7.pdf Go Classes gateoverflow Linear Algebra

7/15

1. Chapter J Solving px =b

\

Aron

2. a E) Linear Algebra coclases

Chapter 2:
Determinant, Inverses, Eigenvalues and
| _— a - 9-7
Eigenvectors
In this chapter all matrices are square matrices] Ayn n
Ann a
We

©...

3. CEM

Determinants

Os

4. E calat

det CA) frank ( A)

f.M—OR

Où
pa l

5. CEM

Property 1
The determinant of the n by n identity matrix is 1.
nn. -
1
1 0
L 0 = i and =1
1

6. y E)
Property 2

The determinant changes sign when two rows ( or two columns) are
exchanged. ——

10 u 0 1 u
act | À =1 and det : o] =-1.
I ST

©...

7. # Ne) Line
Property 3

Linearity for one row at a time

ta tb mL b

6 ud yield

ata’ b+b a b| la y
; cd i e d

CE?

Gr

8. Ate we
4 m,
e
| a
= 7

il pas

+ A '

ef

ag he

L WA
gu

9. G CLASSES

14. o Lin

Tru e/ False Use only property 1, 2 or 3 to answer this question
See
ata’ b+b'|_|a db] lav
ete! dd led ce! d’
false
—

15. nl Lin

Tru e/ False Use only property 1, 2 or 3 to answer this questions

©...

16. de (oo det + dette)

A YO © > |
\ | © %6
pre = [ ]

17. CEM

Tru e/ False Use only property 1, 2 or 3 to answer this question

Ë
E

oo»
oO m 00
mm 00
ll
>
oor
oar NN
rr N

Os

18. CEM

Tru e/ False Use only property 1, 2 or 3 to answer this question

ta tb} _,|a@ 6
te dl le a

pS

Os

19. o Lin

Tru e/Fa Ise Use only property 1, 2 or 3 to answer this question

20. CEM

Tru e/ False Use only property 1, 2 or 3 to answer this question

_

If two rows of A are equal, then det(A) =0.

i)
b
yp RR, 11

Gr

22. ) u

Tru e/Fa Ise Use only property 1, 2 or 3 to answer this question
_ —

Subtracting a multiple of one row from another row leaves det A unchanged.

ee

a a RN
c=la d-b| |c d
Se ABs

AS R 2 xi
(Ou ocissozin

27. N
5 AM

Tru e/ False Use only property 1, 2 or 3 to answer this question

Subtracting a multiple of one row from another row leaves det A unchanged.

a Deia 1D
c-la d=tb| le d

Determinant is unchanged when we have R; > R¡ + KR;

©...

28. “Lines

Tru e/ False Use only property 1, 2 or 3 to answer this questions
—__ 7

A matrix with a row of zeros has det A = 0.

©...

29. $

wi“ A
0 0
® d =0
> a:

|. A —

BI
2 so so
pe

30. CEM

Tru e/ False Use only property 1, 2 or 3 to answer this question

Determinant of diagonal Matrix is product of diagonal elements

det ra = (411) (422) +++ (Ann).

0 nn

32. N
e E)

Tru e/ False Use only property 1, 2 or 3 to answer this questions
=>

Determinant of upper/lower Matrix is product of diagonal elements

A A, 9% a, O O
‘a \
| à Le O a, INT
O . O as
o 0 E) Oo 9
RED RAE Ay 02083
wee eye NO

Jwww.goclassesin

33. 14. [1 pt] Suppose

au
an
a31
Gat

A=

If det A = 3, what is det B?

a12
22
a32
az

13
023
Q33
Ga3

Qi
Qs
Q34
Gaa

and B=

au
an
a31
Qaı

013
03
Q33
043

au + 3013
G24 + 3093
Ga + 3033
Gas + 3043

34. 14. [1 pt] Suppose
au

a
A= 21
a31

Gat

If det A = 3, what is det B?

