CHAPTER 1 : MATRIX
1.1 Matrix
Definition A matrix (plural: matrices, not matrixes) is a rectangular array of numbers
in rows and columns. Rows run horizontally and columns run vertically
enclosed in brackets :
A =
11 12 13 1n
21 22 23 2n
m1 m2 m3 mn
a a a a
a a a a
a a a a
 
 
 
 
 
 
K
K
M
K
[ ]11 12 13 1na a a aK , [ ]21 22 23 2na a a aK , [ ]m1 m2 m3 mna a a aK
are called the rows of A.
11
21
m1
a
a
a
 
 
 
 
 
 
M
,
12
22
m2
a
a
a
 
 
 
 
 
 
M
,
1n
2n
mn
a
a
a
 
 
 
 
 
 
M
are called the columns of A.
Since A has m rows and n columns , it is said to be an m by n matrix , written m×n.
m× n is called the dimension of the matrix (size of the matrix).

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Example :
0 1
2 3
 
 
 
is 2×2 matrix.
[ ]1 2 3 is 1×3 matrix.
[ ]20 is 1×1 matrix.
1
2
70
 −
 
 
 
 
is 3×1 matrix.
1 1
2
− 
 
 
is not matrix.(since there are not the element in row2 and column2.
Some matrices have special names because of what they look like.
a) Row matrix : only has 1 row.
[ ]1 2 3 , [ ]1 0− , [ ]1 1 2 3− , [ ]11 12 13 1na a a aK
b) Column matrix : only has 1 column.
1
2
 
 
 
,
1
2
3
 
 
 
 
 
,
11
21
m1
a
a
a
 
 
 
 
 
 
M
c) Square matrix : has the same number of rows and columns.
[ ]1 ,
1 2
3 4
 
 
 
,
1 2 3
4 5 6
7 8 9
 
 
 
 
 
,
11 12 13 1m
21 22 23 2m
m1 m2 m3 mm
a a a a
a a a a
a a a a
 
 
 
 
 
 
K
K
M
K
d) Zero matrix : contains all zeros , denote by m n0 × or 0.
[ ]0 , [ ]0 0 ,
0 0
0 0
 
 
 
, [ ]0 0 0 ,
0 0 0
0 0 0
 
 
 
,
0 0 0
0 0 0
0 0 0
 
 
 
 
 

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The notation A = ij m×n
a   is used to denote matrix A , which A is m×n matrix ,
aij is the element of A in row i and column j ; i∈{1 , 2 , 1 , m}and j∈{1 , 2 , 1 , n}
( the first subscript tells the row of the element and the second subscript tells the column
of the element.)
Example : Let A =
1 2 3
4 5 6
− 
 − − 
Solution Since A has 2 rows and 3 columns so the dimension of matrix is 2×3.
The elements of A : a11 = 1 a12 = 2 a13 = −3
a21 = −4 a22 = 5 a23 = −6
Definition Let A = ij m×n
a   and B = ij m×n
b  
A is equal to B if and only if aij = bij for all i∈{1 , 2 , 1 , m}and
j∈{1 , 2 , 1 , n}, denote by A = B.
If A and C have not same dimension then A is not equal to C. And if A and B
have same dimension but there exist their corresponding elements are not equal
then A is not equal to C. Denote A is not equal to C by A ≠ C.
Example : Which of the following matrices , if any , are equal ?
A =
4 1
2 3
 
 
 
, B =
2 3
4 1
 
 
 
, C =
4 2
1 3
 
 
 
, D =
2 3
4 1
 
 
 
Solution Although all four matrices have the same dimension and contain the scalar 1 , 2 ,
3 , 4 , only B and D contain equal corresponding elements.Thus only B = D.
Example : Find the value of x and y if
2x + y 2
=
x + 2y 11
   
   − −   
.
Solution Since
2x + y 2
=
x + 2y 11
   
   − −   
So 2x + y = 2  (1)
−x + 2y = −11  (2)

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The solution of the system of equations is x = 3 and y = −4
Therefore the value of x and y such that
2x + y 2
=
x + 2y 11
   
   − −   
are 3 and −4 , respectively.
Example : Find the value of x and y if
x + 2y 3
=
2x + 4y 6
   
   
   
.
Solution Since
x + 2y 3
=
2x + 4y 6
   
   
   
So x + 2y = 3  (1)
2x + 4y = 6  (2)
The solution of the system of equations is {(x , y) |x + 2y = 3}
= {(3−2y , y) | y∈R}
= {(x ,
3 x
2
−
) | x∈R}
Therefore the value of x and y such that
x + 2y 3
=
2x + 4y 6
   
   
   
are
(x , y) ∈ {(a , b) |a+ 2b = 3}
or (x , y) ∈ {(3−2b , b) | b∈R}
or (x , y) ∈ {(a ,
3 a
2
−
) | a∈R}
Definition Let A = ij m×n
a   and B = ij m×n
b  
A add B is C = ij m×n
c   , cij = aij + bij for all i∈{1 , 2 , 1 , m}
and j∈{1 , 2 , 1 , n} , denote by A + B.

