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Киевский национальный университет им. Т. Шевченко

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electrodynamics / Classical Electrodynamics for Undergraduates - H. Norbury

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10	CHAPTER 1. MATRICES
	(1.20)

Notice that the ij matrix element of the inverse is given by the ji cofactor.

||||||||||||||||||||||||||||||||{

Example 1.3.2 Find the inverse of			1		1	! and check your
Example 1.3.2 Find the inverse of			1	°	1	! and check your
answer.				°
Solution Let's write A =	A11	A12	!	=		1		1	!.
Solution Let's write A =	A21	A22	!	=		1	°	1	!.
							°

Now cof(A11) ¥ (°)1+1jA22j = A22 = °1. Notice that the determinant in the cofactor is just a single number for 2 £ 2 matrices. The other cofactors are

cof(A12) ¥ (°)1+2jA21j = °A21 = °1

cof(A21) ¥ (°)2+1jA12j = °A12 = °1

cof(A22) ¥ (°)2+2jA11j = A11 = 1 .

The determinant of A is jAj = A11A22 ° A21A12 = °1 ° 1 = °2. Thus from (1.20)

(A°1)11 ¥ °12 cof(A11) = 12 , (A°1)12 ¥ °12 cof(A21) = 12 , (A°1)21 ¥ °12 cof(A12) = 12 , (A°1)22 ¥ °12 cof(A22) = °12 .

Thus A°1	= 1	1	1	.
Thus A°1	= 1	1	°1	.
	2

We can check our answer by making sure that AA°1 = I as

follows,	1		1 !		1		1	! = 2	1		1	! 1		1 !
AA°1 =	1		1 !	2	1		1	! = 2	1		1	! 1		1 !
	1	°	1	1	1	°	1	1	1	°	1	1	°	1
= 21 0	2 !	°	0		1 !.	°				°			°
= 21 0	2 !	=	0		1 !.
2	0		1		0

Thus we are now sure that our calculation of A°1 is correct. Also having veriØed this gives justiØcation for believing in all the formulas for cofactors and inverse given above. (do Problems 1.4 and 1.5)

||||||||||||||||||||||||||||||||{

1.4. SOLUTION OF COUPLED EQUATIONS

1.4Solution of Coupled Equations

By now we have developed quite a few skills with matrices, but what is the point of it all ? Well it allows us to solve simultaneous (coupled) equations

(such as (1.5)) with matrix methods.						Writing (1.6)) as AX = Y where
A =	1	1	! ; X =	x	! ; Y =	2	!, we see that the solution we
A =	1	1	! ; X =	y	! ; Y =	0	!, we see that the solution we
		°

want is the value for x and y. In other words we want to Ønd the column

matrix X. We have
AX = Y	(1.21)
so that
A°1AX = A°1Y:	(1.22)
Thus
X = A°1Y	(1.23)

where we have used (1.18).

||||||||||||||||||||||||||||||||{

Example 1.4.1 Solve the set of coupled equations (1.5) with matrix methods.

Solution Equation (1.5) is re-written in (1.6) as AX = Y with

A = 1

! ; X =

! ; Y

We want X =

A°1Y . From Example 1.3.2 we have A°1

1 !. Thus

X = A°1Y means

0 !

= 21

2 !

1 !.

y ! = 21

1 1 !

Thus x = 1 and y = 1.

||||||||||||||||||||||||||||||||{

1.5Summary

This concludes our discussion of matrices. Just to summarize a little, Ørstly it is more than reasonable to simply think of matrices as another way of writing and solving simultaneous equations. Secondly, our invention of how to multiply matrices in (1.12) was invented only so that it reproduces the

12	CHAPTER 1. MATRICES

coupled equations (1.5). For multiplying two square matrices equation (1.14) is just (1.12) with another index tacked on. Thirdly, even though we did not prove mathematically that the inverse is given by (1.20), nevertheless we can believe in the formula becasue we always found that using it gives AA°1 = I. It doesn't matter how you get A°1, as long as AA°1 = I you know that you have found the right answer. A proof of (1.20) can be found in mathematics books [3, 4].

1.6. PROBLEMS

1.6Problems

1.1Show that Bi = Aikxk gives (1.11).

1.2Show that Cij = AikBkj gives (1.13).

1.3Show that matrix multiplication is non-commutative, i.e. AB 6= BA.

1.4 Find the inverse of	1	1	and check your answer.
	0	1

	B	1	1
1.5 Find the inverse of	B	2	0
1.5 Find the inverse of	0	1	°1
	@

1 A and check your answer.

1.6Solve the following simultaneous equations with matrix methods: x + y + z = 4

x ° y + z = 2 2x + z = 4

14	CHAPTER 1. MATRICES

1.7Answers

1.4		1	°1	!
		0	1
		1	°	1
		°1 2	°	1
1.5	B		°	2
	B	2	°	2
	0	2		2
	@	1	1
		1	1

0 A

°1

1.6 x = 1; y = 1; z = 2.

1.8. SOLUTIONS

1.8Solutions

1.1

Bi = Aikxk. We simply evaluate each term. Thus

B1 = A1kxk = A11x1 + A12x2

B2 = A2kxk = A21x1 + A22x2.

1.2

Cij = AikBkj. Again just evaluate each term. Thus

C11 = A1kBk1 = A11B11 + A12B21

C12 = A1kBk2 = A11B12 + A12B22

C21 = A2kBk1 = A21B11 + A22B21

C22 = A2kBk2 = A21B12 + A22B22 .

1.3

This can be seen by just multiplying any two matrices, say
3	4 !	7	8 !	=	33	50	! .
1	2	5	6		19	22
	5	6	!	1	2	=	23	34	! .
Whereas 7		8	!	3	4 !	=	31	46	! .

showing that matrix multiplication does not commute.