Answer: —3

a12
22
a32
az

13
023
Q33
Ga3

Qi
Qs
Q34
Gaa

and B

a
a22

Ga2

au
an
a31
Qaı

013
03
Q33
043

au + 3013
G24 + 3093
Ga + 3033
Gas + 3043

35. Gi 2 41,3 CD

12. Suppose that |az1 dz2 @23| = —2. Calculate the following two determinants.
93,1 432 033

21 42,2 023
aıı @2 Q13[= 9

431 032 033

ar 2a12 013 — 201,1
ai 2092 2,3 — 2a2,1| = -Y
as1 2432 033 — 203,1

36. a2,ı
ai
43,1

a1
a2,1
as,

a, 2 41,3
12. Suppose that |az1 dz2 @23| = —2. Calculate the following two determinants.

23,1 43,2 43,3

a2 G23
az 44,3|= Solution: 2
43,2 3,3

201,2 013 — 201,1
2a22 0233 — 2021] = Solution: —4
2a32 033 — 243,1

37. 4. If det |az a2 = 3, what is det |2a3ı 2a9 209 |?

Qu 412 an 31 432 03
a31 432 an Qu a12 Qı3

Answer:

~b
it~”

38. Qu G2 3 Ga 43 03
4. If det |a21 as» Qa23| = 3, what is det |2a7, 2a9 2a23|?
31 G32 033 aı 1 013

Answer: -6

39. Example 12.5. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. If B is
obtained from A by interchanging rows 2 and 4, what is det B? —

S -4

—

—

40. Example 12.5. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. If B is
obtained from A by interchanging rows 2 and 4, what is det B?

Solution. Interchanging (or swapping) rows changes the sign of the determinant. Therefore,

det B = —11

41. Example 12.6. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let
a,, az, az, ay denote the rows of A. If B is obtained from A by replacing row az by 3a; + a3,
what is det B?

A Ay > 4443
_ A2 TKK<—K<<
n= Gs

An

42. Example 12.6. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let
a,, &, az, ay denote the rows of A. If B is obtained from A by replacing row az by 3a; + a3,
what is det B?

Solution. This is a Type 3 elementary row operation, which preserves the value of the de-
terminant. Therefore,
det B = 11.

==
=> q =

43. Example 12.7. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let

a

A), &, az, ay denote the rows of A. If B is obtained from A by replacing row ay by 3a; +7ay,
what is det B? —
" >
Ri RULERS
[ei 2, ZZ (TR Lu 3 a, A) L
— rn
Ki a y
— a, = a
4, a
CA a, 31% TAg

as
a

44. | g. — kK RL KBs
t
k \A)

ie Ry
O

45. Example 12.7. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let
a,, Ag, az, ay denote the rows of A. If B is obtained from A by replacing row ag by 3a) +7a3,
what is det B?

a

Solution. This is not quite a Type 3 elementary row operation because ag is multiplied by
7. The third row of B is bs = 3a; + 7az. Therefore, expanding det B along the third row

det B = (3a; + 7a3) cl
= 3a, cl +72, cl
= 7(as- ci)
= 7det A

=77

46. nl

Example 12.8. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let
1, 42, as, a4 denote the rows of A. If B is obtained from A by replacing row az by 4a, +5a2,
what is det B?

454%
ar a
N à t

47. a,
a % =
a +54,
+
<—

=;
Ay
=—

A —_
a, —
a)

=
i
bn

=>

a

— ae

—
SA,

— Ñ

48. CE :

Example 12.8. Suppose that A is a 4 x 4 matrix and suppose that det A = 11. Let
a;, 42, az, ay denote the rows of A. If B is obtained from A by replacing row az by 4a, +5a2,
what is det B?