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Example : Let A =
1 2
1 3
 
 − 
and B =
1 0
2 1
− 
 
 
. Find A + B.
Solution A + B =
1 2
1 3
 
 − 
+
1 0
2 1
− 
 
 
=
1 ( 1) 2 0
1 2 3 1
+ − + 
 − + + 
=
0 2
1 4
 
 
 
Definition Let A = ij m×n
a   and c is a constant.
The product of c and A is B = ij m×n
b   , bij = caij for all
i∈{1 , 2 , 1 , m}and j∈{1 , 2 , 1 , n} , denote by cA.
The negative of an m×n matrix A = ij m×n
a   is the m×n matrix −A = ij m×n
a−  
Example : Find the negative of A =
1 3 4 7
2 5 0 8
− 
 − − 
and B =
2 3
6 1
− 
 − 
.
Solution −−−−A =
1 3 4 7
2 5 0 8
− − − 
 − 
and −B =
2 3
6 1
− 
 − 
Example : Let A =
1 1 0
2 1 3
 
 − 
and B =
1 1 2
2 1 3
− 
 − − 
.
Find 2A , A + 2B and 2A + (−3)B .
Solution 2A = 2
1 1 0
2 1 3
 
 − 
=
2 1 2 1 2 0
2 2 2 ( 1) 2 3
⋅ ⋅ ⋅ 
 ⋅ ⋅ − ⋅ 
=
2 2 0
4 2 6
 
 − 

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A + 2B =
1 1 0
2 1 3
 
 − 
+ 2
1 1 2
2 1 3
− 
 − − 
=
1 1 0
2 1 3
 
 − 
+
2 2 4
4 2 6
− 
 − − 
=
1 2 1+2 0+4
2 4 1 2 3+6
− 
 − − − 
=
1 3 4
2 3 9
− 
 − − 
2A + (−3)B = 2
1 1 0
2 1 3
 
 − 
+ (−3)
1 1 2
2 1 3
− 
 − − 
=
2 2 0
4 2 6
 
 − 
+
3 3 6
6 3 9
− − 
 − 
=
2 3 2 3 0 6
4 6 2 3 6 9
+ − − 
 + − + − 
=
5 1 6
10 1 3
− − 
 − 
Definition Let A = ij m×n
a   , B = ij m×n
b   and α and β are constants.
αA −βB is C = ij m×n
c   , cij = αaij −β bij for all i∈{1 , 2 , 1 , m}
and j∈{1 , 2 , 1 , n}.
From definition αA −βB = αA +(−β)B
And A −B = A +(−1)B

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Example : Let A =
1 0
2 1
3 3
 −
 
 
 
 
and B =
1 1
2 1
0 2
 
 
− 
 
 
. Find A − 2B and 3A − 4B.
Solution A − 2B =
1 0
2 1
3 3
 −
 
 
 
 
− 2
1 1
2 1
0 2
 
 
− 
 
 
=
1 (2 1) 0 (2 1)
2 (2 2) 1 (2 ( 1))
3 (2 0) 3 (2 2)
 − − ⋅ − ⋅
 
− ⋅ − ⋅ − 
 − ⋅ − ⋅ 
=
3 2
2 3
3 1
 − −
 
− 
 − 
3A −4B = 3
1 0
2 1
3 3
 −
 
 
 
 
− 4
1 1
2 1
0 2
 
 
− 
 
 
=
(3 ( 1)) (4 1) (3 0) (4 1)
(3 2) (4 2) (3 1) (4 ( 1))
(3 3) (4 0) (3 3) (4 2)
 ⋅ − − ⋅ ⋅ − ⋅
 
⋅ − ⋅ ⋅ − ⋅ − 
 ⋅ − ⋅ ⋅ − ⋅ 
=
3 4 0 4
6 8 3 4
9 0 9 8
 − − −
 
− + 
 − − 
=
7 4
2 7
9 1
 − −
 
− 
 
 

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Property for matrices A , B , C , 0 which be m ×××× n matrix.
1. the dimension of A + B is m × n.
2. A + B = B + A
3. A + (B + C) = (A + B) + C
4. A + 0 = 0 + A = A
5. A + (−A) = (−A) + A = 0
6. c(A + B) = cA + cB ; c is constant.
7. (c + d)A = cA + dA ; c , d are constant.
8. (cd)A = c(dA) ; c , d are constant.
9. 1A = A
10. 0A = 0
0 is the addition identity matrix. (A + 0 = 0 + A = A)
−A is the addition inverse matrix . (A + (−A) = (−A) + A = 0)
Example : Let A =
2 4 6
2 0 2
 
 − 
. Find matrix X , which 2(A + X) = X +
1
2
A.
Solution Let X is 2 × 3 matrix , which 2(A + X) = X +
1
2
A.
So 2A + 2X =
1
2
A + X
2A + 2X − X =
1
2
A + X −X
2A + X =
1
2
A
2A + X − 2A =
1
2
A − 2A
X =
3
2
− A
Therefore X =
3
2
− A =
3
2
−
2 4 6
2 0 2
 
 − 
=
3 6 9
3 0 3
− − − 
 − 

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Definition Let A = ij m×n
a   and B = ij n×r
b   .
AB = ij m×r
c   , which cij = ai1b1j + ai2b2j + 1 + ainbmj
For all i∈{1 , 2 , 1 , m}and j∈{1 , 2 , 1 , r}.
11 12 1n
21 22 2n
m1 m2 mn
a a a
a a a
a a a
 
 
 
 
 
 
K
K
M
K
11 12 1r
21 22 2r
n1 n2 nr
b b b
b b b
b b b
 
 
 
 
 
 
K
K
M
K
=
Which
n
ik kj
k = 1
a b∑ = ai1b1j + ai2b2j + 1 + ainbmj
Example : Let A =
1 2
3 4
 
 
 
, B =
5 6 7
8 9 10
 
 
 
and C =
11
12
13
 
 
 
 
 
Find AB , BC and A(BC) .
Solution AB =
1 2
3 4
 
 
 
5 6 7
8 9 10
 
 
 
=
(1 5) (2 8) (1 6) (2 9) (1 7) (2 10)
(3 5) (4 8) (3 6) (4 9) (3 7) (4 10)
⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ 
 ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ 
=
5 16 6 18 7 20
15 32 18 36 21 40
+ + + 
 + + + 
=
21 24 27
47 54 61
 
