Chapt_08

Prévia do material em texto

88 Matrices
EXERCISES 8.1
Matrix Algebra
1. 2× 4 2. 3× 2 3. 3× 3 4. 1× 3 5. 3× 4
6. 8× 1 7. Not equal 8. Not equal 9. Not equal 10. Not equal
11. Solving x = y − 2, y = 3x− 2 we obtain x = 2, y = 4.
12. Solving x2 = 9, y = 4x we obtain x = 3, y = 12 and x = −3, t = −12.
13. c23 = 2(0)− 3(−3) = 9; c12 = 2(3)− 3(−2) = 12
14. c23 = 2(1)− 3(0) = 2; c12 = 2(−1)− 3(0) = −2
15. (a) A + B =
(
4− 2 5 + 6
−6 + 8 9− 10
)
=
(
2 11
2 −1
)
(b) B−A =
(−2− 4 6− 5
8 + 6 −10− 9
)
=
(−6 1
14 −19
)
(c) 2A + 3B =
(
8 10
−12 18
)
+
(−6 18
24 −30
)
=
(
2 28
12 −12
)
16. (a) A−B =
−2− 3 0 + 14− 0 1− 2
7 + 4 3 + 2
 =
−5 14 −1
11 5

(b) B−A =
 3 + 2 −1− 00− 4 2− 1
−4− 7 −2− 3
 =
 5 −1−4 1
−11 −5

(c) 2(A + B) = 2
 1 −14 3
3 1
 =
 2 −28 6
6 2

17. (a) AB =
(−2− 9 12− 6
5 + 12 −30 + 8
)
=
(−11 6
17 −22
)
(b) BA =
(−2− 30 3 + 24
6− 10 −9 + 8
)
=
(−32 27
−4 −1
)
(c) A2 =
(
4 + 15 −6− 12
−10− 20 15 + 16
)
=
(
19 −18
−30 31
)
373
8.1 Matrix Algebra
(d) B2 =
(
1 + 18 −6 + 12
−3 + 6 18 + 4
)
=
(
19 6
3 22
)
18. (a) AB =
 −4 + 4 6− 12 −3 + 8−20 + 10 30− 30 −15 + 20
−32 + 12 48− 36 −24 + 24
 =
 0 −6 5−10 0 5
−20 12 0

(b) BA =
(−4 + 30− 24 −16 + 60− 36
1− 15 + 16 4− 30 + 24
)
=
(
2 8
2 −2
)
19. (a) BC =
(
9 24
3 8
)
(b) A(BC) =
(
1 −2
−2 4
) (
9 24
3 8
)
=
(
3 8
−6 −16
)
(c) C(BA) =
(
0 2
3 4
) (
0 0
0 0
)
=
(
0 0
0 0
)
(d) A(B + C) =
(
1 −2
−2 4
) (
6 5
5 5
)
=
(−4 −5
8 10
)
20. (a) AB = ( 5 −6 7 )
 34
−1
 = (−16)
(b) BA =
 34
−1
 ( 5 −6 7 ) =
 15 −18 2120 −24 28
−5 6 −7

(c) (BA)C =
 15 −18 2120 −24 28
−5 6 −7

 1 2 40 1 −1
3 2 1
 =
 78 54 99104 72 132
−26 −18 −33

(d) Since AB is 1× 1 and C is 3× 3 the product (AB)C is not defined.
21. (a) ATA = ( 4 8 −10 )
 48
−10
 = (180)
(b) BTB =
 24
5
 ( 2 4 5 ) =
 4 8 108 16 20
10 20 25

(c) A + BT =
 48
−10
 +
 24
5
 =
 612
−5

22. (a) A + BT =
(
1 2
2 4
)
+
(−2 5
3 7
)
=
(−1 7
5 11
)
(b) 2AT −BT =
(
2 4
4 8
)
−
(−2 5
3 7
)
=
(
4 −1
1 1
)
374
8.1 Matrix Algebra
(c) AT (A−B) =
(
1 2
2 4
) (
3 −1
−3 −3
)
=
(−3 −7
−6 −14
)
23. (a) (AB)T =
(
7 10
38 75
)T
=
(
7 38
10 75
)
(b) BTAT =
(
5 −2
10 −5
) (
3 8
4 1
)
=
(
7 38
10 75
)
24. (a) AT + B =
(
5 −4
9 6
)
+
(−3 11
−7 2
)
=
(
2 7
2 8
)
(b) 2A + BT =
(
10 18
−8 12
)
+
(−3 −7
11 2
)
=
(
7 11
3 14
)
25.
(−4
8
)
−
(
4
16
)
+
(−6
9
)
=
(−14
1
)
26.
 63
−3
 +
−5−5
15
 +
−6−8
10
 =
 −5−10
22

27.
(−19
18
)
−
(
19
20
)
=
(−38
−2
)
28.
−717
−6
 +
−11
4
−
 28
−6
 =
−1010
4

29. 4× 5 30. 3× 2
31. AT =
(
2 −3
4 2
)
; (AT )T =
(
2 4
−3 2
)
= A 32. (A + B)T =
(
6 −6
14 10
)
= AT + BT
33. (AB)T =
(
16 40
−8 −20
)T
=
(
16 −8
40 −20
)
; BTAT =
(
4 2
10 5
) (
2 −3
4 2
)
=
(
16 −8
40 −20
)
34. (6A)T =
(
12 −18
24 12
)
= 6AT
35. B = AAT =
 2 16 3
2 5
 ( 2 6 2
1 3 5
)
=
 5 15 915 39 27
9 27 29
 = BT
36. Using Problem 33 we have (AAT )T = (AT )TAT = AAT , so that AAT is symmetric.
37. Let A =
(
1 0
0 0
)
and B =
(
0 0
0 1
)
. Then AB = 0.
38. We see that A �= B, but AC =
 2 3 44 6 8
6 9 12
 = BC.
39. Since (A+B)2 = (A+B)(A+B) �= A2+AB+BA+B2, and AB �= BA in general, (A+B)2 �= A2+2AB+B2.
40. Since (A + B)(A−B) = A2 −AB + BA−B2, and AB �= BA in general, (A + B)(A−B) �= A2 −B2.
41. a11x1 + a12x2 = b1; a21x1 + a22x2 = b2
42.
 2 6 11 2 −1
5 7 −4

x1x2
x3
 =
 7−1
9

375
8.1 Matrix Algebra
43. (x y )
(
a b/2
b/2 c
) (
x
y
)
=( ax + by/2 bx/2 + cy )
(
x
y
)
=( ax2 + bxy/2 + bxy/2 + cy2 )=( ax2 + bxy + cy2 )
44.
 0 −∂/∂z ∂/∂y∂/∂z 0 −∂/∂x
−∂/∂y ∂/∂x 0

PQ
R
 =
−∂Q/∂z + ∂R/∂y∂P/∂z − ∂R/∂x
−∂P/∂y + ∂Q/∂x
 = curl F
45. (a) MY
xy
z
 =
 cos γ sin γ 0− sin γ cos γ 0
0 0 1

xy
z
 =
 x cos γ + y sin γ−x sin γ + y cos γ
z
 =
xYyY
zY

(b) MR =
 cosβ 0 − sinβ0 1 0
sinβ 0 cosβ
; MP
 1 0 00 cosα sinα
0 − sinα cosα

(c) MP
 11
1
 =
 1 0 00 cos 30◦ sin 30◦
0 − sin 30◦ cos 30◦

 11
1
 =
 1 0 00 √32 12
0 − 12
√
3
2

 11
1
 =
 112 (√3 + 1)
1
2 (
√
3− 1)

MRMP
 11
1
 =
 cos 45
◦ 0 − sin 45◦
0 1 0
sin 45◦ 0 cos 45◦

 112 (√3 + 1)
1
2 (
√
3− 1)
 =

√
2
2 0 −
√
2
2
0 1 0√
2
2 0
√
2
2

 112 (√3 + 1)
1
2 (
√
3− 1)

=

1
4 (3
√
2−√6 )
1
2 (
√
3 + 1)
1
4 (
√
2 +
√
6 )

MY MRMP
 11
1
 =
 cos 60
◦ sin 60◦ 0
− sin 60◦ cos 60◦ 0
0 0 1


1
4 (3
√
2−√6 )
1
2 (
√
3 + 1)
1
4 (
√
2 +
√
6 )

=

1
2
√
3
2 0
−
√
3
2
1
2 0
0 0 1


1
4 (3
√
2−√6 )
1
2 (
√
3 + 1)
1
4 (
√
2 +
√
6 )
 =

1
8 (3
√
2−√6 + 6 + 2√3 )
1
8 (−3
√
6 + 3
√
2 + 2
√
3 + 2)
1
4 (
√
2 +
√
6 )

46. (a) LU =
(
1 0
1
2 1
) (
2 −2
0 3
)
=
(
2 −2
1 2
)
= A
(b) LU =
(
1 0
2
3 1
) (
6 2
0 − 13
)
=
(
6 2
4 1
)
= A
(c) LU =
 1 0 00 1 0
2 10 1

 1 −2 10 1 2
0 0 −21
 =
 1 −2 10 1 2
2 6 1
 = A
(d) LU =
 1 0 03 1 0
1 1 1

 1 1 10 −2 −1
0 0 1
 =
 1 1 13 1 2
1 −1 1
 = A
47. (a) AB =
(
A11 A12
A21 A22
) (
B1
B2
)
=
(
A11B1 + A12B2
A21B1 + A22B2
)
=
 17 433 75
−14 51

376
8.2 Systems of Linear Algebraic Equations
since
A11B1 + A12B2 =
(
13 25
−9 49
)
+
(
4 18
12 26
)
=
(
17 43
3 75
)
and
A21B1 + A22B2 = (−24 34 ) + ( 10 17 ) = (−14 51 ) .
(b) It is easier to enter smaller strings of numbers and the chance of error is decreased. Also, if the large matrix
has submatrices consisting of all zeros or diagonal matrices, these are easily entered without listing all of
the entries.
EXERCISES 8.2
Systems of Linear Algebraic Equations
1.
(
1 −1 11
4 3 −5
)
−4R1+R2−−−−−−→
(
1 −1 11
0 7 −49
)
1
7R2−−−−−−→
(
1 −1 11
0 1 −7
)
R3+R1−−−−−−→
(
1 0 4
0 1 −7
)
The solution is x1 = 4, x2 = −7.
2.
(
3 −2 4
1 −1 −2
)
R12−−−−−−→
(
1 −1 −2
3 −2 4
)
−3R1+R2−−−−−−→
(
1 −1 −2
0 1 10
)
R2+R1−−−−−−→
(
1 0 8
0 1 10
)
The solution is x1 = 8, x2 = 10.
3.
(
9 3 −5
2 −1 −1
)
1
9R1−−−−−−→
(
1 13 − 59
2 1 −1
)
−2R1+R2−−−−−−→
(
1 13 − 59
0 13
1
9
)
3R2−−−−−−→
(
1 13 − 59
0 1 13
)
− 13R2+R1−−−−−−→
(
1 0 − 23
0 1 13
)
The solution is x1 = − 23 , x2 = 13 .
4.
(
10 15 1
3 2 −1
)
1
10R1−−−−−−→
(
1 32
1
10
3 2 −1
)
−3R1+R2−−−−−−→
(
1 32
1
10
0 − 52 − 1310
)
− 25R2−−−−−−→
(
1 32
1
10
0 1 1325
)
− 32R2+R1−−−−−−→
(
1 0 − 1725
0 1 1325
)
The solution is x1 = − 1725 , x2 = 1325 .
5.
 1 −1 −1 −32 3 5 7
1 −2 3 −11
 −2R1+R2−−−−−−→−R1+R3
 1 −1 −1 −30 5 7 13
0 −1 4 −8
 15R2−−−−−−→
1 −1 −1 −30 1 75 135
0 −1 4 −8

R2+R1−−−−−−→
R2+R3
 1 0
2
5 − 25
0 1 75
13
5
0 0 275 − 275
 527R3−−−−−−→
 1 0
2
5 − 25
0 1 75
13
5
0 0 1 −1
 − 25R3+R1−−−−−−→
− 75R3+R2
 1 0 0 00 1 0 4
0 0 1 −1

The solution is x1 = 0, x2 = 4, x3 = −1.
377
8.2 Systems of Linear Algebraic Equations
6.
 1 2 −1 02 1 2 9
1 −1 1 3
 −2R1+R2−−−−−−→−R1+R3
 1 2 −1 00 −3 4 9
0 −3 2 3
 − 13R2−−−−−−→
 1 2 −1 00 1 − 43 −3
0 −3 2 3

−2R2+R1−−−−−−→
3R2+R3
 1 0
5
3 6
0 1 − 43 −3
0 0 −2 −6
 − 12R3−−−−−−→
 1 0
5
3 6
0 1 − 43 −3
0 0 1 3
 − 53R3+R1−−−−−−→
4
3R3+R2
 1 0 0 10 1 0 1
0 0 1 3

The solution is x1 = 1, x2 = 1, x3 = 3.
7.
(
1 1 1 0
1 1 3 0
)
−R1+R2−−−−−−→
(
1 1 1 0
0 0 2 0
)
Since x3 = 0, setting x2 = t we obtain x1 = −t, x2 = t, x3 = 0.
8.
(
1 2 −4 9
5 −1 2 1
)
−5R1+R2−−−−−−→
(
1 2 −4 9
0 −11 22 −44
)
− 111R2−−−−−−→
(
1 2 −4 9
0 1 −2 4
)
−2R2+R1−−−−−−→
(
1 0 0 1
0 1 −2 4
)
If x3 = t, the solution is x1 = 1, x2 = 4 + 2t, x3 = t
9.
 1 −1 −1 81 −1 1 3
−1 1 1 4
 row−−−−−−→
operations
 1 −1 −1 80 0 2 −5
0 0 0 12

Since the bottom row implies 0 = 12, the system is inconsistent.
10.
 3 1 44 3 −3
2 −1 11
 row−−−−−−→
operations
 1
1
3
4
3
0 1 −5
0 0 0

The solution is x1 = 3, x2 = −5.
11.
 2 2 0 0−2 1 1 0
3 0 1 0
 row−−−−−−→
operations
 1 1 0 00 1 13 0
0 0 1 0

The solution is x1 = x2 = x3 = 0.
12.
 1 −1 −2 02 4 5 0
6 0 −3 0
 row−−−−−−→
operations
 1 −1 −2 00 1 32 0
0 0 0 0

The solution is x1 = 12 t, x2 = − 32 t, x3 = t.
13.
 1 2 2 21 1 1 0
1 −3 −1 0
 row−−−−−−→
operations
 1 2 2 20 1 1 2
0 0 1 4

The solution is x1 = −2, x2 − 2, x3 = 4.
14.
 1 −2 1 23 −1 2 5
2 1 1 1
 row−−−−−−→
operations
 1 −2 1 20 1 − 15 − 15
0 0 0 −2

Since the bottom row implies 0 = −2, the system is inconsistent.
378
8.2 Systems of Linear Algebraic Equations
15.
 1 1 1 31 −1 −1 −1
3 1 1 5
 row−−−−−−→
operations
 1 1 1 30 1 1 2
0 0 0 0

If x3 = t the solution is x1 = 1, x2 = 2− t, x3 = t.
16.
 1 −1 −2 −1−3 −2 1 −7
2 3 1 8
 row−−−−−−→
operations
 1 −1 −2 −10 1 1 2
0 0 0 0

If x3 = t the solution is x1 = 1 + t, x2 = 2− t, x3 = t.
17.