Solution. Again, this is not a Type 3 elementary row operation. The third row of B is
bs = 4a; + 5a2. Therefore, expanding det B along the third row

det B = (4a; + 5a2) - cy,
= 4a, «cl +5ag-cF

=0+0

49. CEM

1.) If A

Il
sae
ree

c a -c b
fl'andB=|d -f el then det(A) = det(B).
i g — h

FALSE

3

50. CE :

aa
mo
on

a -c b
1) FA= adB=|d -f e| then det(A) = det(B).

g -ih

o
en

TRUE FALSE

Solution: This is true. The matrix B results from the matrix A by swapping one column
and then multiplying column 2 by —1. As det(A) = det(47) and a row swap changes the
determinant by a factor of —1, so does a column swap. The same is true for multiplying a
column by —1. So in total det(A) = (—1)(—1) det(B) = det(B).

©...

51. CEM

(b) Let A = [ai a, as] and B = [b, by bs] be two 3 x 3-matrices. Suppose that det(A) = 5
and b; = aj, b = a; + 2a2, b3 = az. What is det(B)?

ea

52. (b) Let A = [ai a, as] and B = [b, by bs] be two 3 x 3-matrices. Suppose that det(A) = 5
and b; = aj, b = a; + 2a2, b3 = az. What is det(B)?

(b) We have:
A 2220202001, q

The first column operation multiplies the determinant by 2, and the second column
operation does not change the value of the determinant. Hence,

det(B) = 2det(A) = 10.

©...

53. au

8. Let A= |
as

ar

a
2
a32
Q42

a3
a23
a33
043

Qa
Q24
a3q
aa

and B =

341 341 3413 3414
431 432 433 a34
aa Q22 dag 94

41 — 2aıı 049 — 2012 043 — 2013 aga — 2014

If det(A) = 5 then what is the value of det(B)?

54. aı Gi2 013 ais 3411 3412 3413 3414

an dy da a as as as a:

8. Let A= [92 22 a 24) Gp 31 32 33 34
31 032 G33 G34 aa Q22 dag 94
a4 42 Qu Ga aa — 2a11 G42 — 2012 043 — 2a13 Qua — 2014

If det(A) = 5 then what is the value of det(B)?
Solution: det(B) = (—3) x 5 = -15.

55. CEM

abe a b c
6.LetA=|d e f|,B= 9 h i 2

ghi 2d-a 2e-b 2f-c
Find det(B) knowing that det(A) = 5.

(A) det(B) = -5.
(B) det(B) = 5.
(C) det(B) = 10.
(D) det(B) = -10.
(E) det(B) .
(E) det(B) = 2.

56. CE :

abe a b c
6.LetA=|d e f|,B= 9 h i 2

ghi 2d-a 2e-b 2f-c
Find det(B) knowing that det(A) = 5.

(A) det(B) = -5.
(B) det(B) = 5.
(C) det(B) = 10.
(D) det(B) = -10.
(E) det(B) = -2.
(E) det(B) = 2.

Solution: (D) det(B) = (-2) x 5 = -10.

57. aı 412 413
A= az an ax,
3 432 33

and det A = 3. If

An An da an Q
B= {a an a3}, and C= fan a»
31 G32 033 31 Qg

what are det B and det C?

413 — 3011
x3 — 3071
433 — 3031

58. and det A = 3. If

an
B= |an
a31

what are det B and det C?

a2
12
32

an
A= laa
azı
Q23
&3l, and
33

Answer: det B = —3, det C = 3

2
42 da

a32

Cc

13

a33

an

azı
a31

>

1
az
a32

a3 — 3011
x3 — 3071
433 — 3031

59. Calculate determinant of 2x2 matrix

KY Use only property 1, 2 or 3 to answer this questions

a 0 0 A
j AN ed
__ a — .
rare
Na,

c Ô
Qc

ue

60. e E)
Calculate determinant of 2x2 matrix

ao
ao |
a bl_fa 0| [o 6 [ls
ce di le d c d
a a 0 0 b 0
AEG
ad - bc g\ =?