 
 
n n n
1k k1 1k k2 1k kr
k = 1 k = 1 k = 1
n n n
2k k1 2k k2 2k kr
k = 1 k = 1 k = 1
n n n
mk k1 mk k2 mk kr
k = 1 k = 1 k = 1
a b a b a b
a b a b a b
a b a b a b
 
 
 
 
 
 
 
 
 
 
∑ ∑ ∑
∑ ∑ ∑
∑ ∑ ∑
K
K
M
K

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BC =
2 3
5 6 7
8 9 10
×
 
 
 
3 1
11
12
13
×
 
 
 
 
 
=
(5 11) (6 12) (7 13)
(8 11) (9 12) (10 13)
⋅ + ⋅ + ⋅ 
 ⋅ + ⋅ + ⋅ 
=
55 72 91
88 108 130
+ + 
 + + 
=
218
326
 
 
 
A(BC) =
1 2
3 4
 
 
 
218
326
 
 
 
=
(1 218) (2 326)
(3 218) (4 326)
⋅ + ⋅ 
 ⋅ + ⋅ 
=
218 652
654 1304
+ 
 + 
=
870
1958
 
 
 
Example : Let A =
1 0
2 1
 
 
 
, B =
1 3
4 5
− 
 
 
. Find AB and BA.
Solution AB =
1 0
2 1
 
 
 
1 3
4 5
− 
 
 
=
(1 1) (0 4) (1 ( 3)) (0 5)
(2 1) (1 4) (2 ( 3)) (1 5)
⋅ + ⋅ ⋅ − + ⋅ 
 ⋅ + ⋅ ⋅ − + ⋅ 
=
1 0 3 0
2 4 6 5
+ − + 
 + − + 
=
1 3
6 1
− 
 − 

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BA =
1 3
4 5
− 
 
 
1 0
2 1
 
 
 
=
1 6 0 3
4 10 0 5
− − 
 + + 
=
5 3
14 5
− − 
 
 
Example : Let A =
1 1
1 1
− 
 − 
, B =
2 1
1 2
 
 
 
. Find AB and BA.
Solution AB =
1 1
1 1
− 
 − 
2 1
1 2
 
 
 
=
2 1 1 2
2 1 1 2
− − 
 − + − + 
=
1 1
1 1
− 
 − 
BA =
2 1
1 2
 
 
 
1 1
1 1
− 
 − 
=
2 1 2 1
1 2 1 2
− − + 
 − − + 
=
1 1
1 1
− 
 − 

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Definition For any positive integer n , let In = jk n n
i ×
   , where j and k are positive
integer which less than n or equal n.
ijk =
1 ; j = k
0 ; j k

 ≠
In is called < identity matrix =,which is n×n matrix and will be denoted by I.
I1 = [ ]1 , I2 =
1 0
0 1
 
 
 
, I 3 =
1 0 0
0 1 0
0 0 1
 
 
 
 
 
If A is n×n matrix then In A = A In = A , for all positive integers.
Example : Let A is 2 × 2 matrix. Show that AI = IA = A , where I =
1 0
0 1
 
 
 
Solution Let A =
a b
c d
 
 
 
is 2 × 2 matrix.
AI =
a b
c d
 
 
 
1 0
0 1
 
 
 
IA =
1 0
0 1
 
 
 
a b
c d
 
 
 
=
a 0 0 b
c 0 0 d
+ + 
 + + 
=
a 0 b 0
0 c 0 d
+ + 
 + + 
=
a b
c d
 
 
 
=
a b
c d
 
 
 
= A = A
Therefore AI = IA = A

C h a p t e r 1 : M a t r i x P a g e | 13
Definition Let A = ij m n
a ×
   , If B = ij n m
b ×
   , where b ij = a ji , for all
i∈{1 , 2 , 1 , n}and j∈{1 , 2 , 1 , m}.
B is called < Transpose of a matrix = of A and denoted B by At
.
If A =
0 1
2 3
 
 
 
then A t
=
0 2
1 3
 
 
 
If D =
1 0 1
3 2 3
− 
 − 
then Dt
=
1 3
0 2
1 3
 − −
 
 
 
 
If A is m × n matrix then At
is n × m matrix , where row i of At
= column i of A.
For all i∈{1 , 2 , 1 , n}.
Example : Let A =
1 1
0 2
− 
 
 
, B =
2 1 1
0 2 3
− 
 − 
. Find (AB)t
and Bt
At
.
Solution AB =
1 1
0 2
− 
 
 
2 1 1
0 2 3
− 
 − 
=
2 0 1 2 1 3
0 0 0 4 0 6
− + − − + 
 + − + 
=
2 3 4
0 4 6
− − 
 − 
(AB)t
=
2 0
3 4
4 6
 −
 
− − 
 
 
Since A =
1 1
0 2
− 
 
 
so At
=
1 0
1 2
− 
 
 

C h a p t e r 1 : M a t r i x P a g e | 14
B =
2 1 1
0 2 3
− 
 − 
so Bt
=
2 0
1 2
1 3
 
 
− 
 − 
Bt
At
=
2 0
1 2
1 3
 
 
− 
 − 
1 0
1 2
− 
 
 
=
2 0 0 0
1 2 0 4
1 3 0 6
 − + +
 
− − − 
 + + 
=
2 0
3 4
4 6
 −
 
− − 
 
 
From example (AB)t
= Bt
At
.
Property for matrix multiplication and transpose of matrix.
If A = ij m n
a ×
   , B = ij n p
b ×
   and C = ij p q
c ×
   then
1. A(BC) = (AB)C
2. r m0 × A = r n0 × , A n p0 × = m p0 ×
3. ImA = A , AIm = A
4. (cA)B = A(cB) = c(AB) ; c is constant.
5. A(B+D) = AB + AD ; D is n ×p matrix.
6. (A + E) = AB + EB ; E is m × n matrix.
7. (A + F)t
= At
+ Ft
; F is m × n matrix.
8. (AB)t
= Bt
At
9. (At
)t
= A
10. (cA)t
= cAt
; c is constant.