16	CHAPTER 1. MATRICES

1.4	A21		! =				! . Now cof(A11) ¥
Let's write A =		A22			0	1
	A11	A12			1	1
(°)1+1jA22j = A22 =		1. Notice		that the determinant in the

cofactor is just a single number for 2 £ 2 matrices. The other cofactors are

cof(A12) ¥ (°)1+2jA21j = °A21 = 0 cof(A21) ¥ (°)2+1jA12j = °A12 = °1 cof(A22) ¥ (°)2+2jA11j = A11 = 1 .

The determinant of A is jAj = A11A22 ° A21A12 = 1. Thus from (1.20)

(A°1)11 ¥ 11 cof(A11) = 1 , (A°1)12 ¥ 11 cof(A21) = °1 , (A°1)21 ¥ 11 cof(A12) = 0 , (A°1)22 ¥ 11 cof(A22) = 1 .

Thus A°1 =	1	°1 .
	0	1

We can check our answer by making sure that AA°1 = I as

follows,			!		°1	! =			! = I. Thus we are
AA°1 =	0	1	!	0	°1	! =	0	1	! = I. Thus we are
	1	1		1	1		1	0

now sure that our answer for A°1 is correct.

1.8. SOLUTIONS

1.5

A21

A22

A23

= 0

1 .

Let's write A =

A11

A12

A13

B A31

A32

A33

C B 2 0 1 C

The cofactors

are

cof(A11) = (°)1+1

A32

A33

= +(A22A33 ° A32A23) = °1 ° 0 = °1

A22

A23

cof(A12) = (°)1+2

A31

A33

= °(A21A33 ° A31A23) = °1 + 2 = 1

A21

A23

cof(A13) = (°)1+3

A31

A32

= +(A21A32 ° A31A22) = 0 + 2 = 2

A21

A22

cof(A21) = (°)2+1

A32

A33

= °(A12A33 ° A32A13) = °1 + 0 = °1

A12

A13

cof(A22) = (°)2+2

A31

A33

= +(A11A33 ° A31A13) = 1 ° 2 = °1

A11

A13

cof(A23) = (°)2+3

A31

A32

= °(A11A32 ° A31A12) = °0 + 2 = 2

A11

A12

cof(A31) = (°)3+1

A22

A23

= +(A12A23 ° A22A13) = 1 + 1 = 2

A12

A13

cof(A32) = (°)3+2

A21

A23

= °(A11A23 ° A21A13) = °1 + 1 = 0

A11

A13

cof(A33) = (°)3+3

A21

A22

= +(A11A22 ° A21A12) = °1 ° 1 = °2

A11

A12

(1.16))

= 2. Thus from (1.20)

The determinant ofØ A is (see equationØ

(A° )11	¥ 2 cof(A11) = °2	,
(A°1)12	¥ 21 cof(A21) = °21	,
(A°1)13	¥ 21 cof(A31) = 1 ,

18																CHAPTER 1.				MATRICES
(A°1)21	¥	21 cof(A12) = 21 ,
(A°1)22	¥	21 cof(A22) =					°	21 .
(A°1)23	¥	1					°
(A°1)23	¥	2 cof(A32) = 0 ,
(A°1)31	¥	21 cof(A13) = 1 ,
(A°1)32	¥	21 cof(A23) = 1 .
	¥	1
(A°1)33 ¥		2 cof(A33) = °1 ,
			1	°		1			1
			°1 2	°		1			1
		B		°		2			0	C
		B	2	°		2				C
Thus A°1 = 0			2			2				1.
		@	1		1			°1		A
			1		1			°1
We can check our answer by making sure that AA°1 = I as follows,
AA°1 =	0	1	°1	1	1 0				2	1	2			0	1	= 0	0	1	0	1 = I. Thus
		1	1	1					°1	2	°1			1			1	0	0
	B				C B						2				C B					C
	B	2 0 1			C B				1		°1		°	1	C B		0 0 1			C
	@				A @							1	°		A	@				A
we are now sure that our answer for A°													is correct.

1.6

AX = Y is written as

= 0 2 1

1 °1

1 0 y

C B

2 0 1

C B z

4 C

A @

Y and we found A°

in problem 1.5.

Thus X = A°

Thus

0 y

= 0

1 0

= 0

°1 2

°1

B z

C B

Thus

°1 °1

A @

the solution is x = 1; y = 1; z = 2.

Chapter 2

VECTORS

In this review chapter we shall assume some familiarity with vectors from fresman physics. Extra details can be found in the references [1].

2.1Basis Vectors

We shall assume that we can always write a vector in terms of a set of basis vectors

^	^	^	+ A2e^2	+ A3e^3	= Aie^i	(2.1)
A = Axi + Ayj + Azk = A1e^1

where the components of the vector A are (Ax; Ay; Az) or (A1; A2; A3) and

^ ^ ^

the basis vectors are (i; j; k) or (^e1; e^2; e^3). The index notation Ai and e^i is preferrred because it is easy to handle any number of dimensions.

In equation (2.1) we are using the Einstein summation convention for

repeated indices which says that
xiyi ¥ Xi	xiyi:	(2.2)

In other words when indices are repeated it means that a sum is always implied. We shall use this convention throughout this book.

With basis vectors it is always easy to add and subtract vectors

A + B = Aie^i + Bie^i = (Ai + Bi)^ei:

(2.3)

Thus the components of A + B are obtained by adding the components of A and B separately. Similarly, for example,

A ° 2B = Aie^i ° 2Bie^i = (Ai ° 2Bi)^ei:

(2.4)

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