1 0 1 −1 1
0 2 1 1 3
1 −1 0 1 −1
1 1 1 1 2
 row−−−−−−→operations

1 0 1 −1 1
0 1 12
1
2
3
2
0 0 1 −5 1
0 0 0 1 0

The solution is x1 = 0, x2 = 1, x3 = 1, x4 = 0.
18.

2 1 1 0 3
3 1 1 1 4
1 2 2 3 3
4 5 −2 1 16
 row−−−−−−→operations

1 12
1
2 0
3
2
0 1 1 −2 1
0 0 1 −1 −1
0 0 0 1 0

The solution is x1 = 1, x2 = 2, x3 = −1, x4 = 0.
19.

1 3 5 −1 1
0 1 1 −1 4
1 2 5 −4 −2
1 4 6 −2 6
 row−−−−−−→operations

1 3 5 −1 1
0 1 1 −1 4
0 0 1 −4 1
0 0 0 0 1

Since the bottom row implies 0 = 1, the system is inconsistent.
20.

1 2 0 1 0
4 9 1 12 0
3 9 6 21 0
1 3 1 9 0
 row−−−−−−→operations

1 2 0 1 0
0 1 1 8 0
0 0 1 −2 0
0 0 0 0 0

If x4 = t the solution is x1 = 19t, x2 = −10t, x3 = 2t, x4 = t.
21.
 1 1 1 4.2800.2 −0.1 −0.5 −1.978
4.1 0.3 0.12 1.686
 row−−−−−−→
operations
 1 1 1 4.280 1 2.333 9.447
0 0 1 4.1

The solution is x1 = 0.3, x2 = −0.12, x3 = 4.1.
22.
 2.5 1.4 4.5 2.61701.35 0.95 1.2 0.7545
2.7 3.05 −1.44 −1.4292
 row−−−−−−→
operations
 1 0.56 1.8 1.04680 1 −6.3402 −3.3953
0 0 1 0.28

The solution is x1 = 1.45, x2 = −1.62, x3 = 0.28.
23. From x1Na + x2H2O → x3NaOH + x4H2 we obtain the system x1 = x3, 2x2 = x3 + 2x4, x2 = x3. We see
that x1 = x2 = x3, so the second equation becomes 2x1 = x1 + 2x4 or x1 = 2x4. A solution of the system is
x1 = x2 = x3 = 2t, x4 = t. Letting t = 1 we obtain the balanced equation 2Na + 2H2O → 2NaOH + H2.
379
8.2 Systems of Linear Algebraic Equations
24. From x1KClO3 → x2KCl+x3O2 we obtain the system x1 = x2, x1 = x2, 3x1 = 2x3. Letting x3 = t we see that
a solution of the system is x1 = x2 = 23 t, x3 = t. Taking t = 3 we obtain the balanced equation
2KClO3 → 2KCl + 3O2.
25. From x1Fe3O4 + x2C → x3Fe + x4CO we obtain the system 3x1 = x3, 4x1 = x4, x2 = x4. Letting x1 = t we
see that x3 = 3t and x2 = x4 = 4t. Taking t = 1 we obtain the balanced equation
Fe3O4 + 4C → 3Fe + 4CO.
26. From x1C5H8 + x2O2 → x3CO2 + x4H2O we obtain the system 5x1 = x3, 8x1 = 2x4, 2x2 = 2x3 + x4. Letting
x1 = t we see that x3 = 5t, x4 = 4t, and x2 = 7t. Taking t = 1 we obtain the balanced equation
C5H8 + 7O2 → 5CO2 + 4H2O.
27. From x1Cu + x2HNO3 → x3Cu(NO3)2 + x4H2O + x5NO we obtain the system
x1 = 3, x2 = 2x4, x2 = 2x3 + x5, 3x2 = 6x3 + x4 + x5.
Letting x4 = t we see that x2 = 2t and
2t = 2x3 + x5
6t = 6x3 + t + x5
or
2x3 + x5 = 2t
6x3 + x5 = 5t.
Then x3 = 34 t and x5 =
1
2 t. Finally, x1 = x3 =
3
4 t. Taking t = 4 we obtain the balanced equation
3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO.
28. From x1Ca3(PO4)2 + x2H3PO4 → x3Ca(H2PO4)2 we obtain the system
3x1 = x3, 2x1 + x2 = 2x3, 8x1 + 4x2 = 8x3, 3x2 = 4x3.
Letting x1 = t we see from the first equation that x3 = 3t and from the fourth equation that x2 = 4t. These
choices also satisfy the second and third equations. Taking t = 1 we obtain the balanced equation
Ca3(PO4)2 + 4H3PO4 → 3Ca(H2PO4)2.
29. The system of equations is
−i1 + i2 − i3 = 0
10− 3i1 + 5i3 = 0
27− 6i2 − 5i3 = 0
or
−i1 + i2 − i3 = 0
3i1 − 5i3 = 10
6i2 + 5i3 = 27
Gaussian elimination gives−1 1 −1 03 0 −5 10
0 6 5 27
 row−−−−−−→
operations
 1 −1 1 00 1 −8/3 10/3
0 0 1 1/3
 .
The solution is i1 =
35
9
, i2 =
38
9
, i3 =
1
3
.
30. The system of equations is
i1 − i2 − i3 = 0
52− i1 − 5i2 = 0
−10i3 + 5i2 = 0
or
i1 − i2 − i3 = 0
i1 + 5i2 = 52
5i2 − 10i3 = 0
380
8.2 Systems of Linear Algebraic Equations
Gaussian elimination gives 1 −1 −1 01 5 0 52
0 5 −10 0
 row−−−−−−→
operations
 1 −1 −1 00 1 1/6 26/3
0 0 1 4
 .
The solution is i1 = 12, i2 = 8, i3 = 4.
31. Interchange row 1 and row in I3. 32. Multiply row 3 by c in I3.
33. Add c times row 2 to row 3 in I3. 34. Add row 4 to row 1 in I4.
35. EA =
 a21 a22 a23a11 a12 a13
a31 a32 a33
 36. EA =
 a11 a12 a13a21 a22 a23
ca31 ca32 ca33

37. EA =
 a11 a12 a13a21 a22 a23
ca21 + a31 ca22 + a32 ca23 + a33

38. E1E2A = E1
 a11 a12 a13a21 a22 a23
ca21 + a31 ca22 + a32 ca23 + a33
 =
 a21 a22 a23a11 a12 a13
ca21 + a31 ca22 + a32 ca23 + a33

39. The system is equivalent to (
1 0
1
2 1
) (
2 −2
0 3
)
X =
(
2
6
)
.
Letting
Y =
(
y1
y2
)
=
(
2 −2
0 3
)
X
we have (
1 0
1
2 1
) (
y1
y2
)
=
(
2
6
)
.
This implies y1 = 2 and 12y1 + y2 = 1 + y2 = 6 or y2 = 5. Then(
2 −2
0 3
) (
x1
x2
)
=
(
2
5
)
,
which implies 3x2 = 5 or x2 = 53 and 2x1 − 2x2 = 2x1 − 103 = 2 or x1 = 83 . The solution is X =
(
8
3 ,
5
3
)
.
40. The system is equivalent to (
1 0
2
3 1
) (
6 2
0 − 13
)
X =
(
1
−1
)
.
Letting
Y =
(
y1
y2
)
=
(
6 2
0 − 13
)
X
we have (
1 0
2
3 1
) (
y1
y2
)
=
(
1
−1
)
.
381
8.2 Systems of Linear Algebraic Equations
This implies y1 = 1 and 23y1 + y2 =
2
3 + y2 = −1 or y2 = − 53 . Then(
6 2
0 − 13
) (
x1
x2
)
=
(
1
− 53
)
,
which implies − 13x2 = − 53 or x2 = 5 and 6x1 + 2x2 = 6x1 + 10 = 1 or x1 = − 32 . The solution is X =
(− 32 , 5).
41. The system is equivalent to  1 0 00 1 0
2 10 1

 1 −2 10 1 2
0 0 −21
X =
 2−1
1
 .
LettingY =
 y1y2
y3
 =
 1 −2 10 1 2
0 0 −21
X
we have  1 0 00 1 0
2 10 1

 y1y2
y3
 =
 2−1
1
 .
This implies y1 = 2, y2 = −1, and 2y1 + 10y2 + y3 = 4− 10 + y3 = 1 or y3 = 7. Then 1 −2 10 1 2
0 0 −21

x1x2
x3
 =
 2−1
7
 ,
which implies −21x3 = 7 or x3 = − 13 , x2 +2x3 = x2− 23 = −1 or x2 = − 13 , and x1− 2x2 +x3 = x1 + 23 − 13 = 2
or x1 = 53 . The solution is X =
(
5
3 ,− 13 ,− 13
)
.
42. The system is equivalent to  1 0 03 1 0
1 1 1

 1 1 10 −2 −1
0 0 1
X =
 01
4
 .
Letting
Y =
 y1y2
y3
 =
 1 1 10 −2 −1
0 0 1
X
we have  1 0 03 1 0
1 1 1

 y1y2
y3
 =
 01
4
 .
This implies y1 = 0, 3y1 + y2 = y2 = 1, and y1 + y2 + y3 = 0 + 1 + y3 = 4 or y3 = 3. Then 1 1 10 −2 −1
0 0 1

x1x2
x3
 =
 01
3
 ,
which implies x3 = 3, −2x2 − x3 = −2x2 − 3 = 1 or x2 = −2, and x1 + x2 + x3 = x1 − 2 + 3 = 0 or x1 = −1.
The solution is X = (−1,−2, 3).
382
8.3 Rank of a Matrix
43. Using the Solve function in Mathematica we find x1 = −0.0717393 − 1.43084c, x2 = −0.332591 + 0.855709c,
x3 = c, where c is any real number
44. Using the Solve function in Mathematica we find x1 = c/3, x2 = 5c/6, x3 = c, where c is any real number
45. Using the Solve function in Mathematica we find x1 = −3.76993, x2 = −1.09071, x3 = −4.50461, x4 = −3.12221
46. Using the Solve function in Mathematica we find x1 = 83 − 73b+ 23c, x2 = 23 − 13b− 13c, x3 = −3, x4 = b, x5 = c,
where b and c are any real numbers.
EXERCISES 8.3
Rank of a Matrix
1.
(
3 −1
1 3
)
row−−−−−−→
operations
(
1 3
0 1
)
; The rank is 2.
2.
(
2 −2
0 0
)
row−−−−−−→
operations
(
1 −1
0 0
)
; The rank is 1.
3.
 2 1 36 3 9
−1 − 12 − 32
 row−−−−−−→
operations
 1
1
2
3
2
0 0 0
0 0 0
; The rank is 1.
4.
 1 1 2−1 2 4
−1 0 3
 row−−−−−−→
operations
 1 1 20 1 5
0 0 1
; The rank is 3.
5.
 1 1 11 0 4
1 4 1
 row−−−−−−→
operations
 1 1 10 1 −3
0 0 1
; The rank is 3.
6.
(
3 −1 2 0
6 2 4 5
)
row−−−−−−→
operations
(
1 − 13 23 0
0 1 0 54
)
; The rank is 2.
7.

1 −2
3 −6
7 −1
4 5
 row−−−−−−→operations

1 −2
0 1
0 0
0 0
; The rank is 2.
8.

1 −2 3 4
1 4 6 8
0 1 0 0
2 5 6 8
 row−−−−−−→operations

1 −2 3 4
0 1 0 0
0 0 1 43
0 0 0 0
; The rank is 3.
383
8.3 Rank of a Matrix
9.

0 2 4 2 2
4 1 0 5 1
2 1 23 3
1
3
6 6 6 12 0
 row−−−−−−→operations

1 12
1
3
3
2
1
6
0 1 43 1 − 13
0 0 1 0 2
0 0 0 0 0
; The rank is 3.
10.

1 −2 1 8 −1 1 1 6
0 0 1 3 −1 1 1 5
0 0 1 3 −1 2 10 8
0 0 0 0 0 1 1 3
1 −2 1 8 −1 1 2 6
 row−−−−−−→operations

1 −2 1 8 −1 1 1 6
0 0 1 3 −1 1 1 5
0 0 0 0 0 1 9 3
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0
; The rank is 4.
11.
 1 2 31 0 1
1 −1 5
 row−−−−−−→
operations
 1 2 30 1 1
0 0 1
;
Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.
12.

2 6 3
1 −1 4
3 2 1
2 5 4
 row−−−−−−→operations

1 −1 4
0 1 − 58
0 0 1
0 0 0

Since the rank of the matrix is 3 and there are 4 vectors, the vectors are linearly dependent.
13.
 1 −1 3 −11 −1 4 2
1 −1 5 7
 row−−−−−−→
operations
 1 −1 3 −10 0 1 3
0 0 0 1

Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.
14.

