Download a −1

Chapter one
Mathematics II
Matrices and determinates
*Definition:
An m×n matrix is a rectangular array of numbers (m rows and n columns)
enclosed in brackets. The numbers are called the elements of the matrix.
Examples:
1. A 2×3 matrix has 2 rows and 3 columns:
[
2. Here’s a 3 × 3 square matrix:
1 2
5 6
7
[3
4
2
5
3
3
]
7
−1
−2]
0
3. Column vectors are matrices with only one column:
1
[ 5]
8
4. Row vectors are matrices which only have one row:
[3
A general real matrix A
−1
7]
Rmxn with m × n elements is of the form:
a11
a21
[ ⋮
am1
a12
a22
⋮
am2
… a1n
… a2n
⋱
⋮ ]
… amn
Each element has two properties:
1. location: The first index represents the row (i) and The second index represents
the column (j), e.g. (a21 is the element in row 2, column 1 of the matrix A)
2. Value: which represents the amount of the element.
1
Mathematics II
Chapter one
Matrices and determinates
*Special matrices
1. The unit matrix I is a square matrix whose only non-zero elements are on the
diagonal and are equal to one, e.g.
𝐼=[
1
0
0
],
1
1
[0
0
0
1
0
0
0] ,
1
1
[0
0
0
0
1
0
0
0 0
0 0]
⋱ 0
0 1
2. Zero matrix 0 : All elements of the zero matrix are equal to zero, e.g.
[
0
0
0
],
0
0
[0
0
0
0
0
0
0] ,
0
0
[0
⋮
0
0
0
⋮
0
⋯
⋯
⋱
⋯
0
0]
⋮
0
3. A diagonal matrix only has non-zero elements on the main diagonal. These
non-zero elements can have any value, e.g.
𝑑
𝐷 = [ 11
0
0
],
𝑑22
𝑑11
[ 0
0
0
𝑑22
0
0
0 ],
𝑑33
𝑑11 0
0 𝑑22
[
⋮ ⋮
0 0
⋯ 0
⋯ 0
]
⋱
⋮
⋯ 𝑑𝑛𝑛
4. Upper triangular matrix: all the elements under the main diagonal are zero
and the others can have any value, e.g.
𝑈=[
2
0
−1
],
5
1
[0
0
3
−5
0
7
−1] ,
2
2
[0
0
0
5
9
0
0
1
7
1
0
−3
6]
4
−2
5. Lower triangular matrix: all the elements above the main diagonal are zero
and the others can have any value, e.g.
L=[
2
1
0
],
−3
1
[2
7
0
−4
3
0
0] ,
5
3
[2
0
4
0
−5
9
−6
0
0
2
1
0
0]
0
7
2
Mathematics II
Chapter one
Matrices and determinates
*Matrix algebra
1. Matrix equality.
Two matrices are equal if they have the same size and if their corresponding
elements are identical , i.e.
A=B
if and only if:
aij = bij , for i = 1,….., m ; j = 1,……, n
Examples:
1
1. If A = [
5
−2
7
3
2
],B=[
−9
3
4
5
4
] then A ≠ B
2
( same size but the corresponding elements are not equal)
1 0
1
2. If A = [3 −5] , B = [
0
9 6
( different size)
1
3. If A = [
−3
2
1
],B=[
7
−3
3
−5
9
] then A ≠ B
6
−2
] then A ≠ B
7
( same size but a12 ≠ b12 )
1
Ex: A = [
−4
0
a
] , B=[
2
c
b
] , given that A = B, find a, b, c, and d?
d
Sol.
A=B
a=1
b=0
c = -4
d=2
3
Mathematics II
Chapter one
Matrices and determinates
2. Matrix addition.
Two matrices can only be added if they have the same size. The result is
another matrix of the same size. We add matrices by adding their corresponding
elements, i.e.
cij = aij +
Ex: if A = [
1
5
2
6
C=A+B
bij for i = 1,….., m ; j = 1,……, n
3
10
],B=[
7
5
1
16
23
] find:
3
1. C = A + B ?
2. D = B – A ?
Sol.
