Chapter 3 Elementary Matrix Operations and Determinants
3-1 Elementary Operations
Elementary row (or column) operations:
1 2
3 4
1. Interchanging any two adjacent rows (or columns). Eg. 



3 4
1 2
2
1 2
1
2. Multiplying any row (or column) by a nonzero constant. Eg. 



3 4
 6  8
3. Adding any constant multiple of a row (or column) to another row (or column).
1 2
  5  6
nd
st
Eg. 


 : Adding minus double of the 2 row to the 1 row.
3
4
3
4




Rank of A, Rank(A): Rank(A)=dim(Range(LA)), where LA: Fn→Fm.
Theorem A  Mm×n(F), P  Mm×m(F), and Q  Mn×n(F). If P and Q are both invertible, then
Rank(A)=Rank(PA)=Rank(AQ)=Rank(PAQ).
Theorem A, B  Mm×n(F), then Rank(A+B)  Rank(A)+ Rank(B). [2000 台大電研]
(Proof) y  Range(LA+LB).  x such that y=(LA+LB)(x)= LA(x)+LB(x)
∵ LA(x)  Range(LA), LB(x)  Range(LB), (LA+LB)(x)  Range(LA)+Range(LB)
 Range( LA  LB )  Range( LA )  Range( LB )  Rank(A+B)  Rank(A)+ Rank(B)
 Rank ( A)
Theorem A  Mm×n(F), B  Mn×p(F), then Rank(AB)  
. [1999 台大電研]
 Rank (B )
(Proof) y  Range(LALB).  x such that y=(LALB)(x)
∵ z=LB(x)  Range(LB), LA(z)  Range(LA), y=(LALB)(x)=LA(LB (x))=LA(z)  Range(LA),
 R( LA LB )  R( LA )  Rank ( AB)  Rank ( A)
Note: Rank(AB)  Rank(B) can be proved by Rank(A)=Rank(At)
Question: How to determine the rank of a matrix?
1 0 1
1 0 1
Eg. Given A= 0 1 1 , Rank(A)=dim(Span{ 0, 1, 1 })=2


    
1 0 1
1 0 1
1
2
Eg. By elementary operation, B= 
1

1
1 0 1
1

0
2 0 2

0
1 0 1


1 0 1
0
1 0 1
0 0 0
, Rank(B)=1
0 0 0

0 0 0
1 0 1 
Eg. For A= 2 1 4 , find Rank(A). [1993 台大電研]


3 0 a 
0 
0 
1 0 1 
1 0
1 0
1. a  3  rank ( A)  2
(Sol.) 2 1 4  2 1
2   2 1
0  ,



2. a  3  rank ( A)  3
3 0 a 
3 0 a  3
3 0 a  3
1  3 2  1 3
Eg. For A= 
 , determine the dimension of N(A) and the basis for N(A). What
 2  5 4  2 4
is Rank(A)? [1993 中山應數所]
1  3 2  1 3
1  3 0 0 3
1 0 0 0 3
1 0 0 0 0
(Sol.) 







 2  5 4  2 4
 2  5 0 0 4
 2  1 0 0 4
2 1 0 0 0
Rank(A)=2, Nullity(A)=dim(V)-Rank(A) =5-2=3.
x
 y
1

3
2

1
3

 
 2  5 4  2 4  z   0 

 
v 
 w
x
 2 1 3
 2 1 3
 y
 0  0   2 
 0  0   2 
 
     
     
 z   r  1   s 0  t 0 . The basis of N(A) is { 1 , 0, 0}
 
     
     
v
 0  1 0
 0  1 0
 w
 0  0 1
 0  0 1
In Matlab language, we can use the following instructions to obtain the rank of a matrix:
>>A=[2 1;4 3];
>>rank(A)
A=
2
1
4
3
ans =
2
3-2 Inverse of a Matrix
Inverse of a matrix A-1: A A-1= A-1A =I
Eg. Show that
dA 1
dA 1
  A 1
A . [交大電子所]
dx
dx
(Proof) A A-1= A-1A =I, A
dA 1
dA 1
dA 1 dA 1

A  0, ∴
  A 1
A
dx
dx
dx
dx
Eg. Show that (AB)-1=B-1A-1.
(Proof) AB(AB)-1=I=(AB)-1AB and ABB-1A-1=I=B-1A-1AB, ∴ (AB)-1=B-1A-1
Augmented matrix: [A|B]=(A(1),…,A(n),B(1),…,B(n))
Method of obtaining the inverse of a matrix A: M=Ep…E2E1,. If (A|I)=(MA|M)=(I|M), then M=A-1.
1  1 2
Eg. A= 3 0 1 , find A-1.