2 1 1 5
2 2 1 1
3 −1 6 1
1 1 1 −1
 row−−−−−−→operations

1 1 1 −1
0 1 1 −7
0 0 1 −3
0 0 0 1

Since the rank of the matrix is 4 and there are 4 vectors, the vectors are linearly independent.
15. Since the number of unknowns is n = 8 and the rank of the coefficient matrix is r = 3, the solution of the
system has n− r = 5 parameters.
16. (a) The maximum possible rank of A is the number of rows in A, which is 4.
(b) The system is inconsistent if rank(A) < rank(A/B) = 2 and consistent if rank(A) = rank(A/B) = 2.
(c) The system has n = 6 unknowns and the rank of A is r = 3, so the solution of the system has n − r = 3
parameters.
17. Since 2v1 + 3v2 − v3 = 0 we conclude that v1, v2, and v3 are linearly dependent. Thus, the rank of A is at
most 2.
18. Since the rank of A is r = 3 and the number of equations is n = 6, the solution of the system has n − r = 3
parameters. Thus, the solution of the system is not unique.
19. The system consists of 4 equations, so the rank of the coefficient matrix is at most 4, and the maximum number
of linearly independent rows is 4. However, the maximum number of linearly independent columns is the same
384
8.4 Determinants
as the maximum number of linearly independent rows. Thus, the coefficient matrix has at most 4 linearly
independent columns. Since there are 5 column vectors, they must be linearly dependent.
20. Using the RowReduce in Mathematica we find that the reduced row-echelon form of the augmented matrix is
1 0 0 0 0 8342215 − 261443
0 1 0 0 0 18182215
282
443
0 0 1 0 0 13443 − 6443
0 0 0 1 0 42142215 − 130443
0 0 0 0 1 − 60792215 677443

.
We conclude that the system is consistent and the solution is x1 = − 226443 − 8342215c, x2 = 282443 − 18182215c,
x3 = − 6443 − 13443c, x4 = − 130443 − 42142215c, x5 = 677433 + 60792215c, x6 = c.
EXERCISES 8.4
Determinants
1. M12 =
∣∣∣∣ 1 2−2 5
∣∣∣∣ = 9 2. M32 = ∣∣∣∣ 2 41 2
∣∣∣∣
3. C13 = (−1)1+3
∣∣∣∣ 1 −1−2 3
∣∣∣∣ = 1 4. C22 = (−1)2+2 ∣∣∣∣ 2 4−2 5
∣∣∣∣ = 18
5. M33 =
∣∣∣∣∣∣∣
0 2 0
1 2 3
1 1 2
∣∣∣∣∣∣∣ = 2 6. M41 =
∣∣∣∣∣∣∣
2 4 0
2 −2 3
1 0 −1
∣∣∣∣∣∣∣ = 24
7. C34 = (−1)3+4
∣∣∣∣∣∣∣
0 2 4
1 2 −2
1 1 1
∣∣∣∣∣∣∣ = 10 8. C23 = (−1)2+3
∣∣∣∣∣∣∣
0 2 0
5 1 −1
1 1 2
∣∣∣∣∣∣∣ = 22
9. −7 10. 2 11. 17 12. −1/2
13. (1− λ)(2− λ)− 6 = λ2 − 3λ− 4 14. (−3− λ)(5− λ)− 8 = λ2 − 2λ− 23
15.
∣∣∣∣∣∣∣
0 2 0
3 0 1
0 5 8
∣∣∣∣∣∣∣ = −3
∣∣∣∣ 2 05 8
∣∣∣∣ = −48 16.
∣∣∣∣∣∣∣
5 0 0
0 −3 0
0 0 2
∣∣∣∣∣∣∣ = 5
∣∣∣∣−3 00 2
∣∣∣∣ = 5(−3)(2) = −30
17.
∣∣∣∣∣∣∣
3 0 2
2 7 1
2 6 4
∣∣∣∣∣∣∣ = 3
∣∣∣∣ 7 16 4
∣∣∣∣ + 2 ∣∣∣∣ 2 72 6
∣∣∣∣ = 3(22) + 2(−2) = 62
385
8.4 Determinants
18.
∣∣∣∣∣∣∣
1 −1 −1
2 2 −2
1 1 9
∣∣∣∣∣∣∣ =
∣∣∣∣ 2 −21 9
∣∣∣∣− 2 ∣∣∣∣−1 −11 9
∣∣∣∣ + ∣∣∣∣−1 −12 −2
∣∣∣∣ = 20− 2(−8) + 4 = 40
19.
∣∣∣∣∣∣∣
4 5 3
1 2 3
1 2 3
∣∣∣∣∣∣∣ = 4
∣∣∣∣ 2 32 3
∣∣∣∣− 5 ∣∣∣∣ 1 31 3
∣∣∣∣ + 3 ∣∣∣∣ 1 21 2
∣∣∣∣ = 0
20.
∣∣∣∣∣∣∣
1
4 6 0
1
3 8 0
1
2 9 0
∣∣∣∣∣∣∣ = 0, expanding along the third column.
21.
∣∣∣∣∣∣∣
−2 −1 4
−3 6 1
−3 4 8
∣∣∣∣∣∣∣ = −2
∣∣∣∣ 6 14 8
∣∣∣∣ + 3 ∣∣∣∣−1 44 8
∣∣∣∣− 3 ∣∣∣∣−1 46 1
∣∣∣∣ = −2(44) + 3(−24)− 3(−25) = −85
22.
∣∣∣∣∣∣∣
3 5 1
−1 2 5
7 −4 10
∣∣∣∣∣∣∣ = 3
∣∣∣∣ 2 5−4 10
∣∣∣∣− 5 ∣∣∣∣−1 57 10
∣∣∣∣ + ∣∣∣∣−1 27 −4
∣∣∣∣ = 3(40)− 5(−45) + (−10) = 335
23.
∣∣∣∣∣∣∣
1 1 1
x y z
2 3 4
∣∣∣∣∣∣∣ =
∣∣∣∣ y z3 4
∣∣∣∣− ∣∣∣∣x z2 4
∣∣∣∣ + ∣∣∣∣x y2 3
∣∣∣∣ = (4y − 3z)− (4x− 2z) + (3x− 2y) = −x + 2y − z
24.
∣∣∣∣∣∣∣
1 1 1
x y z
2 + x 3 + y 4 + z
∣∣∣∣∣∣∣ =
∣∣∣∣ y z3 + y 4 + z
∣∣∣∣− ∣∣∣∣ x z2 + x 4 + z
∣∣∣∣ + ∣∣∣∣ x y2 + x 3 + y
∣∣∣∣
= (4y + yz − 3z − yz)− (4x + xz − 2z − xz) + (3x + xy − 2y − xy) = −x + 2y − z
25.
∣∣∣∣∣∣∣∣∣
1 1 −3 0
1 5 3 2
1 −2 1 0
4 8 0 0
∣∣∣∣∣∣∣∣∣ = 2
∣∣∣∣∣∣∣
1 1 −3
1 −2 1
4 8 0
∣∣∣∣∣∣∣ = 2(4)
∣∣∣∣ 1 −3−2 1
∣∣∣∣− 2(8) ∣∣∣∣ 1 −31 1
∣∣∣∣ = 8(−5)− 16(4) = −104
26.
∣∣∣∣∣∣∣∣∣
2 1 −2 1
0 5 0 4
1 6 1 0
5 −1 1 1
∣∣∣∣∣∣∣∣∣ = 5
∣∣∣∣∣∣∣
2 −2 1
1 1 0
5 1 1
∣∣∣∣∣∣∣ + 4
∣∣∣∣∣∣∣
2 1 −2
1 6 1
5 −1 1
∣∣∣∣∣∣∣ = 5(0) + 4(80) = 320
27. Expanding along the first column in the original matrix and each succeeding minor,we obtain 3(1)(2)(4)(2) = 48.
28. Expanding along the bottom row we obtain
−1
∣∣∣∣∣∣∣∣∣
2 0 0 −2
1 6 0 5
1 2 −1 1
2 1 −2 3
∣∣∣∣∣∣∣∣∣ +
∣∣∣∣∣∣∣∣∣
2 2 0 0
1 1 6 0
1 0 2 −1
2 0 1 −2
∣∣∣∣∣∣∣∣∣ = −1(−48) + 0 = 48.
29. Solving λ2 − 2λ− 15− 20 = λ2 − 2λ− 35 = (λ− 7)(λ + 5) = 0 we obtain λ = 7 and −5.
30. Solving −λ3 + 3λ2 − 2λ = −λ(λ− 2)(λ− 1) = 0 we obtain λ = 0, 1, and 2.
386
8.5 Properties of Determinants
EXERCISES 8.5
Properties of Determinants
1. Theorem 8.11 2. Theorem 8.14
3. Theorem 8.14 4. Theorem 8.12 and 8.11
5. Theorem 8.12 (twice) 6. Theorem 8.11 (twice)
7. Theorem 8.10 8. Theorem 8.12 and 8.9
9. Theorem 8.8 10. Theorem 8.11 (twice)
11. detA = −5 12. detB = 2(3)(5) = 30
13. detC = −5 14. detD = 5
15. detA = 6( 23 )(−4)(−5) = 80 16. detB = −a13a22a31
17. detC = (−5)(7)(3) = −105 18. detD = 4(7)(−2) = −56
19. detA = 14 = detAT 20. detA = 96 = det T
21. detAB =
∣∣∣∣∣∣∣
0 −2 2
10 7 23
8 4 16
∣∣∣∣∣∣∣ = −80 = 20(−4) = detA detB
22. From Problem 21, (detA)2 = detA2 = det I = 1, so detA = ±1.
23. Using Theorems 8.14, 8.12, and 8.9, detA =
∣∣∣∣∣∣∣
a 1 2
b 1 2
c 1 2
∣∣∣∣∣∣∣ = 2
∣∣∣∣∣∣∣
a 1 1
b 1 1
c 1 1
∣∣∣∣∣∣∣ = 0.
24. Using Theorems 8.14 and 8.9,
detA =
∣∣∣∣∣∣∣
1 1 1
x y z
x + y + z x + y + z x + y + z
∣∣∣∣∣∣∣ = (x + y + z)
∣∣∣∣∣∣∣
1 1 1
x y z
1 1 1
∣∣∣∣∣∣∣ = 0.
25.
∣∣∣∣∣∣∣
1 1 5
4 3 6
0 −1 1
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
1 1 5
0 −1 −14
0 −1 1
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
1 1 5
0 −1 −14
0 0 15
∣∣∣∣∣∣∣ = 1(−1)(15) = −15
26.
∣∣∣∣∣∣∣
2 4 5
4 2 0
8 7 −2
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
2 4 5
0 −6 −10
0 −9 −22
∣∣∣∣∣∣∣ = −2
∣∣∣∣∣∣∣
2 4 5
0 3 5
0 −9 −22
∣∣∣∣∣∣∣ = −2
∣∣∣∣∣∣∣
2 4 5
0 3 5
0 0 −7
∣∣∣∣∣∣∣ = −2(2)(3)(−7) = 84
387
8.5 Properties of Determinants
27.
∣∣∣∣∣∣∣
−1 2 3
4 −5 −2
9 −9 6
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
−1 2 3
0 3 10
0 9 33
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
−1 2 3
0 3 10
0 0 3
∣∣∣∣∣∣∣ = −1(3)(3) = −9
28.
∣∣∣∣∣∣∣
−2 2 −6
5 0 1
1 −2 2
∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣
1 −2 2
5 0 1
−2 2 −6
∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣
1 −2 2
0 10 −9
0 −2 −2
∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣
1 −2 2
0 10 −9
0 0 − 195
∣∣∣∣∣∣∣ = −1(10)(−
19
5
) = 38
29.
∣∣∣∣∣∣∣∣∣
1 −2 2 1
2 1 −2 3
3 4 −8 1
3 −11 12 2
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 −2 2 1
0 5 −6 1
0 10 −14 −2
0 −5 6 −1
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 −2 2 1
0 5 −6 1
0 0 −2 −4
0 0 0 0
∣∣∣∣∣∣∣∣∣ = 1(5)(−2)(0) = 0
30.
∣∣∣∣∣∣∣∣∣
0 1 4 5
2 5 0 1
1 2 2 0
3 1 3 2
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 2 2 0
2 5 0 1
0 1 4 5
3 1 3 2
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 2 2 0
0 1 −4 1
0 1 4 5
0 −5 −3 2
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 2 2 0
0 1 −4 1
0 0 8 4
0 0 −23 7
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 2 2 0
0 1 −4 1
0 0 8 4
0 0 −23 372
∣∣∣∣∣∣∣∣∣
= −(1)(1)(8)(37
2
) = −148
31.
∣∣∣∣∣∣∣∣∣
1 2 3 4
1 3 5 7
2 3 6 7
1 5 8 20
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 2 3 4
0 1 2 3
0 −1 0 −1
0 3 5 16
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 2 3 4
0 1 2 3
0 0 2 2
0 0 −1 7
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 2 3 4
0 1 2 3
0 0 2 2
0 0 0 8
∣∣∣∣∣∣∣∣∣ = 1(1)(2)(8) = 16
32.
∣∣∣∣∣∣∣∣∣
2 9 1 8
1 3 7 4
0 1 6 5
3 1 4 2
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 3 7 4
2 9 1 8
0 1 6 5
3 1 4 2
∣∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣∣
1 3 7 4
0 3 −13 0
0 1 6 5
0 −8 −17 −10
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 3 7 4
0 1 6 5
0 3 −13 0
0 −8 −17 −10
∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣
1 3 7 4
0 1 6 5
0 0 −31 −15
0 0 31 30
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 3 7 4
0 1 6 5
0 0 −31 −15
0 0 0 15
∣∣∣∣∣∣∣∣∣ = 1(1)(−31)(15) = −465
33. We first use the second row to reduce the third row. Then we use the first row to reduce the second row.∣∣∣∣∣∣∣
1 1 1
a b c
0 b2 − ab c2 − ac
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
1 1 1
0 b− a c− a
0 b(b− a) c(c− a)
∣∣∣∣∣∣∣ = (b− a)(c− a)
∣∣∣∣∣∣∣
1 1 1
0 1 1
0 b c
∣∣∣∣∣∣∣ .
Expanding along the first row gives (b− a)(c− a)(c− b).
34. In order, we use the third row to reduce the fourth row, the second row to reduce the third row, and the first
row to reduce the second row. We then pull out a common factor from each column.∣∣∣∣∣∣∣∣∣
1 1 1 1
a b c d
a2 b2 c2 d2
a3 b3 c3 d3
∣∣∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
1 1 1 1
0 b− a c− a d− a
0 b2 − ab c2 − ac d2 − ac
0 b3 − ab2 c3 − ac2 d3 − ad2
∣∣∣∣∣∣∣∣∣ = (b− a)(c− a)(d− a)
∣∣∣∣∣∣∣∣∣
1 1 1 1
0 1 1 1
0 b c d
0 b2 c2 d2
∣∣∣∣∣∣∣∣∣ .
Expanding along the first column and using Problem 33 we obtain (b− a)(c− a)(d− a)(c− b)(d− b)(d− c).
388
8.6 Inverse of a Matrix
35. Since C11 = 4, C12 = 5, and C13 = −6, we have a21C11 + a22C12 + a23C13 = (−1)(4) + 2(5) + 1(−6) = 0. Since
C12 = 5, C22 = −7, and C23 = −3, we have a13C12 + a23C22 + a33C32 = 2(5) + 1(−7) + 1(−3) = 0.
36. Since C11 +−7, C12 = −8, and C13 = −10 we have a21C11 + a22C12 + a23C13 = −2(−7) + 3(−8)− 1(−10) = 0.
Since C12 = −8, C22 = −19, and C32 = −7 we have a13C12 + a23C22 + a33C32 = 5(−8)− 1(−19)− 3(−7) = 0.
37. det(A + B) =
∣∣∣∣ 10 00 −3
∣∣∣∣ = −30; detA + detB = 10− 31 = −21
38. det(2A) = 25 detA = 32(−7) = −224
39. Factoring −1 out of each row we see that det(−A) = (−1)5 detA = −detA. Then −detA = det(−A) =
detAT = detA and detA = 0.
40. (a) Cofactors: 25! ≈ 1.55(1025); Row reduction: 253/3 ≈ 5.2(103)
(b) Cofactors: about 90 billion centuries; Row reduction: about 110 second
EXERCISES 8.6
Inverse of a Matrix
1. AB =
(
3− 2 −1 + 1
6− 6 −2 + 3
)
=
(
1 0
0 1
)
2. AB =
 2− 1 −1 + 1 −2 + 26− 6 −3 + 4 6− 6
2 + 1− 3 −1− 1 + 2 2 + 2− 3
 =
 1 0 00 1 0
0 0 1