C h a p t e r 1 : M a t r i x P a g e | 15
1.2 Inverse Matrix
Definition Let A is n × n matrix if B is n × n matrix which
AB = BA = In
then B is called < inverse of a matrix = of A and denoted B by A−1
.
If A =
a b
c d
 
 
 
and ad ? bc ≠ 0 then A has inverse and
A −1
=
d b1
c aad bc
− 
 −−  
Example : Find A −1
, where A =
2 2
4 1
− 
 
 
.
Solution Since (2×1) ? (4×(−2)) = 10 ≠ 0
Therefore A −1
=
1 21
4 210
 
 − 
Definition Let A = ij 1 1
a ×
   , a is called < determinant = of A .
Definition Let A = ij n n
a ×
   ; n ≥ 2 < minor of aij= is determinant of matrix which
removed row i and column j from A , Denoted minor of aij by Mij(A).
Example : Let A = ij 2 2
a ×
   , find minor of all elements of A.
Solution Since A =
11 12
21 22
a a
a a
 
 
 
So M11(A) = a22 , M12(A) = a21
M21(A) = a12 , M22(A) = a11

C h a p t e r 1 : M a t r i x P a g e | 16
Definition Let A = ij n n
a ×
   ; n ≥ 2 < cofactor of aij= is product of (−1)I + j
and Mij(A),
Denoted cofactor of aij by Cij(A).
Cij(A) = (−1)I + j
Mij(A)
Example : Let A = ij 2 2
a ×
   , find cofactor of all elements of A.
Solution Since A =
11 12
21 22
a a
a a
 
 
 
and M11(A) = a22 , M12(A) = a21
M21(A) = a12 , M22(A) = a11
so C11(A) = (−1)1 + 1
M11(A) = (−1)1 + 1
a22 = a22
C12(A) = (−1)1 + 2
M12(A) = (−1)1 + 2
a21 = −a21
C21(A) = (−1)2 + 1
M21(A) = (−1)2 + 1
a12 = −a12
C22(A) = (−1)2 + 2
M22(A) = (−1)2 + 2
a11 = a11
Example : Let A = ij 2 2
a ×
   , find
1) a11C11(A) + a12C12(A)
2) a21C21(A) + a22C22(A)
3) a11C11(A) + a21C21(A)
4) a12C12(A) + a22C22(A)
Solution 1) a11C11(A) + a12C12(A) = a11a22 − a12a21 = a11a22 − a21a12
2) a21C21(A) + a22C22(A) = −a21a12 − a22a11 = a11a22 − a21a12
3) a11C11(A) + a21C21(A) = a11a22 − a21a12
4) a12C21(A) + a22C22(A) = −a12a21 + a22a11 = a11a22 − a21a12

C h a p t e r 1 : M a t r i x P a g e | 17
Definition Let A = ij n n
a ×
   ; n ≥ 2 < determinant of A = is
a11C11(A) + a12C12(A) + 1 + a1nC1n(A)
Denoted determinant of A by det(A) or
11 12 1n
21 22 2n
n1 n2 nn
a a a
a a a
a a a
K
K
M
K
From definition if A =
11 12
21 22
a a
a a
 
 
 
then det(A) = a11a22 − a21a12 and from example
the value of (1) = (2) = (3) = (4) = det(A).
Example : Let A = ij 3 3
a ×
   , find det(A).
Solution det(A) =
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
= a11C11(A) + a12C12(A) + a13C13(A)
= a11M11(A) − a12M12(A) + a13M13(A)
= a11
22 23
32 33
a a
a a
− a12
21 23
31 33
a a
a a
+ a13
21 22
31 32
a a
a a
= a11(a22a33 − a32a23) − a12(a21a33 − a31a23) + a13(a21a32 − a31a22)
= (a11a22a33 + a12a23a31 + a13a21a32) − (a31a22a13 + a32a23a11 + a33a21a12)

C h a p t e r 1 : M a t r i x P a g e | 18
notice : If A = ij 3 3
a ×
   then det(A) is difference of 2 numbers h , k which
a31a22a13 a32a23a11 a33a21a12
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
11 12
21 22
31 32
a a
a a
a a
a11a22a33 a12a23a31 a13a21a32
Let h = a31a22a13 + a32a23a11 + a33a21a12
k = a11a22a33 + a12a23a31 + a13a21a32
det(A) = h − k
Example : Find det(A) where A =
1 1 2
3 2 4
0 1 3
 
 
− 
 − 
.
Solution Method 1 : By definition of determinant ;
det(A) = a11C11(A) + a12C12(A) + a13C13(A)
= C11(A) + C12(A) + 2C13(A)
= (−1)1 + 1
M11(A) + (−1)1 + 2
M12(A) + 2(−1)1 + 3
M13(A)
= M11(A) − M12(A) + 2M13(A)
=
2 4
1 3−
−
3 4
0 3
−
+ 2
3 2
0 1
−
−
= (6−(−4))−(−9−0) + 2(3−0)
= 10 + 9 + 6
= 25

C h a p t e r 1 : M a t r i x P a g e | 19
Method 2 : Write column 1 , 2 , 3 , 1 , 2
0 −4 −9
1 1 2
3 2 4
0 1 3
 
 
− 
 − 
1 1
3 2
0 1
−
−
6 0 6
det(A) = (6 + 0 + 6) − ( 0 − 4 − 9)
= 12 + 13
= 25
Example : Find det(A) where A =
1 1 2
2 3 4
1 0 3
 