61. sses NES

mal > a, © o O % ha oO © Ae
225 Gn Mee] ae te Ot E ay On Os
” Ass Gy In As Ay Gen ds
| m 6 2
ay © © 2 o
L pd 0 2 we Li o © =
31 Ag an 8
Az 0,43 a *

LS

© goetessesin

62. oo Ag
+] a, Qi Os
Au Uds

|

63. G CLASSES
3

S =
— each ro ms
aı 412 413 3 “nee choices at
an An of a 4 ot hich alement
3 n

YY 8 ‘ e pris 15% ¿houcd bo
he el ACCES pasert

4 a, 0 © ay © 0 00 42

d | Oo An 0 + oy 5 r a0 0

> a

Og, 0 0 a O Aged

_

64. an 412
a 42
431 432

f a ser) colamn we
ab erst One Non ZE

413 5 1 i
423 3 det wih Bus ve
3

433

722 aus) (O

a © O =?
oe © — 2

el —ı

ha vo

Amen Hot

66. Aer ms

comi un

N
WED 5
ml, aes

=)

nf we ex pond ”
a |

68. 411 412 413 an 0 0 ay 0 0 0 ap 0
M1 42 43 = 0 ay 0!+| 0 O ag |+|an 0 0
431 432 43 0 0 az 0 ap 0 0 0 23
0 ay 0 0 0 ay 0 0 a3

+ 0 0 a3 |+| an 0 0 |+ 0 an 0

431 0 0 0 az 0 a31 0 0

Il

411422433 — 411423433 — 412421433
[phone] + 413421432 — 413422431.

69. CEA

Ste
nen au Peermalectiong
£y
dé Lee Tan) yes

<<

70. 5

0 1

0 1 \0 A+) =0

1 0 lo

1 0

= pet BAERS
Gar) —

71. a
NX G ER: ] Er elanentyy 70°
= — a

SRR N°
R: OER ul Hp ly

[= gear ST Dei LE

how fin der
kK

72. 1287
cl

73. Problem 6. Let A be an 4 by 4 matrix with entries a;;. In the expression of the
determinant of A what is the sign of the product a13422034041 ?

74. Or:
x
a Y à
Go.
xy + b >
TZ
p

+C%4
cdl

75. - A249

76. Problem 6. Let A be an 4 by 4 matrix with entries a;;. In the expression of the
determinant of A what is the sign of the product a13a22a34@41 ?

Solution. The corresponding permutation is (3,2,4,1). One needs two switches to
get (1,2,3,4) from it: switch the first and the fourth numbers, then switch the third and
the fourth numbers. The number of switches is even, so the sign is +.

77. ses ES

Properties of determinant

* Determinant evaluated across any row or column is same.

+ If all the elements of a row (or column) are zeros, then the value of
the determinant is zero

+ Determinant of transpose is same as original matrix.

Let A and B be two matrix, then det(AB) = det(A)*det(B).

If A be a matrix then, 1471 = (Apr.

Determinant of Inverse of matrix can be defined as ¡A I ‘A
Determinant of diagonal matrix, triangular matrix (upper triangular or lower
triangular matrix) is product of element of the principle diagonal

Gr

78. CE :

Since determinant of transpose is same as original matrix we can
column operations too.

If A is singular then det A = 0. If A is invertible then det A AH 0.

The determinant of AB is det A times det B: |AB| = | A| |B|-

©...

79. CEM

Is Det(A B 47?) = Det(B) ?

Os

80. T F Aand B = 2Aare n x n matrices. If det(A) = 4, then det(B) = 8.

81. T F Aand B = 2Aare n x n matrices. If det(A) = 4, then det(B) = 8.

False: Matrix B is obtained by scaling each row of A by two. Therefore, the deter-
minant doubles at each of these scalings: det(B) = 2" det(A) = 4-2".

82. 2. Circle the letters which correspond to the statements which are true for
all square 5 x 5 matrices A, B,C.

A. det (-A) = —det (A). .
B. det (3A) = 3det (A). H Ww

©. det (A + B) = det (A) + det (B). a
D. det (ABC) = det (A) det (B) det (C).