3. detA = 9. A is nonsingular. A−1 =
1
9
(
1 1
−4 5
)
=
(
1
9
1
9
− 49 59
)
4. detA = 5. A is nonsingular. A−1 =
1
5
(
3 1
−4 13
)
=
(
3
5
1
5
− 45 115
)
5. detA = 12. A is nonsingular. A−1 =
1
12
(
2 0
3 6
)
=
(
1
6 0
1
4
1
2
)
6. detA = −3π2. A is nonsingular. A−1 = − 1
3π2
(
π π
π −2π
)
=
(− 13π − 13π
− 13π 23π
)
7. detA = −16. A is nonsingular. A−1 = − 1
16
 8 −8 −82 −4 6
−6 4 −2
 =
−
1
2
1
2
1
2
− 18 14 − 38
3
8 − 14 18

8. detA = 0. A is singular.
9. detA = −30. A is nonsingular. A−1 = − 1
30
−14 13 16−2 4 −2
−4 −7 −4
 =

7
15 − 1330 − 815
1
15 − 215 115
2
15
7
30
2
15

389
8.6 Inverse of a Matrix
10. detA = 78. A is nonsingular. A−1 =
1
78
 8 20 2−2 −5 19
12 −9 3


4
39
10
39
1
39
− 139 − 578 1978
2
13 − 326 126

11. detA = −36. A is nonsingular. A−1 = − 1
36
−12 0 00 −6 0
0 0 18
 =

1
3 0 0
0 16 0
0 0 − 12

12. detA = 16. A is nonsingular. A−1 =
1
16
 0 0 28 0 0
0 16 0
 =
 0 0
1
8
1
2 0 0
0 1 0

13. detA = 27. A is nonsingular. A−1 =
1
27

6 21 −9 −36
−1 1 6 −3
10 17 −6 −51
4 −4 3 12
 =

2
9
7
9 − 13 − 43
− 127 127 29 − 19
10
27
17
27 − 29 − 179
4
27 − 427 19 49

14. detA = −6. A is nonsingular. A−1 = −1
6

0 1 −3 3
0 1 3 −9
0 −2 0 0
−6 −1 −3 15
 =

0 − 16 12 − 12
0 − 16 − 12 32
0 13 0 0
1 16
1
2 − 52

15.
(
6 −2 1 0
0 4 0 1
)
1
6R1−−−−−−→
1
4R2
(
1 − 13 16 0
0 1 0 14
)
1
3R2+R1−−−−−−→
(
1 0 16
1
12
0 1 0 14
)
; A−1 =
(
1
6
1
12
0 14
)
16.
(
8 0 1 0
0 12 0 1
)
1
8R1−−−−−−→
2R2
(
1 0 18 0
0 1 0 2
)
; A−1 =
( 1
8 0
0 2
)
17.
(
1 3 1 0
5 3 0 1
)
−5R1+R2−−−−−−→
(
1 3 1 0
0 −12 −5 1
)
− 112R2−−−−−−→
(
1 3 1 0
0 1 512 − 112
)
−3R2+R1−−−−−−→
(
1 0 − 14 14
0 1 512 − 112
)
;
A−1 =
(− 14 14
5
12 − 112
)
18.
(
2 −3 1 0
−2 4 0 1
)
1
2R1−−−−−−→
(
1 − 32 12 0
−2 4 0 1
)
2R1+R2−−−−−−→
(
1 − 32 12 0
0 1 1 1
)
3
2R2+R1−−−−−−→
(
1 0 2 32
0 1 1 1
)
;
A−1 =
(
2 32
1 1
)
19.
 1 2 3 1 0 04 5 6 0 1 0
7 8 9 0 0 1
 row−−−−−−→
operations
 1 2 3 1 0 00 12 43 − 13 0
0 0 0 1 −2 1
; A is singular.
20.
 1 0 −1 1 0 00 −2 1 0 1 0
2 −1 3 0 0 1
 row−−−−−−→
operations
 1 0 0
5
9 − 19 29
0 1 0 − 29 − 59 19
0 0 1 − 49 − 19 29
; A−1 =

5
9 − 19 29
− 29 − 59 19
− 49 − 19 29

21.
 4 2 3 1 0 02 1 0 0 1 0
−1 −2 0 0 0 1
 R13−−−−−−→
−1 −2 0 0 0 12 1 0 0 1 0
4 2 3 1 0 0
 row−−−−−−→
operations
 1 0 0 0
2
3
1
3
0 1 0 0 − 13 − 23
0 0 1 13 − 23 0
;
390
8.6 Inverse of a Matrix
A−1 =
 0
2
3
1
3
0 − 13 − 23
1
3 − 23 0

22.
 2 4 −2 1 0 04 2 −2 0 1 0
8 10 −6 0 0 1
 row−−−−−−→
operations
 1 2 −1
1
2 0 0
0 1 − 13 13 − 16 0
0 0 0 −2 −1 1
; A is singular.
23.
−1 3 0 1 0 03 −2 1 0 1 0
0 1 2 0 0 1
 row−−−−−−→
operations
 1 −3 0 −1 0 00 1 1 1 1 0
0 0 1 −1 −1 1
 row−−−−−−→
operations
 1 0 0 5 6 −30 1 0 2 2 −1
0 0 1 −1 −1 1
;
A−1 =
 5 6 −32 2 −1
−1 −1 1

24.
 1 2 3 1 0 00 1 4 0 1 0
0 0 8 0 0 1
 row−−−−−−→
operations
 1 0 0 1 −2
5
8
0 1 0 0 1 − 12
0 0 1 0 0 18
; A−1 =
 1 −2
5
8
0 1 − 12
0 0 18

25.

1 2 3 1 1 0 0 0
−1 0 2 1 0 1 0 0
2 1 −3 0 0 0 1 0
1 1 2 1 0 0 0 1
 row−−−−−−→operations

1 2 3 1 1 0 0 0
0 1 52 1
1
2
1
2 0 0
0 0 1 − 23 13 −1 − 23 0
0 0 0 1 − 12 1 12 12

row−−−−−−→
operations

1 0 0 0 − 12 − 23 − 16 76
0 1 0 0 1 13
1
3 − 43
0 0 1 0 0 − 13 − 13 13
0 0 0 1 − 12 1 12 12
; A−1 =

− 12 − 23 − 16 76
1 13
1
3 − 43
0 − 13 − 13 13
− 12 1 12 12

26.

1 0 0 0 1 0 0 0
0 0 1 0 0 1 0 0
0 0 0 1 0 0 1 0
0 1 0 0 0 0 0 1
 row−−−−−−→interchange

1 0 0 0 1 0 0 0
0 1 0 0 0 0 0 1
0 0 1 0 0 1 0 0
0 0 0 1 0 0 1 0
; A−1 =

1 0 0 0
0 0 0 1
0 1 0 0
0 0 1 0

27. (AB)−1 = B−1A−1 =
(− 13 13
−1 103
)
28. (AB)−1 = B−1A−1 =
−1 −4 202 6 −30
3 6 −32

29. A = (A−1)−1 =
(−2 3
3 −4
)
30. AT =
(
1 2
4 10
)
; (AT )−1 =
(
5 −1
−2 12
)
; A−1 =
(
5 −2
−1 12
)
; (A−1)T =
(
5 −1
−2 12
)
31. Multiplying
(
4 −3
x −4
) (
4 −3
x −4
)
=
(
16− 3x 0
0 16− 3x
)
we see that x = 5.
391
8.6 Inverse of a Matrix
32. A−1 =
(
sin θ − cos θ
cos θ sin θ
)
33. (a) AT =
(
sin θ − cos θ
cos θ sin θ
)
= A−1 (b) AT =

1√
3
1√
3
1√
3
0 1√
2
− 1√
2
− 2√
6
1√
6
1√
6
 = A−1
34. Since detA · detA−1 = detAA−1 = det I = 1, we see that detA−1 = 1/detA. If A is orthogonal, detA =
detAT = detA−1 = 1/detA and (detA)2 = 1, so detA = ±1.
35. Since A and B are nonsingular, detAB = detA · detB �= 0, and AB is nonsingular.
36. Suppose A is singular. Then detA = 0, detAB = detA · detB = 0, and AB is singular.
37. Since detA · detA−1 = detAA−1 = det I = 1, detA−1 = 1/detA.
38. Suppose A2 = A and A is nonsingular. Then A2A−1 = AA−1, and A = I. Thus, if A2 = A, either A is
singular or A = I.
39. If A is nonsingular, then A−1 exists, and AB = 0 implies A−1AB = A−10, so B = 0.
40. If A is nonsingular, A−1 exists, and AB = AC implies A−1AB = A−1AC, so B = C.
41. No, consider A =
(
1 0
0 0
)
and B =
(
0 0
0 1
)
.
42. A is nonsingular if a11a22a33 = 0 or a11, a22, and a33 are all nonzero.
A−1 =
 1/a11 0 00 1/a22 0
0 0 1/a33

For any diagonal matrix, the inverse matrix is obtaining by taking the reciprocals of the diagonal entries and
leaving all other entries 0.
43. A−1 =
(
1
3
1
3
2
3 − 13
)
; A−1
(
4
14
)
=
(
6
−2
)
; x1 = 6, x2 = −2
44. A−1 =
(
2
3
1
6
− 13 16
)
; A−1
(
2
−5
)
=
( 1
2
− 32
)
; x1 =
1
2
, x2 = −32
45. A−1 =
(
1
16
3
8
− 18 14
)
; A−1
(
6
1
)
=
(
3
4
− 12
)
; x1 =
3
4
, x2 = −12
46. A−1 =
(−2 1
3
2 − 12
)
; A−1
(
4
−3
)
=
(−11
15
2
)
; x1 = −11, x2 = 152
47. A−1 =
−
1
5
1
5
1
5
−1 1 0
6
5 − 15 − 15
; A−1
−40
6
 =
 24
−6
; x1 = 2, x2 = 4, x3 = −6
48. A−1 =

5
12 − 112 14
− 23 13 0
− 112 512 − 14
; A−1
 12
−3
 =
−
1
2
0
3
2
; x1 = −12 , x2 = 0, x3 = 32
392
8.6 Inverse of a Matrix
49. A−1 =
−2 −3 214 − 14 0
5
4
7
4 −1
; A−1
 1−3
7
 =
 211
−11
; x1 = 21, x2 = 1, x3 = −11
50. A−1 =

2 −1 1 1
−1 2 −1 −1
1 −1 1 1
1 −1 1 0
; A−1

2
1
−5
3
 =

1
2
−1
−4
; x1 = 1, x2 = 2, x3 = −1, x4 = −4
51.
(
7 −2
3 2
) (
x1
x2
)
=
(
b1
b2
)
; A−1 =
(
1
10
1
10
− 320 720
)
; X = A−1
(
5
4
)
=
(
9
10
13
20
)
; X = A−1
(
10
50
)
=
(
6
16
)
;
X = A−1
(
0
−20
)
=
(−2
−7
)
52.
 1 2 52 3 8
−1 1 2

x1x2
x3
 =
 b1b2
b3
; A−1 =
 2 −1 −112 −7 −2
−5 3 1
; X = A−1
−14
6
 =
−12−52
23
;
X = A−1
 33
3
 =
 09
−3
; = A−1
 0−5
4
 =
 127
−11

53. detA = 18 �= 0, so the system has only the trivial solution.
54. detA = 0, so the system has a nontrivial solution.
55. detA = 0, so the system has a nontrivial solution.
56. detA = 12 �= 0, so the system has only the trivial solution.
57. (a)
 1 1 1−R1 R2 0
0 −R2 R3

 i1i2
i3
 =
 0E2 − E1
E3 − E2

(b) detA = R1R2 + R1R3 + R2R3 > 0, so A is nonsingular.
(c) A−1 =
1
R1R2 + R1R3 + R2R3
R2R3 −R2 −R3 −R2R1R3 R3 −R1
R1R2 R2 R1 + R2
;
A−1
 0E2 − E1
E3 − E2
 = 1
R1R2 + R1R3 + R2R3
 R2E1 −R2E3 + R3E1 −R3E2R1E2 −R1E3 −R3E1 + R3E2
−R1E2 + R1E3 −R2E1 + R2E3

58. (a) We write the equations in the form
−4u1 + u2 + u4 = −200
u1 − 4u2 + u3 = −300
u2 − 4u3 + u4 = −300
u1 + u3 − 4u4 = −200.
In matrix form this becomes