1 0 2
1  1 2 1 0 0
1  1 2 1 0 0
1  1 2 1 0 0 






0 1 0 1 
(Sol.) 3 0 1 0 1 0  0 3  5  3 1 0  0 1
1 0 2 0 0 1
0 1
0 0  5 0 1  3
0  1 0 1

2
1  1 2 1 0 0 
1  1 0 1
5


 0 1
0  1 0 1   0 1 0  1 0

1
0 0  5 0 1  3
0 0 1 0 
5

6
 
1 0 0 0
2 5  1 5
5


0
1 ,
1   0 1 0  1

3
0 0 1 0  1 5 3 5 

5 
 0 2 5  1 5
∴ A   1
0
1 
 0  1 5 3 5 
1
0 2 4
Eg. A= 2 4 2 , find A-1.
3 3 1

0 2 4 1 0 0 2 4 2 0 1 0 1 2 1 0

 

2 4 2 0 1 0    0 2 4 1 0 0   0 2 4 1


(Sol.)
3 3 1 0 0 1 3 3 1 0 0 1 3 3 1 0



1
1 2
 0 1
2

0  3  2


0
1
2

1

1 0 0 81
 0 1 0
4

0 0 1 3

8

0

1  
1
0 1 2 1 0
0
2
2
0 0   0 2 4 1 0 0 
 
3 
0 1  0  3  2 0
1
2
 





1
1
0
1
0
1


2
2

1 0  3 1

1 0  3 1
0 0   0 1 2
0 0   0 1 2
0
2
2




3 
3
0 0 4 3  3 
0 0 1 3
1
1




2
2
2
8
8




1
2
5
8
3
4
3
8
3
 1 5
4 
8
8
1

1 3
1
2 , A  
4
1
4
3

3

4 
 8
8
3
4
 1

2
1
4 

0

0

1
4 
1  4 4 
Eg. (a) Find the inverse of A, where A= 0 1  1 . (b) Find the null space of A. [交大控制、光


1  2 1 
1 4 0 
0 
電研究所] (Ans.) A-1= 1 3  1 , N(A)={ 0  }


 
1 2  1
0 
1 4 7 
Eg. Compute the inverse matrix of A= 2 5 8  by some elementary operations. [台科大電研]


3 6 10
 a b 0
Eg. Compute the inverse matrix of A=  c d 0 . [2003 台科大電研]


 0 0 1
 d /   b /  0
(Ans.) A-1=  c /  a /  0 , where △=ad-bc


 0
0
1
In Matlab language, we can use the following instructions to obtain the inverse of a matrix:
>>A=[2,1;4,3]
A=
2
1
4
3
>>inv(A)
ans =
1.5000 -0.5000
-2.0000
1.0000
3-3 System of Linear Equations
Matrix representation for system of linear equations:
 x1 
2 x1  3x2  x3  1 2 3 1   1 
Eg. 

   x 2   6  Ax  B .
x

x

2
x

6
1

1
2

 x   
2
3
 1
 3
Theorem A  Mn×n(F). AX=B has only one solution A-1B if A is invertible. Conversely, if the
system has exactly one solution, then A is invertible.
Theorem AX=B has at least one solution  Rank(A)=Rank(A|B).
Gaussian elimination method:
 x1  2 x 2  x3  1

Eg. Solve  2 x1  2 x 2  x3  1 .
3x  5 x  2 x  1
2
3
 1
1 2  1

(Sol.) 2 2 1
3 5  2
1 0 1

 0 1  1
0 0 1
 1

1
 1
3

 2 
 1 
1 2  1

0  2 3
0  1 1
1 0 0

0 1 0
0 0 1
1 2  1
 1


3   0  1 1
0  2 3
2 
 1

2
3 
1 2  1

0 1  1
0  2 3
 1

 2
3 
4

 3 , ∴ x1=4, x2=-3, x3=-1.
 1
In Matlab language, we can use the following instructions to solve a linear system of equations:
>>A=[1,3,5;2,4,6;1,2,1];
>>B=[2,1,3]'; % [2,1,3]’ is the transpose of [2,1,3]
 x  3 y  5z  2

>>rref([A,B]) % solve 2 x  4 y  6 z  1.
 x  2y  z  3

ans =
1.0000
0
0
0
1.0000
0
0
0
1.0000
-3.7500
4.0000
-1.2500
Homogeneous linear system: AX =0.
1  5
0 0
Eg. Solve AX=0 for A= 
0 0