1 + 10
C=A+B=[
5+5
2+1
6 + 16
3 + 23
11
]=[
7+3
10
10 − 1
D=B–A=[
5−5
1−2
16 − 6
23 − 3
9
]=[
3−7
0
3
22
−1
10
26
]
10
20
]
−4
3. Multiplication of a matrix by a scalar.
A matrix is multiplied by a scalar ( a number) by multiplying each element
of the matrix by that scalar. The result is a matrix of the same size.
A=αB
bij = α aij for i = 1,….., m ; j = 1,……, n
3
Ex: If A = [4
3
5
2
−4
−2
4 ] find 3A?
3
Sol.
3×3
3A = [3 × 4
3×3
3×5
3×2
3 × −4
3 × −2
9
3 × 4 ] = [12
3×3
9
15
6
−12
−6
12 ]
9
4
Mathematics II
Chapter one
Matrices and determinates
4. Matrix – Matrix multiplication.
The product of an m × n matrix A and an n × p matrix B produces an m × p
matrix C, i.e.
This implies that the number of columns of the first matrix must be equal
to the number of rows in the second matrix.
Here are some examples:
1
A= [
3
5
0
−2
],
7
2
B = [4
4
2×3
1 2 3 2
C = [ 5 −7 6 −2 ] ,
−9 0 −1 7
3×4
−1
5
−3
3
9]
6
3×3
2
1
6]
D = [5
9 −2
3
7
4×2
We can form :
AB : the result is a 2 × 3 matrix
AC : the result is a 2 × 4 matrix
BC : the result is a 3 × 4 matrix
CD : the result is a 3 × 2 matrix
But, It is not possible to form:
BA or AD
The product matrix of two matrices is obtained by taking dot products of
the rows of the left matrix with the columns of the right matrix.
5
Mathematics II
Chapter one
1
Ex: Find AB if A = [
3
2
4
] ,B = [
4
2
3
1
Matrices and determinates
2
]?
1
Sol.
3
2
2) ( ) ( 1 2) ( )
1
1 ]
AB = [
(3
(3 4) (3) (3 4) (2)
1
1
(1 × 4 + 2 × 2) (1 × 3 + 2 × 1) (1 × 2 + 2 × 1)
= [
]
(3 × 4 + 4 × 2) (3 × 3 + 4 × 1) (3 × 2 + 4 × 1)
(1
=[
4
2) ( )
2
4
4) ( )
2
8
20
5
13
(1
4
]
10
For practical computation Falk’s scheme could be used to evaluate AB in
tabular form as follows:
4
3
2
2
1
1
1
2
(1×4 + 2×2)
(1×3 + 2×1)
(1×2 + 2×1)
3
4
(3×4 + 4×2)
(3×3 + 4×1)
(3×2 + 4×1)
8
AB = [
20
5
13
4
]
10
Properties of matrices multiplication.
1. Matrix multiplication is not commutative (AB ≠ BA)
2 2
1 2
] ,B = [
]
4 1
3 1
1×2+2×3 1×2+2×1
8
AB = [
] = [
4×2+1×3 4×2+1×1
11
Proof: A = [
BA = [
2×1+2×4
3×1+1×4
2×2+2×1
10
]= [
3×2+1×1
7
4
]
9
6
]