−4 1 0 1
1 −4 1 0
0 1 −4 1
1 0 1 −4


u1
u2
u3
u4
 =

−200
−300
−300
−200
.
393
8.6 Inverse of a Matrix
(b) A−1 =

− 724 − 112 − 124 − 112
− 112 − 724 − 112 − 124
− 124 − 112 − 724 − 112
− 112 − 124 − 112 − 724
; A−1

−200
−300
−300
−200
 =

225
2
275
2
275
2
225
2
; u1 = u4 = 2252 , u2 = u3 = 2752
EXERCISES 8.7
Cramer’s Rule
1. detA = 10, detA1 = −6, detA2 = 12; x1 = −610 = − 35 , x2 = 1210 = 65
2. detA = −3, detA1 = −6, detA2 = −6; x1 = −6−3 = 2, x2 = −6−3 = 2
3. detA = 0.3, detA1 = 0.03, detA2 = −0.09; x1 = 0.030.3 = 0.1 , x2 = −0.090.3 = −0.3
4. detA = −0.015, detA1 = −0.00315, detA2 = −0.00855; x1 = −0.00315−0.015 = 0.21, x2 = −0.00855−0.015 = 0.57
5. detA = 1, detA1 = 4, detA2 = −7; x = 4, y = −7
6. detA = −70, detA1 = −14, detA2 = 35; r = −14−70 = 15 , s = 35−70 = − 12
7. detA = 11, detA1 = −44, detA2 = 44, detA3 = −55; x1 = −4411 = −4, x2 = 4411 = 4, x3 = −5511 = −5
8. detA = −63, detA1 = 173, detA2 = −136, detA3 = − 612 ; x1 = − 17363 , x2 = 13663 , x3 = 61126
9. detA = −12, detA1 = −48, detA2 = −18, detA3 = −12; u = 4812 = 4, v = 1812 = 32 , w = 1
10. detA = 1, detA1 = −2, detA2 = 2, detA3 = 5; x = −2, y = 2, z = 5
11. detA = 6− 5k, detA1 = 12− 7k, detA2 = 6− 7k; x1 = 12− 7k6− 5k , x2 =
6− 7k
6− 5k . The system is inconsistent
for k = 6/5.
12. (a) detA = �− 1, detA1 = �− 2, detA2 = 1; x1 = �− 2
�− 1 =
�− 1− 1
�− 1 = 1−
1
�− 1 , x2 =
1
�− 1
(b) When � = 1.01, x1 = −99 and x2 = 100. When � = 0.99, x1 = 101 and x2 = −100.
13. detA ≈ 0.6428, detA1 ≈ 289.8, detA2 ≈ 271.9; x1 ≈ 289.80.6428 ≈ 450.8, x2 ≈ 271.90.6428 ≈ 423
14. We have (sin 30◦)F + (sin 30◦)(0.5N) + N sin 60◦ = 400 and (cos 30◦)F + (cos 30◦)(0.5N)−N cos 60◦ = 0.The
system is
(sin 30◦)F + (0.5 sin 30◦ + sin 60◦)N = 400
(cos 30◦)F + (0.5 cos 30◦ − cos 60◦)N = 0.
detA ≈ −1, detA1 ≈ −26.795, detA2 ≈ −346.41; F ≈ 26.795, N ≈ 346.41
15. The system is
i1 + i2 − i3 = 0
r1i1 − r2i2 = E1 − E2
r2i2 + Ri3 = E2
detA = −r1R− r2R− r1r2, detA3 = −r1E2, −r2E1; i3 = r1E2 + r2E1
r1R + r2R + r1r2
394
8.8 The Eigenvalue Problem
EXERCISES 8.8
The Eigenvalue Problem
1. K3 since
(
4 2
5 1
) (−2
5
)
=
(
2
−5
)
= (−1)
(−2
5
)
; λ = −1
2. K1 and K2 since
(
2 −1
2 −2
) (
1
2−√2
)
=
( √
2
−2 + 2√2
)
=
√
2
(
1
2−√2
)
, λ =
√
2(
2 −1
2 −2
) (
2 +
√
2
2
)
=
(
2 + 2
√
2
2
√
2
)
=
√
2
(
2 +
√
2
2
)
; λ =
√
2
3. K3 since
(
6 3
2 1
) (−5
10
)
=
(
0
0
)
= 0
(−5
10
)
; λ = 0
4. K2 since
(
2 8
−1 −2
) (
2 + 2i
−1
)
=
(−4 + 4i
−2i
)
= 2i
(
2 + 2i
−1
)
; λ = 2i
5. K2 and K3 since
 1 −2 2−2 1 −2
2 2 1

 4−4
0
 =
 12−12
0
 = 3
 4−4
0
; λ = 3
 1 −2 2−2 1 −2
2 2 1

−11
1
 =
−11
1
; λ = 1
6. K2 since
−1 1 01 2 1
0 3 −1

 14
3
 =
 312
9
 = 3
 14
3
; λ = 3
7. We solve det(A− λI) =
∣∣∣∣−1− λ 2−7 8− λ
∣∣∣∣ = (λ− 6)(λ− 1) = 0.
For λ1 = 6 we have
(−7 2 0
−7 2 0
)
=⇒
(
1 −2/7 0
0 0 0
)
so that k1 = 27k2. If k2 = 7 then K1 =
(
2
7
)
. For λ2 = 1 we have
(−2 2 0
−7 7 0
)
=⇒
(
1 −1 0
0 0 0
)
so that k1 = k2. If k2 = 1 then K2 =
(
1
1
)
.
8. We solve det(A− λI) =
∣∣∣∣ 2− λ 12 1− λ
∣∣∣∣ = λ(λ− 3) = 0.
For λ1 = 0 we have
(
2 1 0
2 1 0
)
=⇒
(
1 1/2 0
0 0 0
)
395
8.8 The Eigenvalue Problem
so that k1 = − 12k2. If k2 = 2 then K1 =
(−1
2
)
. For λ2 = 3 we have
(−1 1 0
2 −2 0
)
=⇒
(
1 −1 0
0 0 0
)
so that k1 = k2. If k2 = 1 then K2 =
(
1
1
)
.
9. We solve det(A− λI) =
∣∣∣∣−8− λ −116 −λ
∣∣∣∣ = (λ + 4)2 = 0.
For λ1 = λ2 = −4 we have
(−4 −1 0
16 4 0
)
=⇒
(
1 1/4 0
0 0 0
)
so that k1 = − 14k2. If k2 = 4 then K1 =
(−1
4
)
.
10. We solve det(A− λI) =
∣∣∣∣ 1− λ 11/4 1− λ
∣∣∣∣ = (λ− 3/2)(λ− 1/2) = 0.
For λ1 = 3/2 we have
(−1/2 1 0
1/4 −1/2 0
)
=⇒
(
1 −2 0
0 0 0
)
so that k1 = 2k2. If k2 = 1 then K1 =
(
2
1
)
. If λ2 = 1/2 then
(
1/2 1 0
1/4 1/2 0
)
=⇒
(
1 2 0
0 0 0
)
so that k1 = −2k2. If k2 = 1 then K2 =
(−2
1
)
.
11. We solve det(A− λI) =
∣∣∣∣−1− λ 2−5 1− λ
∣∣∣∣ = λ2 + 9 = (λ− 3i)(λ + 3i) = 0.
For λ1 = 3i we have
(−1− 3i 2 0
−5 1− 3i 0
)
=⇒
(
1 −(1/5) + (3/5)i 0
0 0 0
)
so that k1 =
(
1
5 − 35 i
)
k2. If k2 = 5 then K1 =
(
1− 3i
5
)
. For λ2 = −3i we have
(−1 + 3i 2 0
−5 1 + 3i 0
)
=⇒
(
1 − 15 − 35 i 0
0 0 0
)
so that k1 =
(
1
5 +
3
5 i
)
k2. If k2 = 5 then K2 =
(
1 + 3i
5
)
.
12. We solve det(A− λI) =
∣∣∣∣ 1− λ −11 1− λ
∣∣∣∣ = λ2 − 2λ + 2 = 0.
For λ1 = 1− i we have
(
i −1 0
1 i 0
)
=⇒
(
i −1 0
0 0 0
)
so that k1 = −ik2. If k2 = 1 then K1 =
(−i
1
)
and K2 = K1 =
(
i
1
)
.
396
8.8 The Eigenvalue Problem
13. We solve det(A− λI) =
∣∣∣∣ 4− λ 80 −5− λ
∣∣∣∣ = (λ− 4)(λ + 5) = 0.
For λ1 = 4 we have
(
0 8 0
0 −9 0
)
=⇒
(
0 1 0
0 0 0
)
so that k2 = 0. If k1 = 1 then K1 =
(
1
0
)
. For λ2 = −5 we have(
9 8 0
0 0 0
)
=⇒
(
1 89 0
0 0 0
)
so that k1 = − 89 k2. If k2 = 9 then K2 =
(−8
9
)
.
14. We solve det(A− λI) =
∣∣∣∣ 7− λ 00 13− λ
∣∣∣∣ = (λ− 7)(λ− 13) = 0.
For λ1 = 7 we have
(
0 0 0
0 6 0
)
=⇒
(
0 1 0
0 0 0
)
so that k2 = 0. If k1 = 1 then K1 =
(
1
0
)
. For λ2 = 13 we have(−6 0 0
0 0 0
)
=⇒
(
1 0 0
0 0 0
)
so that k1 = 0. If k2 = 1 then K2 =
(
0
1
)
.
15. We solve det(A− λI) =
∣∣∣∣∣∣∣
5− λ −1 0
0 −5− λ 9
5 −1 −λ
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
4− λ −1 0
4− λ −5− λ 9
4− λ −1 −λ
∣∣∣∣∣∣∣ = λ(4− λ)(λ + 4) = 0.
For λ1 = 0 we have
 5 −1 0 00 −5 9 0
5 −1 0 0
 =⇒
 1 0 −9/25 00 1 −9/5 0
0 0 0 0

so that k1 = 925k3 and k2 =
9
5k3. If k3 = 25 then K1 =
 945
25
. If λ2 = 4 then
 1 −1 0 00 −9 9 0
5 −1 −4 0
 =⇒
 1 0 −1 00 1 −1 0
0 0 0 0

so that k1 = k3 and k2 = k3. If k3 = 1 then K2 =
 11
1
. If λ3 = −4 then
 9 −1 0 00 −1 9 0
5 −1 4 0
 =⇒
 1 0 −1 00 1 −9 0
0 0 0 0

so that k1 = k3 and k2 = 9k3. If k3 = 1 then K3 =
 19
1
.
397
8.8 The Eigenvalue Problem
16. We solve det(A− λI) =
∣∣∣∣∣∣∣
3− λ 0 0
0 2− λ 0
4 0 1− λ
∣∣∣∣∣∣∣ = (3− λ)(2− λ)(1− λ) = 0.
For λ1 = 1 we have
 2 0 0 00 1 0 0
4 0 0 0
 =⇒
 1 0 0 00 1 0 0
0 0 0 0

so that k1 = 0 and k2 = 0. If k3 = 1 then K1 =
 00
1
. If λ2 = 2 then
 1 0 0 00 0 0 0
4 0 −1 0
 =⇒
 1 0 0 00 0 1 0
0 0 0 0

so that k1 = 0 and k3 = 0. If k2 = 1 then K2 =
 01
0
. If λ3 = 3 then
 0 0 0 00 −1 0 0
4 0 −2 0
 =⇒
 1 0 −1/2 00 1 0 0
0 0 0 0

so that k1 = 12k3 and k2 = 0. If k3 = 2 then K3 =
 10
2
.
17. We solve det(A− λI) =
∣∣∣∣∣∣∣
−λ 4 0
−1 −4− λ 0
0 0 −2− λ
∣∣∣∣∣∣∣ = −(λ + 2)3 = 0.
For λ1 = λ2 = λ3 = −2 we have
 2 4 0 0−1 −2 0 0
0 0 0 0
 =⇒
 1 2 0 00 0 0 0
0 0 0 0

so that k1 = −2k2. If k2 = 1 and k3 = 1 then
K1 =
−21
0
 and K2 =
 00
1
 .
18. We solve det(A− λI) =
∣∣∣∣∣∣∣
1− λ 6 0
0 2− λ 1
0 1 2− λ
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣
1− λ 6 0
0 3− λ 3− λ
0 1 2− λ
∣∣∣∣∣∣∣ = (3− λ)(1− λ)2 = 0.
For λ1 = 3 we have
−2 6 0 00 0 0 0
0 1 −1 0
 =⇒
 1 0 −3 00 1 −1 0
0 0 0 0

398
8.8 The Eigenvalue Problem
so that k1 = 3k3 and k2 = k3. If k3 = 1 then K1 =
 31
1
. For λ2 = λ3 = 1 we have
 0 6 0 00 1 1 0
0 1 1 0
 =⇒
 0 1 0 00 0 1 0
0 0 0 0

so that k2 = 0 and k3 = 0. If k1 = 1 then K2 =
 10
0
.
19. We solve det(A− λI) =
∣∣∣∣∣∣∣
−λ 0 −1
1 −λ 0
1 1 −1− λ
∣∣∣∣∣∣∣ = −(λ + 1)(λ2 + 1) = 0.
For λ1 = −1 we have
 1 0 −1 01 1 0 0
1 1 0 0
 =⇒
 1 0 −1 00 1 1 0
0 0 0 0

so that k1 = k3 and k2 = −k3. If k3 = 1 then K1 =
 1−1
1
. For λ2 = i we have
−i 0 −1 01 −i 0 0
1 1 −1− i 0
 =⇒
 1 0 −i 00 1 −1 0
0 0 0 0

so that k1 = ik3 and k2 = k3. If k3 = 1 then K2 =
 i1
1
 and K3 = K2 =
−i1
1
.
20. We solve
det(A− λI) =
∣∣∣∣∣∣∣
2− λ −1 0
5 2− λ 4
0 1 2− λ
∣∣∣∣∣∣∣ = −λ3 + 6λ2 − 13λ + 10 = (λ− 2)(−λ2 + 4λ− 5)
= (λ− 2)(λ− (2 + i))(λ− (2− i)) = 0.
For λ1 = 2 we have
 0 −1 0 05 0 4 0
0 1 0 0
 =⇒
 1 0 4/5 00 1 0 0
0 0 0 0

so that k1 = − 45k3 and k2 = 0. If k3 = 5 then K1 =
−40
5
. For λ2 = 2 + i we have
−i −1 0 05 −i 4 0
0 1 −i 0
 =⇒
 1 −i 0 00 1 −i 0
0 0 0 0