0 0
1  5
0 0
(Sol.) 
0 0

0 0
0 2 0  4
1 0 0  3
0 0 4 8 

0 0 0 0
0 2 0  4
1 0 0  3
. [2003 台科大電研]
0 0 4 8 

0 0 0 0
 x1 
 x  0 
 2  
 x 3  = 0  ,
 x  0 
 x4   
 5  0 
 x 6 
 x1   4    2  5 
x       
 2   0   0  1 
 x 3  =r  3  +s  0  +t 0 
 x   0   1  0 
 x 4    2  0  0 
 5      
 x 6   1   0  0 
Theorem A  Mm×n(F). AX=0 has a nontrivial solution if m<n and Rank(A)=m.
Theorem Let K be the set of all solutions to AX=B, and let KH be the set of the solutions of AX=0.
Then for any solution S to AX=B, K={S}+ KH={S+h: h  KH }.
 x1 
1 2 1    7 
Eg. Solve 
  x2     .
1  1  1  x   4
 3
 x1 
1 
1 2 1    0
 
(Sol.) 
  x 2   0  K H  {t  2, t  F }
1

1

1

x   
3 
 3
1 
 x1 
1  1 
1
2
1
7




 
   
S  1  fulfills 
  x2    4  K  {1   t  2, t  F }
1

1

1

 x   
4
4 3 
 3
 x1  4 x2  x3  x4  3

Eg. Solve the system of linear equations:  2 x1  8 x2  x3  4 x4  9 . [中正電研]
 x  4 x  2 x  5 x  6
2
3
4
 1
 1  4 1 1

(Sol.)  2  8 1  4
 1 4  2 5
1  4  1 1

 0 0
1 2
0 0  1 2
3

1
 1
3

9
 6
1  4  1 1

 0 0
3 6
0 0  3 6
1  4  1 1

 0 0
1 2
0 0
0
0
3

3
 3
3

1
0
 x1  4 x2  x3  x4  3

 
x3  2 x 4  1  x3  1  2 x 4  x1  4x2  x4  4

00

 x1  4 4 1
 x   0  1   0 
Let x2=r, x4=s   2      r    s  
 x3  1 0 2
       
 x 4   0   0  1 
3-4 Determinants of Matrices
n
Determinant: det(A)=   1
i 1
i j
 
~
~
Aij  det Aij , where Aij is the (n-1)×(n-1) matrix obtained from A by
deleting row i and column j.
1
1
Eg. det( 
0

1
  1
1 2
1 0 1
4
0 1 1
~
i2
)=   1 Ai 2  det Ai 2
0 1 1 i 1

1 1 1
 
1 1 1
1 0 1
1 0 1
1 0 1
2 2
3 2
4 2






 1  det 0 1 1   1  0  det 0 1 1   1  0  det 1 1 1   1  1  det 1 1 1
1 1 1
1 1 1
1 1 1
0 1 1
Theorem A  Mn×n(F) and Rank(A)<n, then det(A)=0.
Theorem
(a) For A, B  Mn×n(F), then det(AB)=det(A)det(B).
(b) If A is invertible, det(A-1)=1/det(A).
1 2
1 3 
(c) det(At)=det(A). Eg. det( 
)=-2= det( 

)
3 4
 2 4
n
(d) det(A)=  Aii
i 1
0
0
 2
if A is a triangular matrix. Eg. det(  11.4  3 0  )=2×(-3)×(-4)=24.


 6.2 5.3  4
Theorem Q is invertible  det(Q)≠0.
Orthogonal matrix B: BtB=BBt=I.


Eg. B= 


1
2
3
2

3

2  is orthogonal because
1 
2 





1
2
3
2

3

2 
1 
2 
 1

 2
 3
 2
3

2  = 1 0 .