7
6
Mathematics II
Chapter one
Matrices and determinates
2. AB = 0 does not imply A = 0, B = 0 or BA = 0.
1 1
] ,B =
2 2
1 × (−1) + 1 × 1
AB = [
2 × (−1) + 2 × 1
Proof: A = [
BA = [
−1
1
1
1
] .[
−1 2
1
−1
1
]
−1
= [
[
−1 1
]
1 −1
1 × 1 + 1 × (−1)
0
]= [
2 × 1 + 2 × (−1)
0
(−1) × 1 + 1 × 2
1
]= [
1 × 1 + (−1) × 2
2
0
]
0
(−1) × 1 + 1 × 2
]
1 × 1 + (−1) × 2
3. AB = AC does not necessarily imply B = C
1
2 1
] , B= [
] , C=
2
2 2
1×2+1×2
1 2 1
] .[
]= [
2×2+2×2
2 2 2
Proof: A = [
1
AB = [
2
AC = [
1
2
1
2
1 3
] .[
2 1
1×3+1×1
0
]= [
2×3+2×1
3
3 0
]
1 3
1×1+1×2
4
] = [
2×1+2×2
8
[
1×0+1×3
4
] = [
2×0+2×3
8
3
]
6
3
]
6
But, other properties are similar to numbers
4. A(B + C) = AB + AC distributive law
3
1 1
2 1
] , B= [
] , C= [
2 2
2 2
1
5 1
B+C= [
]
3 5
8
6
1 1 5 1
A(B+C) = [
] .[
]= [
]
2 2 3 5
16 12
4 3
4 3
AB = [
] , AC = [
]
8 6
8 6
4 3
4 3
8
6
AB + AC = [
]+ [
]= [
]
8 6
8 6
16 12
Proof: A = [
0
]
3
∴ A(B + C) = AB + AC O.K.
5. A(BC) = (AB)C associative law
7
Mathematics II
Chapter one
Matrices and determinates
*Transpose of a matrix.
The transpose of a matrix is obtained by interchanging its rows and columns
aTij = aji
for i = 1,….., m ; j = 1,……, n
The transpose is denoted by a superscript T and the general matrix becomes:
a11 a21
a12 aa22
AT = [ ⋮
⋮
a1n a2n
1
Ex: Find AT if A = [
5
2
6
⋯ am1
⋯ am2
⋱
⋮ ]
… anm
3
]?
7
Sol.
1
A = [2
3
T
5
6]
7
Hint: If AT = A then A is a symmetric matrix, e.g.
3
A=[ 2
−1
2
7
0
−1
3
T
0] , A = [2
8
−1
2
7
0
−1
0]
8
Properties of matrices transpose:
1. (AT)T = A for any matrix A.
2. (A + B)T = AT + BT. for matrices A and B with compatible dimensions.
3. (A.B)T = BT. AT . for matrices A and B with compatible dimensions.
4. If AAT = ATA = I then A is orthogonal matrix.
8
Mathematics II
Chapter one
Matrices and determinates
*Determinant of a matrix
1. The determinant of a 2 × 2 matrix
a11
A = [a
21
˗
+
a11
is written det. A or |A| = |a
21
Ex: Find det. A if A = [
1
4
a12
a22 ]
a12
a22 | = 𝑎11 . 𝑎22 − 𝑎12 . 𝑎21
2
]?
−7
Sol.
|A| = |
1
4
2
| = 1 × (−7) − 2 × 4 = −15
−7
2.The determinant of a 3 × 3 matrix is written as:
a11
|A| = |a21
a31
a12
a22
a32
a13
a
a23 | = a11 . | 22
a32
a33
a23
a21
|
−
a
.
|
12 a
a33
31
a23
a21
|
+
a
.
|
13 a
a33
31
a22
a32 |
|A| = 𝑎11 (𝑎22 𝑎33 − 𝑎23 𝑎32 ) − 𝑎12 (𝑎21 𝑎33 − 𝑎23 𝑎31 ) + 𝑎13 (𝑎21 𝑎32 − 𝑎22 𝑎31 ).
3
Ex: Find det. A if A = [2
1
3
Sol. |A| = |2
1
= 3|
1
0
−5
1
0
−5
1
0
3
−1] ?
4
3
−1|
4
−1
2
| − (−5) |
4
1
2
−1
| + 3|
4
1
1
|
0
= 3(4) + 5(9) + 3(-1)
= 54
9
Mathematics II
Chapter one
Matrices and determinates
There is another method and to compute the determinant of a 3×3 matrix.