E. Determinant of A does not change if the rows A are rearranged in the
opposite order.

83. 2. Circle the letters which correspond to the statements which are true for
all square 5 x 5 matrices A, B,C.

A. det (—A) = -det (A).

B. det (34) = 3det (A).

C. det (A + B) = det (A) + det (B).

D. det (ABC) = det (A) det (B) det (C).

E. Determinant of A does not change if the rows A are rearranged in the
opposite order.

Ans.: A,D,E.

Solution. A. True. Multiplying A on —1 is the same as multiplying each row
on —1, so determinant is multiplied on (-1)?
B. No. The correct formula is det (3A)

C. No. For example, it is easy to write J, whose determinant is 1, as
a sum of two diagonal matrices with zeros and ones on the main diagonal;
their determinants are 0.

D. True. In general det (AB) = det (A)det (B), this is a theorem which
was proved in class. Applying it twice we obtain the formula in D.

E, True. This depends on the sign of the permutation (n,n — 1.
For n = 5 this permutation has 2 transpositions so it is even.

84. Zr)

A
A

Calgatade — Aekteinart

Ust na Par mutehion
Convert gie madrix de echelon form
A a dingalat )

omd Jedse product of pivots

Card using ca faces

86. ) u

The Pivot Formula a
echan dem

aT
After converting A to upper triangular matrix, the pivots d;, d,, ... dy
are on diagonal. —

If no row exchanges are involved, multiply those pivots to find the

determinant: det A = d:d,d3 ... dy
_ MS
In general : detA = +d,d2d3...d,

©...

87. G
voy .> ft *
A=|0 2 3 o 2
4 5 6
hh 2 pul
PA
magie A

©...

88. CE :

Roo
ane
Du
oo»
ona
Hwa

| Odd |
Odd_
row exchanges

| det A = —(4)(2)(1) = -8.

©...

89. nl

The Cofactor Formula

The determinant is the dot product of any row (or column) i of A with its
cofactors using other rows.

é DOI A

< 13
det A = a,1C11 + 012C12 + 01333 + + QinCin
Csfecetes of di;

90. y E) Line e

Howmany terms are there in Cofactor expansion ?

©...

92. al. =)

¢
o Y ’
N yal + Vela
{

À (2 Y 45 (2°) =

93. CEM

Write down determinant of A along expansion of 2"4 row

Be Zur A ro | +)

©...

94. CEM

Write down determinant of A along expansion of 1% column
_

95. 5 1 Cu =7 C2=3 G3=-2
0 2 3 Ca=-11 C32=-7 G3=2

| 1 -1 i Cu = 13 Ci2=9 C3 = -6
=>

@ along the 1st row:

det(A) = 1-13 + (-1)-9+2-(-6) =13-9-12=-8
e along the 2nd row:

det(A) = (-3):7+5:3+1:(—2) =-214+15-2=-8
@ along the 3rd row:

det(A) =0:(-11)+2:(-7)+3:2=0-—14+6 = -8
@ along the 1st column:

det(A) = 1-13 + (—3)-7+0-(-11) =13-21+0=-8
e along the 2nd column:

det(A) = (-1)-9+5-3+2-(-7)=-9+15-14=-8
e along the 3rd column:

det(A) = 2-(—6) +1-(—2)

96. Az

a

cH

is

Az
2 Ass

> “en
sin

CANE a

“ to~ Jack” ?

- Op (

+ om (l

97. CE :

Question

An di Az
IfA= @, 42 43| and A, is Cofactors of a,, then value of A is given by

93, 42 Az
(A) a,A,ta,Ay+a,A, (B) Ata, A, + ai Ay
©) a, Aj? a, Ay +a, Ay (DY a, Ayt a, Ay, + 43, Ay,
m

y Pa

98. coithod Siqn > ani not
__ BA
- cache ;
ysith Sg ón sec vi mA
MA 27

Er

| C1 | | J. @fact

99. CEM

True/False~

If we replace element a, to 9 then C,, changes false

Y 3 3
7 a 1 ra)
Sry 7
©

-1 1 0
-1 0 1

©...