1  0 1 
2 
Unitary matrix B: BB*=I, where (B*)ij=Bji*.
Theorem (a) If B is orthogonal, then det(B)=±1.
(b) If B is unitary, then |det(B)|=1.
(Proof) (a) det(I)=1=det(BBt)=det(B) det(Bt)=[det(B)]2  det(B)=±1.
(b) det(I)=1=det(BB*)=det(B)det(B*)=det(B)det(B)*=|det(B)|2  |det(B)|=1.
Summary of properties of the determinants:
1. If B is obtained by exchanging two adjacent rows (or two adjacent columns) of A, then
1 2
3 4
det(B)=-det(A). Eg. det( 
)=-2 and det( 

 )=2.
3 4
1 2
2. If B is obtained by multiplying each entry of some row (or column) of A by c, then det(B)=c．det(A).
2 
1 2
1
Eg. det( 
)=-2 and det( 

 )=(-12)-(-18)=6=(-3)×(-2).
3 4
 9  12
3. If B is obtained from A by adding a multiple of row i (or column i) to row j (or column j), where i≠j,
1 2
  5  6
then det(A)=det(B). Eg. det( 
)=-2 and det( 
)=-2.

4 
3 4
3
4. det(I)=1.
0 0
 1

if A is a triangular matrix. Eg. det( 11.4  3 0 )=(1)×(-3)×(2)=-6.


 6.2 5.3 2
n
5. det(A)=  Aii
i 1
4 2
2 0
0 1
1
2
Eg. Find the determinants of A= 
3

1
(Sol.) Triangularize A, we obtain
det(A)=
1 4 2
3
2 2
0
4
3 0
1
2
1 2
2
3

2
2
1
4
2
3
0
6
4
2
0  12
5
7
2
4
6
0


 and A-1.


 3
3
4
2
1

4
0 6
0
0
0
0
2
3
4
2
3
3
8 3  16
3
1

4
0 6
0
0
0
0
∴ det(A-1)= -1/144
0
3
1 1
 2 1 1 1 
 , det(A)=?
Eg. A= 
 1 2 3  1


 3 1 1 2 
0
3
0
3 
0
3 
1 1
1 1
1 1
0  1  1  5 
0  1  1  5 
0  1  1  5 



(Sol.) A  
0 3
0 0 0  13
0 0
3
2
3
13 






0  13
0  4  1  7 
0 0 3 13 
0 0
det(A)=-[1．(-1)．3．(-13)]=-39
2
3
4
2
3 3
0
8
=-144,
Eg. Digits 1 to 9 can be arranged into 3×3 matrices in 9! ways. Find the sum of the determinants
of these matrices. [台大電研]
a b
(Sol.) [det(  d e

 g h
c
d e
f  )+det(  a b
 g h
i 
f
a
c  )]+[det( b
 c
i 
x2
x3
1  x1
 x
1  x2
x3
 1
-1
Eg. Find det(A ) for A=  x1
x2
1  x3

.....
.....
 .....
 x1
x2
x3
(Sol.)
d
e
f
g
a
h  )+det( b
 c
i 
g
h
i
d
e  )]+…=0.
f 
.....
xn 
.....
x n 
.....
x n  . [1990 中央土木所]

..... ..... 
..... 1  x n 
x2 
1 
1 
1  x1
1
1
det ( 
)

det(
)

det(
x 1  x 
0 1  x  x )  1  x1  x2 ,
1  x2 
2
1
2
 x1
 1

x2
x3 
1
0 
1  x1
1
1




det(  x1
1  x2
x3  )  det(  x1 1  x2
x3  )  det(  0
 x1
 x1
 x1
x2
1  x3 
x2
1  x3 
1
1
x2
0 
 1  )
1  x3 
1
0 
0
1
1  1




)  1  x  x  x ,…, ∴ det A 1   1  1
1
 1 )  det( 0 1
1
= det( 0
1
2
3
n

det  A
1   xi
0 x1  x2 1  x3 
0 0 1  x1  x2  x3 
i 1
 2 1
 1 3
Eg. Triangularize the matrix: 
3
0

 2 2
0 3
2 4 
. [台大電研]
2  1

4 6
In Matlab language, we can use the following instructions to obtain the determinant of a matrix:
>>A=[1,3,0;-1,5,2;1,2,1];
>>det(A)
ans =
10
Eg. Let f1(t), f2(t), …, fn(t) be the polynomials of degrees  n-2. Prove that det(A)=0
 f1 a1 
 .....

A=  .....

 .....
 f n a1 
f1 a 2 
.....
.....
.....
f n a 2 
f1 a3 
.....
.....
.....
f n a3 
.....
.....
.....
.....
.....
f1 a n 
..... 
.....  , where a1, a2, …, an are arbitrary constants. [1991 台大

..... 
f n a n 
電研]

(Sol.) i  n, f i t   ai 0  ai1t  .....  ai ,n2 t n2  Span 1, t , ...., t n2



∵ dim Span 1, t , ...., t n  2  n  1  n , ∴ { f1(t), f2(t), …, fn(t)} is linearly dependent.
That is,  j
f j t   c1 f1 t   c 2 f 2 t   ...  c j 1 f j 1 (t )  c j 1 f j 1 (t )  ...  c n f n t , 1  j  n
s.t
 for arbitrary ai , f j ai   c1 f1 ai   c 2 f 2 ai   ...  c j 1 f j 1 (ai )  c j 1 f j 1 (ai )  ...  c n f n ai 
 f1 a1 
 .....