It is named Sarrus' rule or Sarrus' scheme.
Consider a 3×3 matrix :
a11
[a21
a31
a12
a22
a32
a13
a23 ]
a33
then its determinant can be computed by the following scheme:
= a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a11 a23 a32 − a12 a21 a33
+
3
|A| = |2
1
+
+
−5
1
0
3 3
−1| 2
4 1
-
-
−5
1
0
= 3×1×4 + (-5)×(-1)×(1) + 3×2×0 - 3×1×1 - 3×(-1)×0 – (-5)×2×4
= 12 + 5 + 0 – 3 – 0 – (-40) = 54
Properties of determinate.
1. if det. A = 0 the matrix A is called a singular matrix.
2. det. A = det. AT
3. det. (AB) = det. A det. B
10
Mathematics II
Chapter one
Matrices and determinates
*Cofactors.
The cofactor cij is defined as the coefficient of aij in the determinant A. It is
given by the formula:
𝑐𝑖𝑗 = (−1)𝑖+𝑗 |𝑚𝑖𝑗 |
where mij is the minor matrix of order (n −1)×(n −1) formed by deleting
the column and row containing aij, e.g.
c11
a11
= (−1)1+1 |m11 | = (1) |a21
a31
c23
a11
= (−1)2+3 |m23 | = (−1) |a21
a31
a12
a22
a32
a13
a
a23 | = | 22
a32
a33
a12
a22
a32
a13
a
a23 | = | 11
a31
a33
a23
a33 | = a22 a33 − a23 a32
a12
a32 | = a11 a32 − a12 a31
2 1 4
Ex1: Find the cofactors matrix of A = [5 2 3] ?
8 7 3
Sol.
5 3
𝑐12 = (−1)1+2 |
| = (−1)[5 × 3 − 3 × 8] = (-1).(-9) = 9
8 3
𝑐22 = (−1)2+2 |
2
8
4
| = (−1)[2 × 3 − 4 × 8] = (1). (−26) = −26 , etc...
3
Ex2: Find the cofactors matrix of A = [
−2
7
3
]?
11
Sol.
𝑐11 = (−1)1+1 |11| = (1). (11) = 11
𝑐21 = (−1)2+1 |3| = (-1).(3) = -3 , etc...
11
Chapter one
Mathematics II
Matrices and determinates
*The matrix inverse
The inverse a−1 of a scalar (number) a is defined by a a−1 = 1. For square
matrices we use a similar definition: the inverse A−1 of a n × n matrix A fulfils
the relation:
AA−1 = I (unit matrix)
The matrix inverse can be computed as follows:
1. Find the determinant det. A.
2. Find the cofactors of all elements in A and form a new matrix C of cofactors,
where each element is replaced by its cofactor.
3. The inverse of A is now given as:
−1
A
CT
=
|A|
Note: the inverse A−1 exists if (and only if) det. A ≠ 0.
1 −1
2
Ex: Find the inverse of A = [−3
1
2 ]?
3 −2 −1
Sol.
+
+ +
1
|A| = |−3
3
−1
1
−2
-
-
2 1 −1
2 | −3 1
−1 3 −2
= 1×1×(-1) + (-1) ×2×3 + 2×(-3) ×(-2) - 2×1×3 - 1×2×(-2) - (-1) ×(-3) ×(-1)
=6
Since the determinant is nonzero an inverse exists.
1
2
|=3
−2 −1
−3 2
= (−1)1+2 |
| = 3 , etc.