399
8.8 The Eigenvalue Problem
so that k1 = ik2 and k2 = ik3. If k3 = i then K2 =
 −i−1
i
. For λ3 = 2− i we have
 i −1 0 05 i 4 0
0 1 i 0
 =⇒
 1 i 0 00 1 i 0
0 0 0 0

so that k1 = −ik2 and k2 = −ik3. If k3 = i then K3 =
−11
i
.
21. We solve det(A− λI) =
∣∣∣∣∣∣∣
1− λ 2 3
0 5− λ 6
0 0 −7− λ
∣∣∣∣∣∣∣ = −(λ− 1)(λ− 5)(λ + 7) = 0.
For λ1 = 1 we have
 0 2 3 00 4 6 0
0 0 −6 0
 =⇒
 0 1 0 00 0 1 0
0 0 0 0

so that k2 = k3 = 0. If k1 = 1 then K1 =
 10
0
. For λ2 = 5 we have
−4 2 3 00 0 6 0
0 0 −12 0
 =⇒
 1 −
1
2 0 0
0 0 1 0
0 0 0 0

so that k3 = 0 and k2 = 2k1. If k1 = 1 then K2 =
 12
0
. For λ3 = −7 we have
 8 2 3 00 12 6 00 0 0 0
 =⇒
 1 0
1
4 0
0 1 12 0
0 0 0 0

so that k1 = − 14 k3 and k2 = − 12 k3. If k3 = 4 then K3 =
−1−2
4
.
22. We solve det(A− λI) =
∣∣∣∣∣∣∣
−λ 0 0
0 −λ 0
0 0 1− λ
∣∣∣∣∣∣∣ = −λ2(λ− 1) = 0.
For λ1 = λ2 = 0 we have
 0 0 0 00 0 0 0
0 0 1 0
 =⇒
 0 0 1 00 0 0 0
0 0 0 0

so that k3 = 0. If k1 = 1 and k2 = 0 then K1 =
 10
0
 and if k1 = 0 and k2 = 1 then K2 =
 01
0
. For λ3 = 1
400
8.8 The Eigenvalue Problem
we have −1 0 0 00 −1 0 0
0 0 0 0
 =⇒
 1 0 0 00 1 0 0
0 0 0 0

so that k1 = k2 = 0. If k3 = 1 then K3 =
 00
1
.
23. The eigenvalues and eigenvectors of A =
(
5 1
1 5
)
are
λ1 = 4, λ2 = 6, K1 =
(
1
−1
)
, K2 =
(
1
1
)
and the eigenvalues and eigenvectors of A−1 =
1
24
(
5 −1
−1 5
)
are
λ1 =
1
4
, λ2 =
1
6
, K1 =
(
1
−1
)
, K2 =
(
1
1
)
.
24. The eigenvalues and eigenvectors of A =
 1 2 −11 0 1
4 −4 5
 are
λ1 = 1, λ2 = 2, λ3 = 3, K1 =
−11
2
 , K2 =
−21
4
 , K3 =
−11
4
 .
and the eigenvalues and eigenvectors of A−1 =
1
6
 4 −6 2−1 9 −2
−4 12 −2
 are
λ1 = 1, λ2 =
1
2
, λ3 =
1
3
, K1 =
−11
2
 , K2 =
−21
4
 , K3 =
−11
4
 .
25. Since detA =
∣∣∣∣ 6 03 0
∣∣∣∣ = 0 the matrix is singular. Now from
det(A− λI) =
∣∣∣∣ 6− λ 03 −λ
∣∣∣∣ = λ(λ− 6)
we see λ = 0 is an eigenvalue.
26. Since detA =
∣∣∣∣∣∣∣
1 0 1
4 −4 5
7 −4 8
∣∣∣∣∣∣∣ = 0 the matrix is singular. Now from
det(A− λI) =
∣∣∣∣∣∣∣
1− λ 0 1
4 −4− λ 5
7 −4 8− λ
∣∣∣∣∣∣∣ = −λ(λ2 − 5λ− 15)
we see λ = 0 is an eigenvalue.
401
8.8 The Eigenvalue Problem
27. (a) Since p + 1− p = 1 and q + 1− q = 1, the first matrix A is stochastic. Since 12 + 14 + 14 = 1, 13 + 13 + 13 = 1,
and 16 +
1
3 +
1
2 = 1, the second matrix A is stochastic.
(b) The matrix from part (a) is shown with its eigenvalues and corresponding eigenvectors.
1
2
1
4
1
4
1
3
1
3
1
3
1
6
1
3
1
2
; eigenvalues: 1, 16 − 112√2 , 16 + 112√2 ;
eigenvectors: (1, 1, 1),
(− 3(−1+√2)−6+√2 , 2(2+√2)−6+√2 , 1), (− 3(1+√2)6+√2 , 2(−2+√2)6+√2 , 1)
Further examples indicate that 1 is always an eigenvalue with corresponding eigenvector (1, 1, 1). To prove
this, let A be a stochastic matrix and K = (1, 1, 1). Then
AK =
 a11 · · · a1n... ...
an1 · · · ann

 1...
1
 =
 a11 + · · ·+ a1n...
an1 + · · ·+ ann
 =
 1...
1
 = 1K,
and 1 is an eigenvalue of A with corresponding eigenvector (1, 1, 1).
(c) For the 3× 3 matrix in part (a) we have
A2 =

3
8
7
24
1
3
1
3
11
36
13
36
5
18
23
72
29
72
 , A3 =

49
144
29
96
103
288
71
216
11
36
79
216
5
16
67
216
163
432
 .
These powers of A are also stochastic matrices. To prove that this is true in general for 2× 2 matrices, we
prove the more general theorem that any product of 2× 2 stochastic matrices is stochastic. Let
A =
(
a11 a12
a21 a22
)
and B =
(
b11 b12
b21 b22
)
be stochastic matrices. Then
AB =
(
a11b11 + a12b21 a11b12 + a12b22
a21b11 + a22b21 a21b12 + a22b22
)
.
The sums of the rows are
a11b11 + a12b21 + a11b12 + a12b22 = a11(b11 + b12) + a12(b21 + b22)
= a11(1) + a12(1) = a11 + a12 = 1
a21b11 + a22b21 + a21b12 + a22b22 = a21(b11 + b12) + a22(b21 + b22)
= a21(1) + a22(1) = a21 + a22 = 1.
Thus, the product matrix AB is stochastic. It follows that any power of a 2× 2 matrix is stochastic. The
proof in the case of an n× n matrix is very similar.
402
8.9 Powers of Matrices
EXERCISES 8.9
Powers of Matrices
1. The characteristic equation is λ2 − 6λ + 13 = 0. Then
A2 − 6A + 13I =
(−7 −12
24 17
)
−
(
6 −12
24 30
)
+
(
13 0
0 13
)
=
(
0 0
0 0
)
.
2. The characteristic equation is −λ3 + λ2 + 4λ− 1. Then
−A3 + A2 + A− I = −
 2 6 134 5 17
1 5 9
 +
 1 2 50 4 5
1 1 4
 + 4
 0 1 21 0 3
0 1 1
 +
 1 0 00 1 0
0 0 1
 =
 0 0 00 0 0
0 0 0
 .
3. The characteristic equation is λ2 − 3λ − 10 = 0, with eigenvalues −2 and 5. Substituting the eigenvalues into
λm = c0 + c1λ generates
(−2)m = c0 − 2c1
5m = c0 + 5c1.
Solving the system gives
c0 =
1
7
[5(−2)m + 2(5)m], c1 = 17[−(−2)
m + 5m].
Thus
Am = c0I + c1A =
(
1
7 [3(−1)m2m+1 + 5m] 37 [−(−2)m + 5m]
2
7 [−(−2)m + 5m] 17 [(−2)m + 6(5)m]
)
and
A3 =
(
11 57
38 106
)
.
4. The characteristic equation is λ2 − 10λ + 16 = 0, with eigenvalues 2 and 8. Substituting the eigenvalues into
λm = c0 + c1λ generates
2m = c0 + 2c1
8m = c0 + 8c1.
Solving the system gives
c0 =
1
3
(2m+2 − 8m), c1 = 16(−2
m + 8m).
Thus
Am = c0I + c1A =
(
1
2 (2
m + 8m) 12 (2
m − 8m)
1
2 (2
m − 8m) 12 (2m + 8m)
)
and
A4 =
(
2056 −2040
−2040 2056
)
.
403
8.9 Powers of Matrices
5. The characteristic equation is λ2 − 8λ− 20 = 0, with eigenvalues −2 and 10. Substituting the eigenvalues into
λm = c0 + c1λ generates
(−2)m = c0 − 2c1
10m = c0 + 10c1.
Solving the system gives
c0 =
1
6
[5(−2)m + 10m], c1 = 112[−(−2)
m + 10m].
Thus
Am = c0I + c1A =
(
1
6 [(−2)m + 2m5m+1] 512 [−(−2)m + 10m]
1
3 [−(−2)m + 10m] 16 [5(−2)m + 10m]
)
and
A5 =
(
83328 41680
33344 16640
)
.
6. The characteristic equation is λ2 + 4λ + 3 = 0, with eigenvalues −3 and −1. Substituting the eigenvalues into
λm = c0 + c1λ generates
(−3)m = c0 − 3c1
(−1)m = c0 − c1.
Solving the system gives
c0 =
1
2
[−(−3)m + 3(−1)m], c1 = 12[−(−3)
m + (−1)m].
Thus
Am = c0I + c1A =
(
(−1)m −(−3)m + (−1)m
0 (−3)m
)
and
A6 =
(
1 −728
0 729
)
.
7. The characteristic equation is −λ3 +2λ2 +λ−2 = 0, with eigenvalues −1, 1, and 2. Substituting the eigenvalues
into λm = c0 + c1λ + c2λ2 generates
(−1)m = c0 − c1 + c2
1 = c0 + c1 + c2
2m = c0 + 2c1 + 4c2.
Solving the system gives
c0 =
1
3
[3 + (−1)m − 2m],
c1 =
1
2
[1− (−1)m],
c2 =
1
6
[−3 + (−1)m + 2m+1].
Thus
404
8.9 Powers of Matrices
Am = c0I + c1A + c2A2 =
 1 −1 + 2
m −1 + 2m
0 13 [(−1)m + 2m+1] − 23 [(−1)m − 2m]
0 13 [−(−1)m + 2m] 13 [2(−1)m + 2m]

and
A10 =
 1 1023 10230 683 682
0 341 342
 .
8. The characteristic equation is −λ3 − λ2 + 2λ + 2 = 0, with eigenvalues −1, −√2 , and √2 . Substituting the
eigenvalues into λm = c0 + c1λ + c2λ2 generates
(−1)m = c0 − c1 + c2
(−
√
2 )m = c0 −
√
2c1 + 2c2
(
√
2 )m = c0 +
√
2c1 + 2c2.
Solving the system gives
c0 = [2− (
√
2 )m−1 − (
√
2 )m−2](−1)m + (
√
2− 1)(
√
2 )m−2,
c1 =
1
2
[1− (−1)m](
√
2 )m−1,
c2 = (−1)m+1 + 12(1 +
√
2 )(−1)m(
√
2 )m−1 +
1
2
(
√
2− 1)(
√
2 )m−1.
Thus Am = c0I + c1A + c2A2 and
A6 =
 1 0 77 8 −7
0 0 8
 .
9. The characteristic equation is −λ3+3λ2+6λ−8 = 0, with eigenvalues −2, 1, and 4. Substituting the eigenvalues
into λm = c0 + c1λ + c2λ2 generates
(−2)m = c0 − 2c1 + 4c2
1 = c0 + c1 + c2
4m = c0 + 4c1 + 16c2.
Solving the system gives
c0 =
1
9
[8 + (−1)m2m+1 − 4m],
c1 =
1
18
[4− 5(−2)m + 4m],
c2 =
1
18
[−2 + (−2)m + 4m].
Thus
Am = c0I + c1A + c2A2 =

1
9 [(−2)m + (−1)m2m+1 + 3 · 22m+1] 13 [−(−2)m + 4m] 0
− 23 [(−2)m − 4m] 13 [(−1)m2m+1 + 4m] 0
1
3 [−3 + (−2)m + 22m+1] 13 [−(−2)m + 4m] 1

405
8.9 Powers of Matrices
and
A10 =
 699392 349184 0698368 350208 0
699391 349184 1
 .
10. The characteristic equation is −λ3 − 32λ2 + 32λ + 1 = 0, with eigenvalues −2, − 12 , and 1. Substituting the
eigenvalues into λm = c0 + c1λ + c2λ2 generates
(−2)m = c0 − 2c1 + 4c2(−12
)m = c0 − 12c1 + 14c2
1 = c0 + c1 + c2.
Solving the system gives
c0 =
1
9
[2−m[(−4)m + 8(−1)m + 2m+1 − (−1)m22m+1],
c1 = −192
−m[(−4)m + 4(−1)m − 5 · 2m],
c2 =
2
9
[1 + (−2)m − (−1)m2m−1].
Thus
Am = c0I + c1A + c2A2
=

1
32
−m[2(−1)m + 2m] 13 [
[−1 + (− 12)m] 0
2
3
[−1 + (− 12)m] 13[2 + (− 12)m] 0
− 192−m[7(−4)m − 6(−1)m − 3 · 2m + (−1)m22m+1] 13
[−1 + (− 12)m] 13 [(−2)m + (−1)m2m+1]

and
A8 =

43
128 − 85256 0
− 85128 171256 0
− 32725128 − 85256 256
 .
11. The characteristic equation is λ2 − 8λ + 16 = 0, with eigenvalues 4 and 4. Substituting the eigenvalues into
λm = c0 + c1λ generates
4m = c0 + 4c1
4m−1m = c1.
Solving the system gives
c0 = −4m(m− 1), c1 = 4m−1m.
Thus
Am = c0I + c1A =
(
4m−1(3m + 4) 3 · 4m−1m
−3 · 4m−1m 4m−1(−3m + 4)
)
and
A6 =
(
22528 18432
−18432 −14336
)
.
406
8.9 Powers of Matrices
12. The characteristic equation is −λ3 − λ2 + 21λ + 45 = 0, with eigenvalues −3, −3, and 5. Substituting the
eigenvalues into λm = c0 + c1λ + c2λ2 generates
(−3)m = c0 − 3c1 + 9c2
(−3)m−1m = c1 − 6c2
5m = c0 + 5c1 + 25c2.
Solving the system gives
c0 =
1
64
[73(−3)m − 2(−1)m3m+2 + 9 · 5m − 40(−3)mm],
c1 =
1
96
[−(−1)m3m+2 + 9 · 5m − 8(−3)mm],
c2 =
1
64
[−(−3)m + 5m − 8(−3)m−1m].
Thus
Am = c0I + c1A + c2A2
=