 
− 
 
 
.
Solution Write column 1 , 2 , 3 , 1 , 2
−6 0 6
1 1 2
2 3 4
1 0 3
 
 
− 
 
 
1 1
2 3
1 0
−
−9 4 0
det(A) = (−9 + 4 + 0) − ( −6 + 0 + 6)
= −5 + 0
= −5

C h a p t e r 1 : M a t r i x P a g e | 20
Example : Find det(A) where A =
2 0 4 0
1 1 1 2
2 3 2 4
1 0 1 3
 
 
 
− 
 − 
.
Solution Method 1 : By definition of determinant ;
det(A) = a11C11(A) + a12C12(A) + a13C13(A) + a14C14(A)
= 2C11(A) + 0C12(A) + 4C13(A) + 0 C14(A)
= 2(−1)1 + 1
M11(A) + 4(−1)1 + 3
M13(A)
= 2M11(A) + 4M13(A)
= 2
1 1 2
3 2 4
0 1 3
−
−
+ 4
1 1 2
2 3 4
1 0 3
−
= 2(25) + 4(−5)
= 50 − 20
= 30
Propertys of determinant of matrix.
Let A = ij n n
a ×
   ; n ≥ 2 and c is the constant.
1. If A = B then det(A) = det(B)
2. . det(AB) = det(A)×det(B) = det(B)×det(B) = det(BA)
3. det(An
) = (det(A))n
4. det(I) = 1
5. If det(A) ≠ 0 then det(A)⋅ det(A−1
) = 1 or det(A−1
) =
1
det(A)
6. det(At
) = det(A)
7. If multiply all elements in any row(column) of A by c , then determinant is
c det(A), where c is the constant.
8. det(cA) = cn
det(A) , where c is constsnt.

C h a p t e r 1 : M a t r i x P a g e | 21
9. If A is n × n matrix then det(−A) = det(A) ; n is even number.
and det(−A) = −det(A) ; n is odd number.
10. If all elements in any row or column of A are 0 then det(A) = 0.
11. If any 2 rows(columns) of A are same or are proportion then det(A) = 0.
12. If A is triangular matrix then det(A) is equal to product of all elements in diagonal.
13. If B is matrix which switch to any 2 rows( columns) of A then det(B) = − det(A)
14. If B is matrix which add c time row i of A to row j of A , where i ≠ j and c is
the constant , such that bjk = caik + ajk for all k ∈{1 , 2 , 1 , n} then
det (B) = det(A)
Example : Find det(A) where A =
1 1 5 2
2 0 1 2
1 3 8 0
1 1 2 1
− 
 
 
− − 
 − 
.
Solution By property of determinant.
det (A) =
1 1 5 2
2 0 1 2
1 3 8 0
1 1 2 1
−
− −
−
=
1 1 5 2
0 2 9 2
1 3 8 0
1 1 2 1
−
− −
− −
−
R2 = −2R1 + R2

C h a p t e r 1 : M a t r i x P a g e | 22
=
1 1 5 2
0 2 9 2
0 4 13 2
1 1 2 1
−
− −
−
−
R3 = R1 + R3
=
1 1 5 2
0 2 9 2
0 4 13 2
0 2 3 3
−
− −
−
− −
R4 = −1R1 + R4
=
2 9 2
4 13 2
2 3 3
− −
−
− −
By column 1 operation.
=
1 9 2
2 2 13 2
1 3 3
− −
−
− −
By property 7.
=
1 9 2
2 0 5 2
1 3 3
− −
− −
− −
R2 = 2R1 + R2
=
1 9 2
2 0 5 2
0 6 1
− −
− −
−
R3 = −1R1 + R3
=
5 2
2
6 1
− −
−
By column 1 operation.
= 2(5 + 12)
= 34

C h a p t e r 1 : M a t r i x P a g e | 23
Let A = ij n n
a ×
   ; n ≥ 2 and aij = 0 ; i > j then det(A) = a11a22a33 1 ann.
Let B = ij n n
b ×
   ; n ≥ 2 and bij = 0 ; i < j then det(B) = b11b22b33 1 bnn.
Example : Let A =
0
0 0
2 1 2 3
3 1 4
3 2
10 0 0
− 
 −
 
− 
 
 
, B =
2
1 1
2 1 4
1 3
0 0 0
0
5 1
0
0
 
 
 
− 
 − − 
det(A) = (2)(−3)(3)(1) = −18
det(B) = (2)(1)(4)(−1) = −8
Example : Find det(A) where A =
1 3 4 1
1 5 5 1
2 6 9 10
1 3 4 4
− 
 
 
 
 − − − 
.
Solution det(A) =
1 3 4 1
1 5 5 1
2 6 9 10
1 3 4 4
−
− − −
=
1 3 4 1
1 5 5 1
2 6 9 10
0 0 0 3
−
R4 = R1 + R4

C h a p t e r 1 : M a t r i x P a g e | 24
=
1 3 4 1
1 5 5 1
0 0 1 12
0 0 0 3
−
R3 = −2R1 + R3
=
1 3 4 1
0 2 1 2
0 0 1 12
0 0 0 3
−
R2 = −1R1 + R2
= 1⋅2⋅1⋅3
= 6
Definition Let A is n × n matrix.
A is singular matrix if det(A) = 0
A is non ? singular matrix if det(A) ≠ 0
Definition Let A is n × n matrix , where n ≥ 2 then adjoint matrix of A is matrix
t
ijC (A)   , denoted adjoint matrix of A by adj(A).
Example : Find det(A) , adj(A) , A adj(A) , adj(A)A , where A =
1 0 1
3 1 2
2 5 8
 −
 