3 −1
c11 = (−1)1+1 |
c12
12
Mathematics II
Chapter one
Matrices and determinates
Cofactors matrix will be:
3
C = [−5
−4
3
C = [3
3
T
3
−7
−8
3
−1 ]
−2
−5
−7
−1
−4
−8]
−2
Finally, the inverse of A is:
−1
A
−1
Check AA
1 3
= [3
6
3
1
2
−5
−7
−1
−4
−8] = [ 12
1
−2
2
−5
6
−7
6
−1
6
−2
3
−4
3
−1
3
]
= I (unit matrix)
−1
AA
1
= [−3
3
−1
1
−2
1
2
2
2 ] [ 12
−1 1
2
−5
6
−7
6
−1
6
−2
3
−4
3
−1
3
1
] = [0
0
0
1
0
0
0] o. k
1
13
*Solution of Ax = B.
Mathematics II
Chapter one
Matrices and determinates
Given:
a11 x1 + a12 x2 + … … … + a1n xn = b1
a21 x1 + a22 x2 + … … … + a2n xn = b2
⋮
⋮
⋱
⋮
⋮
an1 x1 + an2 x2 + … … … + ann xn = bn
In matrices:
a11
a21
[ ⋮
an1
a12
a22
⋮
an2
⋯ a1n x1
b1
⋯ a2n x2
b2
]
{
}
=
[
]
⋱
⋮
⋮
⋮
⋯ ann xn
bn
There are many methods:
1. By inversion of coefficients matrix.
Ax = B → x = A−1 B
Ex: solve the following system of equations:
x + 3y = 5
-x + 2y = 2
Sol.
1 3 x
5
[
] {y} = [ ]
−1 2
2
|𝐴| = 1 × 2 − 3 × (−1) = 5
𝑐11 = (−1)1+1 |2| = 2 , 𝑐12 = 1 , 𝑐21 = −3 , 𝑐22 = 1
C= [
−1
A
2
−3
1
2
] , CT = [
1
1
CT
=
=
|A|
2
𝑥
{𝑦} = [51
5
1
5
−3
5
1
5
2
[
1
−3
]
1
2
−3
] = [51
1
5
−3
5
1
5
]
4
5
] . [ ] = [57 ]
2
5
14
Ex: solve the following system of equations:
Mathematics II
Chapter one
Matrices and determinates
8x1 + 2x2 + 3x3 = 30
x1 − 9x2 + 2x3 = 1
2x1 + 3x2 + 6x3 = 31
Sol.
8
[1
2
c11
c13
3 x1
30
2] . {x 2 } = [ 1 ]
6 x3
31
−9 2
1
= (−1)1+1 . |
| = −60 , c12 = (−1)1+2 |
3 6
2
1 −9
2 3
= (−1)4 |
| = 21 , c21 = (−1)3 |
|=
2 3
3 6
2
−9
3
−60 −2
C = [ −3
42
31 −13
+ + +
8
|A| = |1
2
−1
A
2
−9
3
x1
2
{x2 } = [ 421
x3
−21
421
−3
42
−20
−3 , etc.
31
−13]
−74
3 8 2
2| 1 −9 = (−432) + (8) + (9)— 54 − (48) − (12) = −421
6 2 3
1 −60
=
[ −2
−421
21
60
421
21
−60
T
−20] , C = [ −2
−74
21
- - -
2
| = −2
6
3
421
−42
421
20
421
−3
42
−20
−31
421
13
421
74
421
31
−13]
−74
30
2
] . [ 1 ] = [ 1]
31
4
15
2. Cramer's rule.
Chapter one
Mathematics II
Matrices and determinates
This method based on replacing the column(s) of variables by the column
vector of results, then the value of each variable can be found by dividing the
determinate of the new matrix to the determinate of the matrix of coefficients.
x1 =
x2 =
xn =
b1
b
| 2
⋮
bn
a12 … a1n
a22 … a2n
|
⋮
⋱
⋮
an2 … ann
|A|
a11
a
| 21
⋮
an1
b1 … a1n
b2 … a2n
|
⋮ ⋱
⋮
bn … ann
|A|
a11
a21
|
⋮
an1
a12
a22
⋮
an2
|A|
… b1
… b2
|
⋱ ⋮
… bn
Ex: solve the following system of equations:
-x + 4y = 1
3x + 12y = 1
16
Ex: solve the following system of equations:
Mathematics II
Chapter one
Matrices and determinates
4.7x1 + 1.3x2 – 1.6x3 = 1.3
x1 – 4.1x2 + 1.1x3 = 4.6
2.1x1 + 1.4x2 + 6.2x3 = 5.2
Sol.