100. y E)
True/False

C;; does not depend on aj; TRo E

Ouen

101. 5 Ve Line:
True/False Rot

In both of given matrices, C11, C12 and Cj3 are same

N 4
œ a
oo
N £6
aun
ond

©...

102. y E) Lineal

di A. 43

\

Let C;; be cofactor of aj;
4% An Ay

41 2 dz

q
a, 23

What could be the corresponding matrices for given determinants? _

a As ass

A Kr Cy (a

= 22011 + 2312 + 02113

OS — Qu Rs @)
Ay Qn Rs

a, 43

VV
=

©...

103. di A 43

7 Es a
a, da dy Let C;; be cofactor of aj; ©) Cay Aza 33

a) & Ax Cup Ay an AS

= CA Gy Fy
What could be the corresponding matrices for given determinants ? —

(a qua S du =0
1. A = 0z2Cj1 + 023C12 + 02113
Qu
0)

2. A= 022012 + d23Cz2 + dar Cae > a ES) 9 C

3. A= Cr + az + 023013 Asi . @) A2 2 Azz Azz

4. A= azılıı + 32012 +033€13 1 Az 9
SE ay %

104. # Lineal
ay 2 43 Let C;; be determinant of aj;
% an Ay @
di M dz ms

——”

Which of the following values evaluates to zero ?

Q Sy As
A = 21C11 + 022C12 + 023C13

2 ez |%ı An da
A = ayıCız + 021C22 + 031Cg2 dx Ay Ay
E AH 22011 + Azılız + Q23Cı3

T
wm

A= azılıı + agzlı2 + Q33C13

105. CE :

20. Determine the value of a11 A12 + @21A22 + a31 432 for the

23 0
matrix A = 4 1 -3 |, where À; is the cofactor of
20 1

each aj; ?

o b) 32 032 d)16 e) -16

106. CEA ...

20. Determine the value of a11 412 + 091 422 + a31 Aza for the

-2 3 0
matrix A = 4 1 -3 |, where À; is the cofactor of
2 0 1
each aj; ?
a) 0 b) -32 c) 32 d) 16 e) -16

107. True/False ?

If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero.

eg: Azılıı + A22C12 + A23C 3 = 0.
—

Ons

u 14

108. Ca

CA

Cay

ED

Ent

Ca

109. 9 % > b Gy q “ut J ss
i f i 4 Con not Sg
¡A

À
EN
Yre a 7

os y

y +

e

Ar

E
+ N Gp 13

ac

110. N
5 AM

Find product of below matrices
Cofades af 5+ WO

11 12 913] Cy, Ca Ca
G21 422 Qz3||Cı Coz 6 =?
G31 G32 MzzllCiz C23 Cas
jai © O
o m 0
ly Mz Am © O ¡Al

111. y E) Linear.

Find product of below matrices

pi o 6 fs6
Cu Ca Ca o | Por
Cı2 Con 6] - ? (€) rol 101

11 A 43
221 A22 A3

31 32 QzgllCiz C23 C33 6 d [A
cad #on A
Q11 12 013 iw) 0 0
| ex a22 a =? 0 SN 0
Q31 Az a
31 32 33 o e yp)

©...

114. G CLASSES

yo ©
o | 0
o o |
in Dad
o | xe) | Ca Cr Cot
oo ! Ci2 Cz2 6

w Ci3 (23 Css
—
A wverse

\\

115. N
5 AM

What will be inverse of A ?

A= |@21 422 Q23

431 432 433

11 Aı2 ca|

116. Adjoint of Matrix: The adjoint of a matrix is the transpose of the cofactor element matrix of the given matrix.

Cir Ca Ca
pdi CPSs [G2 Coz Coz
Als 1, Adj A Ci3 C23 C33
A
¿AS
==
cy Cu Cig
ce Cay ca @3
CA Cy Css

Jwww.goclassesin

117. CEM

118. CE :

ae
— 4 -2 1
Example: Find the inverse of the matrix A = ( 5 0) 5) ;
-1 2 6

Os

119. ses ES

10 31 15 31,,15 Ol
lates SM, sr 2!