− − 
 
 
.
Solution det(A) = 1
1 2
5 8
− −
− 0 + (−1)
3 1
2 5
−
= (−8 −(− 10)) − (15 −(− 2))
= −15

C h a p t e r 1 : M a t r i x P a g e | 25
adj(A) =
t
11 12 13
21 22 23
31 32 33
C (A) C (A) C (A)
C (A) C (A) C (A)
C (A) C (A) C (A)
 
 
 
 
 
=
t
1 2 3 2 3 1
5 8 2 8 2 5
0 1 1 1 1 0
5 8 2 8 2 5
0 1 1 1 1 0
1 2 3 2 3 1
− − − − 
− 
 
− − 
− − 
 
 − −
 −
− − − −  
=
t
2 28 17
5 10 5
1 1 1
 −
 
− − 
 − − − 
=
2 5 1
28 10 1
17 5 1
 − −
 
− − 
 − − 
A adj(A) =
1 0 1
3 1 2
2 5 8
 −
 
− − 
 
 
2 5 1
28 10 1
17 5 1
 − −
 
− − 
 − − 
=
15 0 0
0 15 0
0 0 15
 −
 
− 
 − 
= det(A) I3

C h a p t e r 1 : M a t r i x P a g e | 26
Adj(A) A =
2 5 1
28 10 1
17 5 1
 − −
 
− − 
 − − 
1 0 1
3 1 2
2 5 8
 −
 
− − 
 
 
=
15 0 0
0 15 0
0 0 15
 −
 
− 
 − 
= det(A) I3
Theorem 1 Let A is n × n matrix , where n ≥ 2 then
(1) A adj(A) = adj(A) = det(A) I3
(2) A has inverse multiplication iff A is non ? singular matrix.
If det(A) ≠ 0 then A−1
=
1
adj(A)
det(A)
Example : Let A =
1 2 4
3 8 0
1 2 1
 
 
− 
 − 
, find A−1
.
Solution Since det(A) = (? 8 + 0 ? 24 ) ? (32 + 0 + 6 ) = ? 70 ≠ 0
So A has inverse multiplication.
A−1
=
1
adj(A)
det(A)

C h a p t e r 1 : M a t r i x P a g e | 27
=
8 0 3 0 3 8
2 1 1 1 1 2
2 4 1 4 1 21
2 1 1 1 1 270
2 4 1 4 1 2
8 0 3 0 3 8
− −
− − −

− − −
− −
−
− −
t



 
 
 
 
 
 
=
t
8 3 14
1
10 5 0
70
32 12 14
 − − −
 
− − 
 − − 
=
8 10 32
1
3 5 12
70
14 0 14
 − −
 
− − − − 
 − 
Example : Let A =
1 2 3 4
0 1 2 3
0 0 1 2
0 0 0 1
 
 
 
 
 
 
, find A−1
.
Solution Since det(A) = 1 ≠ 0 So A has inverse multiplication.
Find Cij(A) for all I , j.
C11(A) =
1 2 3
0 1 2
0 0 1
= 1 C12(A) =
0 2 3
0 1 2
0 0 1
− = 0

C h a p t e r 1 : M a t r i x P a g e | 28
C13(A) =
0 1 3
0 0 2
0 0 1
= 0 C14(A) =
0 1 2
0 0 1
0 0 0
− = 0
C21(A) =
2 3 4
0 1 2
0 0 1
− = −2 C22(A) =
1 3 4
0 1 2
0 0 1
= 1
C23(A) =
1 2 4
0 0 2
0 0 1
− = 0 C24(A) =
1 2 3
0 0 1
0 0 0
= 0
C31(A) =
2 3 4
1 2 3
0 0 1
= 1 C32(A) =
1 3 4
0 2 3
0 0 1
− = −2
C33(A) =
1 2 4
0 1 3
0 0 1
= 1 C34(A) =
1 2 3
0 1 2
0 0 0
− = 0
C41(A) =
2 3 4
1 2 3
0 1 2
− = 0 C42(A) =
1 3 4
0 2 3
0 1 2
= 1
C43(A) =
1 2 4
0 1 3
0 0 2
− = −2 C44(A) =
1 2 3
0 1 2
0 0 1
= 1

C h a p t e r 1 : M a t r i x P a g e | 29
So A−1
=
t
1 0 0 0
2 1 0 01
1 2 1 01
0 1 2 1
 
 −
 
− 
 − 
=
1 2 1 0
0 1 2 1
0 0 1 2
0 0 0 1
− 
 −
 
− 
 
 
Theorem 2 Let A and B are n × n matrices ,
det(AB) = det(A) det(B)
Remark ;
det (In) = 1
If det(A) ≠ 0 then det(A-1
) =
( )
1
det A
Example : Let A and B are 2 × 2 matrix.
2A + B =
1 4
5 2
 
 
 
and A − B =
2 2
1 4
 
 
 
. Find det(2A− 1
B).
Solution Since 2A + B =
1 4
5 2
 
 
 
and A − B =
2 2
1 4
 
 
 
So (2A + B) + (A − B) =
1 4
5 2
 
 
 
+
2 2
1 4
 
 
 
3A =
3 6
6 6
 
 
 
A =
1 2
2 2
 
 
 

C h a p t e r 1 : M a t r i x P a g e | 30
And B = A −
2 2
1 4
 
 
 
=
1 2
2 2
 
 
 
−
2 2
1 4
 
 
 
=
1 0
1 2
− 
 − 
So det(A) = −2 and det(B) = 2
Therefore det(2A− 1
B) = 2 2
det(A− 1
)det(B)
= 4( )1
2−
2
= −4
1.3 Solve linear equation system by matrix.
Define linear equation system which has m equations and n variables;
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2
M
am1x1 + am2x2 + . . . + amnxn = bm
We can write linear equation system by matrix ;
{ {
11 12 1n 1 1
21 22 1n 2 2
m1 m2 mn n n
X BA
a a a x b
a a a x b
a a a x b
    