 A   f j a1 

 .....
 f n a1 
 f1 a1 

.....

  ci f i a1 
i j

.....

 f n a1 
f1 a 2 
.....
f j a 2 
.....
f n a 2 
f1 a3 
.....
.....
.....
f n a3 
f1 a 2 
.....
 ci f i a2 
.....
.....
.....
.....
.....
f1 a n 
..... 
f j a n 

..... 
f n a n 
f1 a3  .....
.....
.....
.....
.....
i j
.....
f n a 2 
for
.....
f n a3 
f1 a n  

.....

 ci f i an   rank A  n  detA  0
i j

.....
.....

.....
f n a n  
3-5 Classical Adjoint Matrix
n
i j
Cofactor: For det(A)=   1
  is the cofactor of A .
 
~
~
i j
Aij det Aij . , the scalar factor  1 det Aij
ij
1
2
1
Eg. For A= 
2

3
1 1
5
1  4  1
, find the cofactor of A41.
0 3 1 

6 1
2
(Sol.) The cofactor of A41 =  1
4 1
5
1 1
det 1  4  1  23
0  3 1 
Classical adjoint matrix: adj(A)= det(A)A-1  C is called the classical adjoint matrix of A if det(A)≠0,
and Cij=cofactor of Aji.
Note: adj(A)‧ A = det(A)‧ I and A-1=adj(A)/det(A)
1 2 3
Eg. For A= 1 3 4 , find adj(A) and A-1.


1 4 3
 3

 4
 1
(Sol.) det(A)=-2, adj(A) = 
 1
 1
 1

4
3
4
3
3
4

2 3
4 3
1 3
1 3
1 2

1 4
 7
 7 6  1  2
adj ( A)
1
1
A1 
  1
0  1  


d e tA
2
2
 1  2 1   1

 2
2 3 

3 4 
 7 6  1
1 3 

0  1
= 1
1 4 
 1  2 1 
1 2  
1 3 
3
0
1
1 
2 
1 

2 
1
 
2 
Theorem A-1 does not exist  det(A)=0.

Eg. For A =  0
 0
0
3

2
  , find β such that A-1 does not exist. [1990 清大物理所]
3 
(Sol.) det(A)=9β-β3=0, β=0, ±3.
Theorem If C=adj(A)  Mn×n(F), then det(C)=[det(A)]n-1.
0 
det  A 0 

0
0 
.....
(Proof) C  adj ( A)  det  AA 1  CA  det  AI n  

 
 
0  det  A
 0
det CA  det C  det  A  det  A  det(C )  det  A
n 1
n
.
2  2 0 
Eg. For adj(A)= 0 2  1 , find det(A) and A. [1990 台大機研]


0 0
1 
(Sol.) n=3, det(adj(A))=[det(A)]n-1=4  det(A)=±2
0 
1  1
1 1 1
1
A 1 
 adj  A  A 1   0 1  0.5  A   0 1 1 .
det  A
0 0
0 0 2
0.5 
 1 c01

1
 1 c1
Vander monde matrix: ... ...

... ...
1 c 1
n

n
2
c0
2
c1
...
...
cn
2
n
... c0 
n
... c1 
... ... 

... ... 
n
... c n 
j 1
Theorem det(M)=  (c j  ci ) if M is the Vander monde matrix.
j 1 i  0
(Proof) If some j=i, then cj=ci  det(M)=0   (cj-ci) factor
n
j 1
Let det(M)= k   (c j  ci ) . For n=1, det(M)= (c1-c0)=k(c1-c0), k=1
j 1 i  0
1 1
1 2
Eg. Let M= 
1  3

1  1
1
1 
4
8 
, determine det(M).
9  27 

1 1 
(Sol.) n=3, c0=1, c1=2, c2=-3, c3=-1,
n
j 1
det(M)=  (c j  ci ) =(c1-c0)(c2-c0)(c2-c1)(c3-c0)(c3-c1)(c3-c2)
j 1 i  0
=(2-1)(-3-1)(-3-2)(-1-1)(-1-2)(-1+3)=240