= 4(0 x 6-3 x 2) + 2(5 x 6 - (-1) x 3) H(5 x 2-0 x (-1))
= 4(0 - 6) + 2(30 + 3) +1(10 - 0)
=-24+66+10

=52

120. Now, we will determine the adjoint of the matrix A by calculating
the cofactors of each element and then taking the transpose of
the cofactor matrix.

-6 14 -6
Adj A= e 25 -7 )
10 -6 10

The inverse of matrix A is given by the formula A! =

1/5 4 -e
aaa 25 2)
52 Lo -6 10
(3/26 7/26 -3/26
= | -33/52 25/52 -7/52 |
5/26 -3/26 5/26 /
(3/26 7/26 -3/26y
Answer: A"! = (asa 25/52 -7/52
5/26 -3/26 5/26

1 adja
IAI

ssesin

121. CEM

Inverse

f 1 =
A= | > has inverse A! = E E À :

Re

y we A
ez Ml a pAjCo) = ©

Os

122. 3. Define the matrices

0 1 0 0 8 9 -1 -1
3 —2 2 1 0 1 0 16
M= 0 15 O0 1 |” Nis 00 2 3
5 5 55 0 0 0 2

Compute the determinants: det(M), det(N), det(MN), det(M~'), det(adj(M)).
Hint: Recall that the adjoint matrix adj(M) satisfies M - adj(M) = det(M)I.

123. 3. Define the matrices

0 1 0 0 8 9 -1 -1
3 —2 2 1 0 1 0 16
M= 0 15 O0 1 |” Nis 00 2 3
5 5 55 0 0 0 2

Compute the determinants: det(M), det(N), det(MN), det(M~'), det(adj(M)).
Hint: Recall that the adjoint matrix adj(M) satisfies M - adj(M) = det(M)I.

Solution.
321 3 2
det(M) = (—1)det | 0 0 1 | =(-1)(-1)det =5,
555 =

det(N)=3x1x2x2=12 (an upper triangular matrix),
det(MN) = det(M) det(N) = 5 x 12 = 60,
1 _ 1 _ı
det(M~') = qa = ®

det(adj(M)) = det ((det(M))M7") = det (5M=!) = 51 det (M7!) = 5° = 125.

124. Li

#
2 -7 -9
4. (15 points) Let A = 2 5 6 |. Find the third column of A~! without computing the other
1 3 4
columns.
ine)
ar

125. y E) Line
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126. True/False ?

If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero.

EE Gu Haney +@2JC: =0.
U On 02
a Ga Uy = O
ly % 93

©...

127. CEM

Find product of below matrices

d
caf te we
11 12 913] Cy, Ca Ca m o à
21 Mz U23||Ci2 Ca Ca] = ? o No
231 G32 Azz| lC13 C3 Css !
o {Al
adjoint naht
„> Em

©...

128. CEM

Find product of below matrices

12
022
032

413
023
033

C31
C32
C33

|

Cy E
Cr
Ci3

411
A21
34

Can
C22
C23

412
422
432

C31
Cl = ?
C33

413
d3|= ?

a33

©...

129. CEM

What will be inverse of A ?

A= |G21 422 An

431 432 433

11 Aı2 cal

130. Adjoint of Matrix: The adjoint of a matrix is the transpose of the cofactor element matrix of the given matrix.

Az

-Adj A

131. CE :

4 -2 1
Example: Find the inverse of the matrix A = ( 5 O 3 ) ;
-1 2 6

Os

132. ses ES

10 31 15 31,,15 Ol
lates SM, sr 2!