    
    =
    
    
    
K
K
M M M
K
1442443
If m = n and det(A) ≠ 0 then X = A− 1
B
So if m = n and det(A) ≠ 0 then we can find the solution by X = A− 1
B

C h a p t e r 1 : M a t r i x P a g e | 31
Example 1 : Solve the linear equation system.
2x y 2z+ + = 3
x y z+ − =
1
2
−
3x 2y 2z+ − = 2−
Solution We can write matrix from the linear equation system ;
2 1 2
1 1 1
3 2 2
 
 
− 
 − 
x
y
z
 
 
 
 
 
=
3
1
2
2
 
 
− 
 
−  
Let A =
2 1 2
1 1 1
3 2 2
 
 
− 
 − 
det (A) = ( ) ( )4 3 4 6 4 2− − + − − − = 3−
1
A−
=
t1 1 1 1 1 1
2 2 3 2 3 2
1 2 2 2 2 11
2 2 3 2 3 23
1 2 2 2 2 1
1 1 1 1 1 1
− − 
− − −
 
 
− − − − −−  
 
 −
− −  
=
t
0 1 1
1
6 10 1
3
3 4 1
 − −
 
− − −
 − 
=
0 6 3
1
1 10 1
3
1 1 1
 −
 
− − − −
 − − 

C h a p t e r 1 : M a t r i x P a g e | 32
So
x
y
z
 
 
 
 
 
=
0 6 3
1
1 10 1
3
1 1 1
 −
 
− − − −
 − − 
3
1
2
2
 
 
− 
 
−  
=
1
2
3
2
 
 −
 
 
 
 
 
Therefore the solution of linear equation system is ( )3
1 , 2 ,
2
−
Solve linear equation system by Cramer/s rule.
Theorem 3 ; Cramer/s rule
Define the linear equation system , which has n equations and n variables such that
AX = B
Let X =
1
2
n
x
x
x
 
 
 
 
 
 
M
and B =
1
2
n
b
b
b
 
 
 
 
 
 
M
If det(A) ≠ 0 then
1
1
det(A )
x =
det(A)
, 2
2
det(A )
x =
det(A)
, … , n
n
det(A )
x =
det(A)
where iA is the matrix which replace column i of A with column of B for all
i ∈ { }1 , 2 , 3 , , nK

C h a p t e r 1 : M a t r i x P a g e | 33
Example 2 : Solve the linear equation system by CramerBs rule.
2x y 2z+ + = 3
x y z+ − =
1
2
−
3x 2y 2z+ − = 2−
Solution We can write matrix from the linear equation system;
A =
2 1 2
1 1 1
3 2 2
 
 
− 
 − 
, X =
x
y
z
 
 
 
 
 
, B =
3
1
2
2
 
 
− 
 
−  
Since det (A) = ( ) ( )4 3 4 6 4 2− − + − − − = 3− ≠ 0
By CramerBs rule ;
x =
3 1 2
1
1 1
2
2 2 2
det(A)
− −
− −
=
( ) ( )6 2 2 4 6 1
3
− + − − − − +
−
=
3
3−
= 1−
y =
2 3 2
1
1 1
2
3 2 2
det(A)
− −
− −
=
( ) ( )2 9 4 3 4 6
3
− − − − + −
−
=
6
3
−
−
= 2

C h a p t e r 1 : M a t r i x P a g e | 34
z =
2 1 3
1
1 1
2
3 2 2
det(A)
−
−
=
( ) ( )
3
4 6 9 2 2
2
3
− − + − − −
−
=
9
2
3
−
−
=
3
2
Therefore the solution of linear equation system is ( )3
1 , 2 ,
2
−
Solve linear equation system by row operation.
Define linear equation system which has m equations and n variables;
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2
M
am1x1 + am2x2 + . . . + amnxn = bm
We can write linear equation system by augmented matrix ;
11 12 1n 1
21 22 1n 2
m1 m2 mn
a a a b
a a a b
a a a bn
 
 
 
 
 
 
K
K
M M
K

C h a p t e r 1 : M a t r i x P a g e | 35
Definition
Let A is m × n matrix . The following operations are called < row operation=
1. switch to row i and j of A , denoted by Rij.
2. multiplied row i by constant c ≠ 0 , denoted by cRi
3. change row i of A by multiply row j (j ≠ i) by c then add to row i . denoted by
Ri + cRj
Example 3 : Solve the linear equation system by row operation.
x y 2z+ + = 9
2y 7z− = 17−
3x 6y 5z+ − = 0
Solution We can write augment matrix from the linear equation system;
1 1 2 9
0 2 7 17
3 6 5 0
 
 
− − 
 − 
∼
1 1 2 9
0 2 7 17
0 3 11 27
 
 
− − 
 − − 
∼
1 1 2 9
7 17
0 1
2 2
0 3 11 27
 
 − −
 
 
− −  
∼
1 1 2 9
7 17
0 1
2 2
1 3
0 0
2 2
 
 
 − −
 
 
− − 
  
− 3R1 + R3
1
2
R2
− 3R2 + R3

C h a p t e r 1 : M a t r i x P a g e | 36
∼
1 1 2 9
7 17
0 1
2 2
0 0 1 3
 
 − −
 
 
  
then x y 2z+ + = 9
7
y z
2
− =
17
2
−
z = 3
So z = 3 , y =
17 7
z
2 2
− + = 2 , x = 9 − y −2z = 1
Therefore the solution of linear equation system is ( )1 , 2 , 3−
Example 4 : Solve the linear equation system by row operation.
x y z 2t+ + + = 1
2x 3y 2z 5t− + + = 3
3x 2y 2z t+ + + = 0
x y 3z t+ − − = 0
Solution We can write augment matrix from the linear equation system;
1 1 1 2 1
2 3 2 5 3
3 2 2 1 0
1 1 3 1 0
− 
 −
 