1
32
[31(−3)m − (−1)m3m+1 + 4 · 5m] 1
16
[−(−3)m − (−1)m3m+1 + 4 · 5m] 1
32
[(−3)m + (−1)m3m+1 − 4 · 5m]
1
16
[−(−3)m − (−1)m3m+1 + 4 · 5m] 1
8
[7(−3)m − (−1)m3m+1 + 4 · 5m] 1
16
[(−3)m + (−1)m3m+1 − 4 · 5m]
3
32
[(−3)m + (−1)m3m+1 − 4 · 5m] 3
16
[(−3)m + (−1)m3m+1 − 4 · 5m] 1
32
[29(−3)m − (−1)m3m+2 + 12 · 5m]

and
A5 =
 178 842 −421842 1441 −842
−1263 −2526 1020
 .
13. (a) The characteristic equation is λ2 − 4λ = λ(λ− 4) = 0, so 0 is an eigenvalue. Since the matrix satisfies the
characteristic equation, A2 = 4A, A3 = 4A2 = 42A, A4 = 42A2 = 43A, and, in general,
Am = 4mA =
(
4m 4m
3(4)m 3(4)m
)
.
(b) The characteristic equation is λ2 = 0, so 0 is an eigenvalue. Since the matrix satisfies the characteristic
equation, A2 = 0, A3 = AA2 = 0, and, in general, Am = 0.
(c) The characteristic equation is −λ3 + 5λ2 − 6λ = 0, with eigenvalues 0, 2, and 3. Substituting λ = 0 into
λm = c0 + c1λ + c2λ2 we find that c0 = 0. Using the nonzero eigenvalues, we find
2m = 2c1 + 4c2
3m = 3c1 + 9c2.
Solving the system gives
c1 =
1
6
[9(2)m − 4(3)m], c2 = 16[−3(2)
m + 2(3)m].
Thus Am = c1A + c2A2 and
Am =
 2(3)
m−1 3m−1 3m−1
1
6 [9(2)
m − 4(3)m] 16 [3(2)m − 2(3)m] 16 [−3(2)m − 2(3)m]
1
6 [−9(2)m + 8(3)m] 16 [−3(2)m + 4(3)m] 16 [3(2)m + 4(3)m]
 .
407
8.9 Powers of Matrices
14. (a) Let
Xn−1 =
(
xn−1
yn−1
)
and A =
(
1 1
0 1
)
.
Then
Xn = AXn−1 =
(
1 1
1 0
) (
xn−1
yn−1
)
=
(
xn−1 + yn−1
xn−1
)
.
(b) The characteristic equation of A is λ2 − λ− 1 = 0, with eigenvalues λ1 = 12 (1−
√
5 ) and λ2 = 12 (1 +
√
5 ).
From λm = c0 + c1λ we get λm1 = c0 + c1λ1 and λ
m
2 = c0 + c1λ2. Solving this system gives
c0 = (λ2λm1 − λ1λm2 )/(λ2 − λ1) and c1 = (λm2 − λm1 )/(λ2 − λ1).
Thus
Am = c0I + c1A
=
1
2m+1
√
5
(
(1 +
√
5 )m+1 − (1−√5 )m+1 2(1 +√5 )m − 2(1−√5 )m
2(1 +
√
5 )m − 2(1−√5 )m (1 +√5 )(1−√5 )m − (1−√5 )(1 +√5 )m
)
.
(c) From part (a), X2 = AX1, X3 = AX2 = A2X1, X4 = AX3 = A3X1, and, in general, Xn = An−1X1.
With
X1 =
(
1
1
)
we have X12 = A11X1 =
(
144 89
89 55
) (
1
1
)
=
(
233
144
)
,
so the number of adult pairs is 233. With
X1 =
(
1
0
)
we have A11X1 =
(
144 89
89 55
) (
1
0
)
=
(
144
89
)
,
so the number of baby pairs is 144. With
X1 =
(
2
1
)
we have A11X1 =
(
144 89
89 55
) (
2
1
)
=
(
377
233
)
,
so the total number of pairs is 377.
15. The characteristic equation of A is λ2− 5λ+ 10 = 0, so A2− 5A+ 10I = 0 and I = − 110A2 + 12A. Multiplying
by A−1 we find
A−1 = − 1
10
A +
1
2
I = − 1
10
(
2 −4
1 3
)
+
1
2
(
1 0
0 1
)
=
(
3
10
2
5
− 110 15
)
.
16. The characteristic equation of A is −λ3+2λ2+λ−2 = 0, so −A3+2A2+A−2I = 0 and I = − 12A3+A2+ 12A.
Multiplying by A−1 we find
A−1 = −1
2
A2 + A +
1
2
I =

3
2
1
2 − 52
1
2
1
2 − 12
1
2
1
2 − 32
 .
17. (a) Since
A2 =
(
1 0
−1 0
)
we see that Am =
(
1 0
−1 0
)
for all integers m ≥ 2. Thus A is not nilpotent.
(b) Since A2 = 0, the matrix is nilpotent with index 2.
(c) Since A3 = 0, the matrix is nilpotent with index 3.
408
8.10 Orthogonal Matrices
(d) Since A2 = 0, the matrix is nilpotent with index 2.
(e) Since A4 = 0, the matrix is nilpotent with index 4.
(f) Since A4 = 0, the matrix is nilpotent with index 4.
18. (a) If Am = 0 for some m, then (detA)m = detAm = det0 = 0, and A is a singular matrix.
(b) By (1) of Section 8.8 we have AK = λK, A2K = λAK = λ2K, A3K = λ2AK = λ3K, and, in general,
AmK = λmK. If A is nilpotent with index m, then Am = 0 and λm = 0.
EXERCISES 8.10
Orthogonal Matrices
1. (a)–(b)
 0 0 −40 −4 0
−4 0 15

 01
0
 =
 0−4
0
 = −4
 01
0
; λ1 = −4
 0 0 −40 −4 0
−4 0 15

 40
1
 =
−40
1
 = (−1)
 40
1
; λ2 = −1
 0 0 −40 −4 0
−4 0 15

 10
−4
 =
 160
−64
 = 16
 10
−4
; λ3 = 16
(c) KT1 K2 = ( 0 1 0 )
 40
1
 = 0; KT1 K3 = ( 0 1 0 )
 10
−4
 = 0; KT2 K3 = ( 4 0 1 )
 10
−4
 = 0
2. (a)–(b)
 1 −1 −1−1 1 −1
−1 −1 1

−21
1
 =
−42
2
 = 2
−21
1
; λ1 = 2
 1 −1 −1−1 1 −1
−1 −1 1

 01
−1
 =
 02
−2
 = 2
 01
−1
; λ2 = 2
 1 −1 −1−1 1 −1
−1 −1 1

 11
1
 =
−1−1
−1
 = (−1)
 11
1
; λ3 = −1
(c) KT1 K2 = (−2 1 1 )
 01
−1
 = 1− 1 = 0; KT1 K3 = (−2 1 1 )
 11
1
 = −2 + 1 + 1 = 0
409
8.10 Orthogonal Matrices
KT2 K3 = ( 0 1 −1 )
 11
1
 = 1− 1 = 0
3. (a)–(b)
 5 13 013 5 0
0 0 −8


√
2
2√
2
2
0
 =
 9
√
2
9
√
2
0
 = 18

√
2
2√
2
2
0
; λ1 = 18
 5 13 013 5 0
0 0 −8


√
3
3
−
√
3
3√
3
3
 =

− 8
√
2
3
8
√
3
3
− 8
√
3
3
 = (−8)

√
3
3
−
√
3
3√
3
3
; λ2 = −8
 5 13 013 5 0
0 0 −8


√
6
6
−
√
6
6
−
√
6
3
 =

− 8
√
6
6
8
√
6
6
8
√
6
3
 = (−8)

√
6
6
−
√
6
6
−
√
6
3
; λ3 = −8
(c) KT1 K2 = (
√
2
2
√
2
2 0 )

√
3
3
−
√
3
3√
3
3
 = √66 −
√
6
6
= 0;
KT1 K3 = (
√
2
2
√
2
2 0 )

√
6
6
−
√
6
6
−
√
6
3
 = √1212 −
√
12
12
= 0
KT2 K3 = (
√
3
3 −
√
3
3
√
3
3 )

√
6
6
−
√
6
6
−
√
6
3
 = √1818 +
√
18
18
−
√
18
9
= 0
4. (a)–(b)
 3 2 22 2 0
2 0 4

−22
1
 =
 00
0
 = 0
−22
1
; λ1 = 0
 3 2 22 2 0
2 0 4

 12
−2
 =
 36
−6
 = 3
 12
−2
; λ2 = 3
 3 2 22 2 0
2 0 4

 21
2
 =
 126
12
 = 6
 21
2
; λ3 = 6
(c) KT1 K2 = (−2 2 1 )
 12
−2
 = −2 + 4− 2 = 0; KT1 K3 = (−2 2 1 )
 21
2
 = −4 + 2 + 2 = 0
KT2 K3 = ( 1 2 −2 )
 21
2
 = 2 + 2− 4 = 0
5. Orthogonal. Columns form an orthonormal set.
410
8.10 Orthogonal Matrices
6. Not orthogonal. Columns one and three are not unit vectors.7. Orthogonal. Columns form an orthonormal set.
8. Not orthogonal. The matrix is singular.
9. Not orthogonal. Columns are not unit vectors.
10. Orthogonal. Columns form an orthogonal set.
11. λ1 = −8, λ2 = 10, K1 =
(
1
−1
)
, K2 =
(
1
1
)
, P =
(
1√
2
1√
2
− 1√
2
1√
2
)
12. λ1 = 7, λ2 = 4, K1 =
(
1
0
)
, K2 =
(
0
1
)
, P =
(
1 0
0 1
)
13. λ1 = 0, λ2 = 10, K1 =
(
3
−1
)
, K2 =
(
1
3
)
, P =
(
3√
10
1√
10
− 1√
10
3√
10
)
14. λ1 =
1
2
+
√
5
2
, λ2 =
1
2
−
√
5
2
, K1 =
(
1 +
√
5
2
)
, K2 =
(
1−√5
2
)
, P =
 1+
√
5√
10+2
√
5
1−√5√
10−2√5
2√
10+2
√
5
2√
10−2√5

15. λ1 = 0, λ2 = 2, λ3 = 1, K1 =
−10
1
, K2 =
 10
1
, K3 =
 01
0
, P =
−
1√
2
1√
2
0
0 0 1
1√
2
1√
2
0

16. λ1 =−1, λ2 = 1−
√
2 , λ3 = 1+
√
2 , K1 =
−10
1
, K2 =
 1−√2
1
, K3 =
 1√2
1
, P =

− 1√
2
1
2
1
2
0 −
√
2
2
√
2
2
1√
2
1
2
1
2

17. λ1 = −11, λ2 = 0, λ3 = 6, K1 =
−31
1
, K2 =
 1−4
7
, K3 =
 12
1
, P =

− 3√
11
1√
66
1√
6
1√
11
− 4√
66
2√
6
1√
11
7√
66
1√
6

18. λ1 = −18, λ2 = 0, λ3 = 9, K1 =
 1−2
2
, K2 =
−21
2
, K3 =
 22
1
, P =

1
3 − 23 23
− 23 13 23
2
3
2
3
1
3

19.
(
3
5 a
4
5 b
)( 3
5
4
5
a b
)
=
(
1 0
0 1
)
implies 925 + a
2 = 1 and 1625 + b
2 = 1. These equations give a = ± 45 , b = ± 35 .
But 1225 + ab = 0 indicates a and b must have opposite signs. Therefore choose a = − 45 , b = 35 .
The matrix
(
3
5 − 45
4
5
3
5
)
is orthogonal.
20.
(
1√
5
b
a 1√
5
)( 1√
5
a
b 1√
5
)
=
(
1 0
0 1
)
implies 15 + b
2 = 1 and a2 + 15 = 1. These give a = ± 2√5 , b = ± 2√5 .
But
a√
5
+
b√
5
= 0 indicates a and b must have opposite signs. Therefore choose a = − 2√
5
, b = 2√
2
.
The matrix
(
1√
5
2√
5
− 2√
5
1√
5
)
is orthogonal.
411
8.10 Orthogonal Matrices
21. (a)–(b) We compute
AK1 =
 0 2 22 0 2
2 2 0

 1−1
0
 =
−22
0
 = −2
 1−1
0
 = −2K1
AK2 =
 0 2 22 0 2
2 2 0

 10
−1
 =
−20
2
 = −2
 10
−1
 = −2K2
AK3 =
 0 2 22 0 2
2 2 0

 11
1
 =
 44
4
 = 4
 11
1
 = 4K3 ,
and observe that K1 is an eigenvector with corresponding eigenvalue −2, K2 is an eigenvector with corre-
sponding eigenvalue −2, and K3 is an eigenvector with corresponding eigenvalue 4.
(c) Since K1 · K2 = 1 �= 0, K1 and K2 are not orthogonal, while K1 · K3 = 0 and K2 · K3 = 0 so K3
is orthogonal to both K1 and K2, To transform {K1,K2} into an orthogonal set we let V1 = K1 and
compute K2 ·V1 = 1 and V1 ·V1 = 2. Then
V2 = K2 − K2 ·V1V1 ·V1 V1 =
 10
−1
− 12
 1−1
0
 =

1
2
1
2
−1
 .
Now, {V1,V2,K3} is an orthogonal set of eigenvectors with
||V1|| =
√
2, ||V2|| = 3√
6
, and ||K3|| =
√
3.
An orthonormal set of vectors is