4.7
[1
2.1
1.3
−4.1
1.4
1.3
|4.6
𝑥1 = 5.2
4.7
|1
2.1
−1.6 x1
1.3
1.1 ] . {x2 } = [4.6]
x3
6.2
5.2
1.3
−4.1
1.4
1.3
−4.1
1.4
−1.6
1.1 |
6.2 = −109.104 = 0.738
−1.6
−147.785
1.1 |
6.2
4.7 1.3 −1.6
| 1 4.6 1.1 |
109.239
𝑥2 = 2.1 5.2 6.2 =
= −0.739
−147.785
−147.785
4.7 1.3 1.3
| 1 −4.1 4.6|
−111.661
𝑥3 = 2.1 1.4 5.2 =
= 0.756
−147.785
−147.785
Check using eq.1
4.7(0.738) + 1.3(-0.739) – 1.6(0.756) = 1.2983 ≈ 1.3 o.k.
17
3. Gauss elimination.
Mathematics II
Chapter one
Matrices and determinates
The solution using this method has three stages. In the first stage the
equations are written in matrix form. In the second stage the matrix equations are
replaced by a system of equations having the same solution but which are in
triangular form. In the final stage the new system is solved by back-substitution.
Ex: Use Gaussian elimination to solve the system of linear equations:
x1 + 5x2 = 7
−2x1 − 7x2 = −5
Sol.
x1
7
1
5
] . {x } = [ ]
−5
2
−2 −7
Row(2) + 2*Row(1)
7
1 5 x1
[
] . {x } = [ ]
9
2
0 3
[
3x2 = 9 → x2 = 3
x1 + 5x2 = 7 → x1 = 7 − 5 × 3 = −8
Ex: Use Gaussian elimination to solve the system of linear equations:
2x1 + x2 − x3 + 2x4 = 5
4x1 + 5x2 − 3x3 + 6x4 = 9
−2x1 + 5x2 − 2x3 + 6x4 = 4
4x1 + 11x2 − 4x3 + 8x4 = 2
Sol.
2 1
[ 4 5
−2 5
4 11
−1
−3
−2
−4
x1
2
5
x
6] . { 2 } = [ 9]
x3
6
4
x4
8
2
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Mathematics II
Chapter one
x1
2
5
2] . {x2 } = [−1]
x3
8
9
x4
4
−8
2
R1
R2 − 2R1 [0
R3 + R1 0
R4 − 2R1 0
1
3
6
9
−1
−1
−3
−2
2
R1
R2
[0
R3 − 2R2 0
R4 − 3R2 0
1
3
0
0
x1
−1 2
5
−1 2 ] . {x2 } = [−1]
x3
−1
4
11
x4
1 −2
−5
R1
2
R2 [0
R3
0
R4 + R3 0
1
3
0
0
−1
−1
−1
0
Matrices and determinates
x1
2
5
2] . {x2 } = [−1]
x3
4
11
x4
2
6
2x4 = 6
x4 = 3
−x3 + 4x4 = 11
−x3 + 4(3) = 11
x3 = 1
3x2 − x3 + 2x4 = −1
3x2 − 1 + 2(3) = −1
x2 = −2
2x1 + x2 − x3 + 2x4 = 5
2𝑥1 + (−2) − 1 + 2(3) = 5
𝑥1 = 1
Check using Eq. 2
4x1 + 5x2 − 3x3 + 6x4 = 9
4(1) + 5(-2) – 3(1) + 6(3) = 9
o.k.
19