 
 − − 
∼
1 1 1 2 1
0 1 0 1 1
0 5 1 5 3
0 2 4 3 1
− 
 −
 
− − − 
 − − − 
∼
1 1 1 2 1
0 1 0 1 1
0 5 1 5 3
0 2 4 3 1
− 
 − −
 
− − − 
 − − − 
−R3
−2R1 + R2
−3R1 + R3
−R1 + R4
−R2

C h a p t e r 1 : M a t r i x P a g e | 37
∼
1 1 1 2 1
0 1 0 1 1
0 0 1 0 2
0 0 4 1 1
− 
 − −
 
− 
 − − 
∼
1 1 1 2 1
0 1 0 1 1
0 0 1 0 2
0 0 4 1 1
− 
 − −
 
− 
 − − 
∼
1 1 1 2 1
0 1 0 1 1
0 0 1 0 2
0 0 0 1 7
− 
 − −
 
− 
 − − 
∼
1 1 1 2 1
0 1 0 1 1
0 0 1 0 2
0 0 0 1 7
− 
 − −
 
− 
 
 
∼
1 1 1 0 13
0 1 0 0 6
0 0 1 0 2
0 0 0 1 7
− − 
 
 
− 
 
 
∼
1 1 0 0 11
0 1 0 0 6
0 0 1 0 2
0 0 0 1 7
− − 
 
 
− 
 
 
−5R2 + R3
−2R2 + R4
−R3
4R3 + R4
−R4
−2R4 + R1
−R3 + R2
−R3 + R1

C h a p t e r 1 : M a t r i x P a g e | 38
∼
1 0 0 0 5
0 1 0 0 6
0 0 1 0 2
0 0 0 1 7
− 
 
 
− 
 
 
Therefore the solution of linear equation system is ( )5 , 6 , 2 , 7− −
Example 5 : Solve the linear equation system by row operation.
x y 3z 7t+ + + = 0
x y 5z 11t+ + + = 2−
2x 2y 4z 11t+ + + = 2
Solution We can write augment matrix from the linear equation system;
1 1 3 7 0
1 1 5 11 2
2 2 4 11 2
 
 
− 
 
 
∼
1 1 3 7 0
0 0 2 4 2
0 0 2 3 2
 
 
− 
 − − 
∼
1 1 3 7 0
0 0 1 2 1
0 0 2 3 2
 
 
− 
 − − 
∼
1 1 3 7 0
0 0 1 2 1
0 0 0 1 0
 
 
− 
 
 
∼
1 1 3 0 0
0 0 1 0 1
0 0 0 1 0
 
 
− 
 
 
R2 + R1
−R1 + R2
−2R1 + R3
1
2
R2
2R2 + R3
−2R3 + R2
−7R3 + R1

C h a p t e r 1 : M a t r i x P a g e | 39
∼
1 1 0 0 3
0 0 1 0 1
0 0 0 1 0
 
 
− 
 
 
So x y = 3+ , z = 1− , t = 0
Therefore the solution set of linear equation system is
( ){ }x , y , z , t x + y = 3 , z = 1 , t = 0− = ( ){ }x , y , 1 , 0 x + y = 3−
= ( ){ }c , 3 c , 1 , 0 c R− − ∈
Example 6 : Solve the linear equation system by row operation.
x 2y z 5t+ + + = 7
z t+ = 1
x 2y 4t+ + = 7
Solution We can write augment matrix from the linear equation system;
1 2 1 5 7
0 0 1 1 1
1 2 0 4 7
 
 
 
 
 
∼
1 2 1 5 7
0 0 1 1 1
0 0 1 1 0
 
 
 
 − − 
∼
1 2 1 5 7
0 0 1 1 1
0 0 0 0 1
 
 
 
 
 
So x 2y z 5t+ + + = 7
0x 0y z t+ + + = 1
0x 0y 0z 0t+ + + = 1
Since the corresponding equation is 0x 0y 0z 0t+ + + = 1 which has no solution,
because no matter what values we substitute for the unknowns x , y , z , t
Therefore the system does not have solution.
3R2 + R1
−R1 + R3
R2 + R3

C h a p t e r 1 : M a t r i x P a g e | 40
Example 7 : Let A =
1 1 0
1 0 1
6 2 3
 −
 
− 
 − − 
. Find A − 1
by row operation.
Solution Since det (A) = ( ) ( )0 6 0 0 2 3+ + − + + = 1 ≠ 0
So we can find inverse of A ;
[ ]3A IM =
1 1 0 1 0 0
1 0 1 0 1 0
6 2 3 0 0 1
 −
 
− 
 − − 
∼
1 1 0 1 0 0
0 1 1 1 1 0
6 4 3 6 0 1
 −
 
− − 
 − − 
∼
1 1 0 1 0 0
0 1 1 1 1 0
0 0 1 6 0 1
 −
 
− − 
 − 
∼
1 1 0 1 0 0
0 1 0 3 3 1
0 0 1 2 4 1
 −
 
− − 
 − − 
∼
1 0 0 2 3 1
0 1 0 3 3 1
0 0 1 2 4 1
 − −
 
− − 
 − − 
Therefore A −1
=
2 3 0
3 3 1
2 4 1
 − −
 
− − 
 − − 
−R1 + R2
−6R1 + R3
−4R2 + R3
R2 + R3
R1 + R2