1√
2
− 1√
2
0
 ,

1√
6
1√
6
− 2√
6
 , and

1√
3
1√
3
1√
3
 ,
and so the matrix
P =

1√
2
1√
6
1√
3
− 1√
2
1√
6
1√
3
0 − 2√
6
1√
3

is orthogonal.
412
8.10 Orthogonal Matrices
22. (a)–(b) We compute
AK1 =

1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1


−1
0
0
1
 =

0
0
0
0
 = 0

−1
0
0
1
 = 0K1
AK2 =

1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1


−1
0
1
0
 =

0
0
0
0
 = 0

−1
0
1
0
 = 0K2
AK3 =

1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1


−1
1
0
0
 =

0
0
0
0
 = 0

−1
1
0
0
 = 0K3
AK4 =

1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1


1
1
1
1
 =

4
4
4
4
 = 4

1
1
1
1
 = 4K4,
and observe that K1 is an eigenvector with corresponding eigenvalue 0, K2 is an eigenvector with corre-
sponding eigenvalue 0, K3 is an eigenvector with corresponding eigenvalue 0, and K4 is an eigenvector with
corresponding eigenvalue 4.
(c) Since K1 ·K2 = 1 �= 0, K1 and K2 are not orthogonal. Similarly , K1 ·K3 = 1 �= 0 and K2 ·K3 = 1 �= 0
so K1 and K3 and K2 and K3 are not orthogonal. However, K1 ·K4 = 0, K2 ·K4 = 0, and K3 ·K4 = 0,
so each of K1,K2, and K3 is orthogonal to K4. To transform {K1,K2,K3} into an orthogonal set we let
V1 = K1 and compute K2 ·V1 = 1 and V1 ·V1 = 2. Then
V2 = K2 − K2 ·V1V1 ·V1 V1 =

−1
0
1
0
− 12

−1
0
0
1
 =

− 12
0
1
− 12
 .
Next, using K3 ·V1 = 1, K3 ·V2 = 12 , and V2 ·V2 = 32 , we obtain
V3 = K3 =
K3 ·V1
V1 ·V1 V1 −
K3 ·V2
V2 ·V2 V2 =

−1
1
0
0
− 12

−1
0
0
1
− 1/23/2

− 12
0
1
− 12
 =

− 13
1
− 13
− 13
 .
Now, {V1,V2,V3,K4} is an orthogonal set of eigenvectors with
||V1|| =
√
2, ||V2|| = 3√
6
, ||K3|| = 2√
3
and ||K4‖ = 2.
An orthonormal set of vectors is
− 1√
2
0
0
1√
2
 ,

− 1√
6
0
2√
6
− 1√
6
 ,

− 1
2
√
3
3
2
√
3
− 1
2
√
3
− 1
2
√
3
 , and

1
2
1
2
1
2
1
2
 ,
413
8.10 Orthogonal Matrices
and so the matrix
P =

− 1√
2
− 1√
6
− 1
2
√
3
1
2
0 0 3
2
√
3
1
2
0 2√
6
− 1
2
√
3
1
2
1√
2
− 1√
6
− 1
2
√
3
1
2

is orthogonal.
23. If we take K1 =
 01
1
 as in Example 4 in the text then we look for a vector K2 =
 ab
c
 such that 1(a) +
1
4 b− 14 c = 0 and K1 ·K2 = 0 or b + c = 0. The last equation implies c = −b so a + 14 b− 14 (−b) = a + 12 b = 0.
If we let b = −2, then a = 1 and c = 2, so a second eigenvector with eigenvalue −9 and orthogonal to K1 is
K2 =
 1−2
2
.
24. The eigenvalues and corresponding eigenvectors of A are
λ1 = λ2 = −1, λ3 = λ4 = 3, and K1 =

−1
1
0
0
 , K2 =

0
0
−1
1
 , K3 =

0
0
1
1
 , K4 =

1
1
0
0
 .
Since K1 ·K2 = K1 ·K3 = K1 ·K4 = K2 ·K3 = K2 ·K4 = K3 ·K4 = 0, the vectors are orthogonal. Using
‖K1‖ = ‖K2‖ = ‖K3‖ = ‖K4‖ =
√
2 , we construct the orthogonal matrix
P =

− 1√
2
0 0 1√
2
1√
2
0 0 1√
2
0 − 1√
2
1√
2
0
0 1√
2
1√
2
0
 .
25. Suppose A and B are orthogonal matrices. Then A−1 = AT and B−1 = BT and
(AB)−1 = B−1A−1 = BTAT = (AB)T .
Thus AB is an orthogonal matrix.
EXERCISES 8.11
Approximation of Eigenvalues
1. Taking X0 =
(
1
1
)
and computing Xi = AXi−1 for i = 1, 2, 3, 4 we obtain
X1 =
(
2
2
)
, X2 =
(
4
4
)
, X3 =
(
8
8
)
, X4 =
(
16
16
)
.
We conclude that a dominant eigenvector is K =
(
1
1
)
with corresponding eigenvalue λ =
AK ·K
K ·K =
4
2
= 2.
414
8.11 Approximation of Eigenvalues
2. Taking X0 =
(
1
1
)
and computing Xi = AXi−1 for i = 1, 2, 3, 4, 5 we obtain
X1 =
(−5
7
)
, X2 =
(
49
−47
)
, X3 =
(−437
439
)
, X4 =
(
3937
−3935
)
, X5 =
(−35429
35431
)
.
We conclude that a dominant eigenvector is K =
1
35439
(−35429
35431
)
≈
(−0.99994
1
)
with corresponding
eigenvalue λ =
AK ·K
K ·K = −8.9998.
3. Taking X0 =
(
1
1
)
and computing AX0 =
(6
16
)
, we define X1 =
1
16
(
6
16
)
=
(
0.375
1
)
. Continuing in this
manner we obtain
X2 =
(
0.3363
1
)
, X3 =
(
0.3335
1
)
, X4 =
(
0.3333
1
)
.
We conclude that a dominant eigenvector is K =
(
0.3333
1
)
with corresponding eigenvalue λ = 14.
4. Taking X0 =
(
1
1
)
and computing AX0 =
(
1
5
)
, we define X1 =
1
5
(
1
5
)
=
(
0.2
1
)
. Continuing in this manner
we obtain
X2 =
(
0.2727
1
)
, X3 =
(
0.2676
1
)
, X4 =
(
0.2680
1
)
, X5 =
(
0.2679
1
)
.
We conclude that a dominant eigenvector is K =
(
0.2679
1
)
with corresponding eigenvalue λ = 6.4641.
5. Taking X0 =
 11
1
 and computing AX0 =
 1111
6
, we define X1 = 111
 1111
6
 =
 11
0.5455
. Continuing in
this manner we obtain
X2 =
 11
0.5045
 , X3 =
 11
0.5005
 , X4 =
 11
0.5
 .
We conclude that a dominant eigenvector is K =
 11
0.5
 with corresponding eigenvalue λ = 10.
6. Taking X0 =
 11
1
 and computing AX0 =
 52
2
, we define X1 = 15
 52
2
 =
 10.4
0.4
. Continuing in this
manner we obtain
X2 =
 10.2105
0.2105
 , X3 =
 10.1231
0.1231
 , X4 =
 10.0758
0.0758
 , X5 =
 10.0481
0.0481
 .
At this point if we restart with X0 =
 10
0
 we see that K =
 10
0
 is a dominant eigenvector with corresponding
eigenvalue λ = 3.
415
8.11 Approximation of Eigenvalues
7. Taking X0 =
(
1
1
)
and using scaling we obtain
X1 =
(
0.625
1
)
, X2 =
(
0.5345
1
)
, X3 =
(
0.5098
1
)
, X4 =
(
0.5028
1
)
, X5 =
(
0.5008
1
)
.
Taking K =
(
0.5
1
)
as the dominant eigenvector we find λ1 = 7. Now the normalized eigenvector is
K1 =
(
0.4472
0.8944
)
and B =
(
1.6 −0.8
−0.8 0.4
)
. Taking X0 =
(
1
1
)
and using scaling again we obtain X1 =(
1
−0.5
)
, X2 =
(
1
−0.5
)
. Taking K =
(
1
−0.5
)
we find λ2 = 2. The eigenvalues are 7 and 2.
8. Taking X0 =
(
1
1
)
and using scaling we obtain X1 =
(
0.3333
1
)
, X2 =
(
0.3333
1
)
. Taking K =
(
1/3
1
)
as the
dominant eigenvector we find λ1 = 10. Now the normalized eigenvector is K1 =
(
0.3162
0.9486
)
and B =
(
0 0
0 0
)
.
An eigenvector for the zero matrix is λ2 = 0. The eigenvalues are 10 and 0.
9. Taking X0 =
 11
1
 and using scaling we obtain
X1 =
 10
1
 , X2 =
 1−0.6667
1
 , X3 =
 1−0.9091
1
 , X4 =
 1−0.9767
1
 , X5 =
 1−0.9942
1
 .
Taking K =
 1−1
1
 as the dominant eigenvector we find λ1 = 4. Now the normalized eigenvector is
K1 =
 0.5774−0.5774
0.5774
and B =
 1.6667 0.3333 −1.33330.3333 0.6667 0.3333
−1.3333 0.3333 1.6667
. If X0 =
 11
1
 is now chosen only one more
eigenvalue is found. Thus, try X0 =
 11
0
. Using scaling we obtain
X1 =
 10.5
−0.5
 , X2 =
 10.2
−0.8
 , X3 =
 10.0714
−0.9286
 , X4 =
 10.0244
−0.9756
 , X5 =
 10.0082
−0.9918
 .
Taking K =
 10
−1
 as the eigenvector we find λ2 = 3. The normalized eigenvector in this case is
K2 =
 0.70710
−0.7071
 and C =
 0.1667 0.3333 0.16670.3333 0.6667 0.3333
0.1667 0.3333 0.1667
. If X0 =
 11
1
 is chosen, and scaling is used we
416
8.11 Approximation of Eigenvalues
obtain X1 =
 0.51
0.5
, X2 =
 0.51
0.5
. Taking K =
 0.51
0.5
 we find λ3 = 1. The eigenvalues are 4, 3, and 1.
The difficulty in choosing X0 =
 11
1
 to find the second eigenvector results from the fact that this vector is a
linear combination of the eigenvectors corresponding to the other two eigenvalues, with 0 contribution from the
second eigenvector. When this occurs the development of the power method, shown in the text, breaks down.
10. Taking X0 =
 11
1
 and using scaling we obtain
X1 =
−0.3636−0.3636
1
 , X2 =
−0.24310.0884
1
 , X3 =
−0.2504−0.0221
1
 , X4 =
−0.2499−0.0055
1
 .
Taking K =
−0.250
1
 as the dominant eigenvector we find λ1 = 16. The normalized eigenvector is
K1 =
−0.24250
0.9701
 and B =
−0.9412 0 −0.23530 −4 0
−0.2353 0 −0.0588
. Taking X0 =
 11
1
 and using scaling we obtain
X1 =
−0.2941−1
−0.0735
 , X2 =
 0.07351
0.0184
 , X3 =
−0.0184−1
−0.0046
 , X4 =
 0.00461
0.0011
 .
Taking K =
 01
0
 as the eigenvector we find λ2 = −4. The normalized eigenvector in this case is K2 = K =
 01
0
 and C =
−0.9412 0 −0.23530 0 0
−0.2353 0 −0.0588
. Taking X0 =
 11
1
 and using scaling we obtain X1 =
 −10
−0.25
,
X2 =
 10
0.25
. Using K =
 10
0.25
 we find λ3 = −1. The eigenvalues are 16, −4, and −1.
11. The inverse matrix is
(
4 −1
−3 1
)
. Taking X0 =
(
1
1
)
and using scaling we obtain
X1 =
(
1
−0.6667
)
, X2 =
(
1
−0.7857
)
, X3 =
(
1
−0.7910
)
, X4 =
(
1
−0.7913
)
.
Using K =
(
1
−0.7913
)
we find λ = 4.7913. The minimum eigenvalue of
(
1 1
3 4
)
is 1/4.7913 ≈ 0.2087.
417
8.11 Approximation of Eigenvalues
12. The inverse matrix is
(
1 3
4 2
)
. Taking X0 =
(
1
1
)
and using scaling we obtain
X1 =
(
0.6667
1
)
, X2 =
(
0.7857
1
)
, . . . , X10 =
(
0.75
1
)
.
Using K =
(
0.75
1
)
we find λ = 5. The minimum eigenvalue of
(−0.2 0.3
0.4 −0.1
)
is 1/5 = 0.2
13. (a) Replacing the second derivative with the difference expression we obtain
EI
yi+1 − 2yi + yi−1
h2
+ Pyi = 0 or EI(yi+1 − 2yi + yi−1) + Ph2yi = 0.
(b) Expanding the difference equation for i = 1, 2, 3 and using h = L/4, y0 = 0, and y4 = 0 we obtain
EI(y2 − 2y1 + y0) + PL
2
16
y1 = 0
EI(y3 − 2y2 + y1) + PL
2
16
y2 = 0
EI(y4 − 2y3 + y2) + PL
2
16
y3 = 0
or
2y1 − y2 = PL
2
16EI
y1
−y1 + 2y2 − y3 = PL
2
16EI
y2
−y2 + 2y3 = PL
2
16EI
y3.
In matrix form this becomes  2 −1 0−1 2 −1
0 −1 2

 y1y2
y3
 = PL216EI
 y1y2
y3
 .
(c) A−1 =
 0.75 0.5 0.250.5 1 0.5
0.25 0.5 0.75

(d) Taking X0 =
 11
1
 and using scaling we obtain
X1 =
 0.751
0.75
 , X2 =
 0.71431
0.7143
 , X3 =
 0.70831
0.7083
 , X4 =
 0.70731
0.7073
 , X5 =
 0.70711
0.7071
 .
Using K =
 0.70711
0.7071
 we find λ = 1.7071. Then 1/λ = 0.5859 is the minimum eigenvalue of A.
(e) Solving
PL2
16EI
= 0.5859 for P we obtain P = 9.3726
EI
L2
. In Example 3 of Section 3.9 we saw
P = π2
EI
L2
≈ 9.8696 EI
L2
.
14. (a) The difference equation is
EIi(yi+1 − 2yi + yi−1) + Ph2yi = 0, i = 1, 2, 3,
418
8.11 Approximation of Eigenvalues
where I0 = 0.00200, I1 = 0.00175, I2 = 0.00150, I3 = 0.00125, and I4 = 0.00100. The system of equations
is
0.00175E(y2 − 2y1 + y0) + PL
2
16
y1 = 0
0.00150E(y3 − 2y2 + y1) + PL
2
16
y2 = 0
0.00125E(y4 − 2y3 + y2) + PL
2
16
y2 = 0
or
0.0035y1 − 0.00175y2 = PL
2
16E
y1
−0.0015y1 + 0.003y2 − 0.0015y3 = PL
2
16E
y2
−0.00125y2 + 0.0025y3 = PL
2
16E
y3.
In matrix form this becomes 0.0035 −0.00175 0−0.0015 0.003 −0.0015
0 −0.00125 0.0025

 y1y2
y3
 = PL216E
 y1y2
y3
 .
(b) The inverse of A is
A−1 =
 428.571 333.333 200285.714 666.667