= 4(0 x 6-3 x 2) + 2(5 x 6 - (-1) x 3) H(5 x 2-0 x (-1))
= 4(0 - 6) + 2(30 + 3) +1(10 - 0)
=-24+66+10

=52

133. Now, we will determine the adjoint of the matrix A by calculating
the cofactors of each element and then taking the transpose of
the cofactor matrix.

-6 14 -6
Adj A= e 25 -7 )
10 -6 10

The inverse of matrix A is given by the formula A! =

1/5 4 -e
aaa 25 2)
52 Lo -6 10
(3/26 7/26 -3/26
= | -33/52 25/52 -7/52 |
5/26 -3/26 5/26 /
(3/26 7/26 -3/26y
Answer: A"! = (asa 25/52 -7/52
5/26 -3/26 5/26

1 adja
IAI

ssesin

134. CEM

Inverse

1 =
A= : | has inverse A7! = | d a |
cd ad—be|-c a

Os

136. —2 -7 -9
4. (15 points) Let A = | 2 5 6 ) . Find the third column of 47! without computing the other

1.3 4 —
columns.
Cc
At =
„Ai ai
A
y Aro
= A o

138. Li

#
—2 -7 -9
. (15 points) Let A = 2 5 6 |. Find the third column of A~! without computing the other
1 3 4

columns.

SOLUTION. To compute the third column of A7!, we only need to solve the system Ax = (0 0 1)",
where the right hand side is the third column of identity Jz. Using Gaussian elimination, we get

-2 -7 -9|0 1 3 alı 13 4/1
2 5 6/0 28, 2 5 6 \o | 24/0 -1 -2]-2
1 3 ali -2 -7 910) Bra lo -1 -1| 2
( J Fo:

R3—Ra

Bso,

0 -2|-2
as Dan

Then applying back substitution, we have 13 = 4, 22 = —6, zı = 3. Therefore, the third column of
AT is (3 -64)7. - a

©...

139. CEM

0 0 0 0
1 0 0 0
4. Compute the determinant /0 0 0 1 0 y, > Y
00300 3 4
00003
-6
12
-12
18
-24

Bunw»

Os

140. CEM

2
0
4. Compute the determinant |0 \
0
0 3 0
A. -6 eo
B. 12 o 1
C. -12 20
-18 A
E. 18 a>
F. -24 —
=-&
—

141. Cramer’s Rule

Mathematics of millionaires

Heayf Exmpuday

Sy omar money
AN

rta 5 ea)
II
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142. Aup pose trot det CA) — 6) which means

pr oxi sts

145. G CLASSES

nd FT ie
A
\ Cig 1
ln __))

ay 4 %
AM Pr Gag
Ag vz a

3

De 3

147. CE JM

at oxist Curie)

qu
x= wb
a,
00) a
== desta) m
de = del (0) ,
= An) a

Fan det (As)

See

fy =
3)
A, =
_

A, =

by An ag
Le O22 Gag
bz 0, 35

0 +0 +0%#0

a bh %
F Pr |

97 vs as
E 4 a
Ay Ayn V2
934 452 bs

148. Ar =b Rolve foe oc (ohm rl exist)
Uns or | > for
"ee
9 Cammırs vale Beck GubShhbin

HF usuel E: fy Auch
u ~ You can Lind pl and EN by L.

149. Use Cramer's Rule to solve for the indicated unknowns.

Se { 2a, — 32,

te for x, and 2,

yo lind

quo” pan fe
14

Solution. € chs

1. Writing this system in matrix form, we find

[5 4] »-[-]

——
To find the matrix Aj, we remove the column of the coefficient matrix A which holds the
coefficients of +, and replace it with the corresponding entries in B. Likewise, we replace the
column of A which corresponds to the coefficients of z, with the constants to form the matrix

Ay. This yields
4-3 4
2 1 -2

Computing determinants, we get det(A) = 17, det (A,) = —2 and det (A,

A

—24, so that

_det(A)_ 2 _det(A) 24

Ted) — 17 #27 qet(4) IT

The reader can check that the solution to the system is (—,