a

MM a a

= + = + − =

= −− +

−

6 1 23 1

6 1 6 1

1 1 23 6

1

111 23 6 1

= −− +

a a

b For a = 1, N P=−

= −

6 52 3

3 12 7

, :and

X X X

X

M N P M P N P N M+ = ⇒ = − ⇒ = −

= −

−−

−( ) 1

3 12 7

6 52 3

−−

= − −

−−

1 23 7

3 64 4

1 23 7

= −−

15 368 20

13 a

det( ) ( ) ( ) ( )Ma

a

a a a=−

= − − + + +1 32 2 1

2 31 6 2 3 6 4 2

det( )

= + −

= ⇒ + − = ⇒ + − =

2 14

7 2 14 7 2 21 0

2

2 2

a a

M a a a a

,⇒ = − =a a1 2

72

3

So, a = −7

2 or a = 3.

b We will use a GDC to find M–1, so we need to rename matrix M as A.

For a1

7

2= − :

For a2 3= :

c The system of equations is:

x y zx y zx y z

+ + =+ + =

− + + =

3 3 7

2 2 35

3 2 3 14

This can be written as:

M Mxyz

xyz

=

⇒

= −73514

71 335

14

The solution for:

and A M a− −= =1 1 3for is:

The solution is: x = −17, y = 61, z = −53.

14 det( ) ( )A =−

= ⋅ − − = +2 34

2 3 4 2 122xx x

x x x x x

det( )A = ⇒ + = ⇒ + − =14 2 12 14 6 7 02 2x x x x,⇒ = − =x x1 27 1

So, x = −7 or x = 1.

15 a M a a a a a2

222

2 12

2 12

2 14 2 2=

−

=−

−

= + −22 2 5a −

M a aa

225 4

4 54 2 2

2 2 55 44 5

= −−

⇒ + −−

= −−

⇒+ =− =

a

a

2 4 5

2 2 −− 4

b

The solution is a = −1.

M− =−

−

− −− −

= − − −− −

=1 1

1 22 1

1 22 1

1

31 22 1

1

31 222 1

4

Chapter 5

The system of equations can be written as:

M Mxy

xy

xy

= −

⇒

= −

⇒

−33

33

1

=

−

=−

=−

13

1 22 1

33

13

33

11 .

Therefore, the solution is x = 1, y = −1.

16 BA A=

=

11 244 8

5 22 0

and

A− = − −

−

1 1

40 22 5

Note: We will leave A−1 in this form as it is easier for subsequent calculations.

B A=

= −

−−

−11 244 8

14

11 244 8

0 22 5

1

= − − −− −

14

4 1216 48

=

1 34 12

17 A B a bc d

a bc d

X X+ = ⇒−

+

=−

3 15 6

4 80 3

⇒ + +− + − +

+

=−

3 35 6 5 6

4 80 3

a c b da c b d

a bc d

⇒ + +− + − +

=−

4 45 7 5 7

4 80 3

a c b da c b d

By comparing the corresponding elements, we obtain two systems of linear equations:

4 4 4 8

5 7 0 5 7 3

2833

2033

593

a c b d

a c b d

a c b

+ = + =− + = − + = −

= = =,33

2833

, d =

So, a = 28

33, b =

59

33, c =

20

33, and d =

28

33.

18 A = −

5 27 1

a A− =+ −

=−

=−

1 1

5 141 27 5

1

191 27 5

1

19

2

197

19

5

119

b i ( ) ( )i A B C A C B C B AX X X+ = ⇒ = − ⇒ = − −1

ii

X = −−

−−

−

5 08 7

6 75 2

119

219

719

519

= − −−

−

119

11 713 9

1 27 5

= −−

= −−

1

1938 5776 19

2 34 1

19 a

A B a bc d c

a bc d c

+ =

+

= + ++ +

1

1 2 1 21

b AB a bc d c

a bd a bcc d c

=

= + ++

1

1 2 23

20 a Using a GDC:

b The system of equations can be written as:

A Axyz

xyz

=

⇒

= −123

123

1

For A as above and

we have:

The solution is x y z= = =65

35

85

, , .

21 Given C Da

Q C D= −

=−

= −2 41 7

5 21

3 2, , ,and

we have:

a

Q C Da

= −( ) = −

−−

= −

1

32

1

32 2 4

1 75 21

1

399 6

3 14

3 2

114

3−

=−

−

a

a

b

CDa

aa

= −

−

= − − − +− +

2 41 7

5 21

10 4 4 45 7 2 7

= − −− +

14 4 42 7 2

aa

c Da a

a− =−

=+

−

1 5 21

1

5 22

1 5