24 jul 2007 kkkq 3013 pengiraan berangka week 3 – systems of linear equations 24 july 2007 8.00 am...
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24 Jul 2007
KKKQ 3013KKKQ 3013PENGIRAAN BERANGKAPENGIRAAN BERANGKA
Week 3 ndash Systems of Linear Equations24 July 2007
800 am ndash 900 am
24 Jul 2007 Week 2 Page 2
Topics
1048713 Introduction
1048713 Elimination Methods
1048713 Decomposition Methods
1048713 Matrix Inverse and Determinant
1048713 Errors Residuals and Condition Number
1048713 Iteration Methods
1048713 Incomplete and Redundant Systems
24 Jul 2007 Week 2 Page 3
Tutorial Example 1 (WITHOUT row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 1200 1 2 3 -0200 1 4 12 0800 -1 -3 -7 030
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 -1 -4 010
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
R4(1) = R4(0) - (11)R1(0)
R3(2) = R3(1) - (11)R2(1)
R4(2) = R4(1) - (-11)R2(1)
R4(3) = R4(2) - (-12)R3(2)
Forward elimination
[A]x = b
[A]
24 Jul 2007 Week 2 Page 4
(i) Determinant
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
det (A) = (1)(1)(2)(05) = 1
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)0(1)(1)(2)(05) = 1
For triangular matrix
det [A] = product of diagonal terms aii
Triangular obtained from [A] WITHOUT any row interchange
Tutorial Example 1 (WITHOUT row interchange)
Alternatively
24 Jul 2007 Week 2 Page 5
(ii) LU decomposition of [A]
a11 a12 a13 a14 1 2 4 8
0 a22(1) a23
(1) a24(1)
0 1 2 3
0 0 a33(2) a34
(2)0 0 2 9
0 0 0 a44(3)
0 0 0 05
1 0 0 0 1 0 0 0a21a11 1 0 0 (01) 1 0 0
a31a11 a32(1)a22
(1)1 0 (01) (11) 1 0
a41a11 a42(1)a22
(1) a43(2)a33
(2)1 (11) (-11) (-12) 1
1 0 0 00 1 0 00 0 1 01 -1 -05 1
=
=
=[U] =
=[L]
Now check if [L][U] = [A]
Tutorial Example 1 (WITHOUT row interchange)
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 2
Topics
1048713 Introduction
1048713 Elimination Methods
1048713 Decomposition Methods
1048713 Matrix Inverse and Determinant
1048713 Errors Residuals and Condition Number
1048713 Iteration Methods
1048713 Incomplete and Redundant Systems
24 Jul 2007 Week 2 Page 3
Tutorial Example 1 (WITHOUT row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 1200 1 2 3 -0200 1 4 12 0800 -1 -3 -7 030
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 -1 -4 010
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
R4(1) = R4(0) - (11)R1(0)
R3(2) = R3(1) - (11)R2(1)
R4(2) = R4(1) - (-11)R2(1)
R4(3) = R4(2) - (-12)R3(2)
Forward elimination
[A]x = b
[A]
24 Jul 2007 Week 2 Page 4
(i) Determinant
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
det (A) = (1)(1)(2)(05) = 1
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)0(1)(1)(2)(05) = 1
For triangular matrix
det [A] = product of diagonal terms aii
Triangular obtained from [A] WITHOUT any row interchange
Tutorial Example 1 (WITHOUT row interchange)
Alternatively
24 Jul 2007 Week 2 Page 5
(ii) LU decomposition of [A]
a11 a12 a13 a14 1 2 4 8
0 a22(1) a23
(1) a24(1)
0 1 2 3
0 0 a33(2) a34
(2)0 0 2 9
0 0 0 a44(3)
0 0 0 05
1 0 0 0 1 0 0 0a21a11 1 0 0 (01) 1 0 0
a31a11 a32(1)a22
(1)1 0 (01) (11) 1 0
a41a11 a42(1)a22
(1) a43(2)a33
(2)1 (11) (-11) (-12) 1
1 0 0 00 1 0 00 0 1 01 -1 -05 1
=
=
=[U] =
=[L]
Now check if [L][U] = [A]
Tutorial Example 1 (WITHOUT row interchange)
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 3
Tutorial Example 1 (WITHOUT row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 1200 1 2 3 -0200 1 4 12 0800 -1 -3 -7 030
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 -1 -4 010
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
R4(1) = R4(0) - (11)R1(0)
R3(2) = R3(1) - (11)R2(1)
R4(2) = R4(1) - (-11)R2(1)
R4(3) = R4(2) - (-12)R3(2)
Forward elimination
[A]x = b
[A]
24 Jul 2007 Week 2 Page 4
(i) Determinant
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
det (A) = (1)(1)(2)(05) = 1
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)0(1)(1)(2)(05) = 1
For triangular matrix
det [A] = product of diagonal terms aii
Triangular obtained from [A] WITHOUT any row interchange
Tutorial Example 1 (WITHOUT row interchange)
Alternatively
24 Jul 2007 Week 2 Page 5
(ii) LU decomposition of [A]
a11 a12 a13 a14 1 2 4 8
0 a22(1) a23
(1) a24(1)
0 1 2 3
0 0 a33(2) a34
(2)0 0 2 9
0 0 0 a44(3)
0 0 0 05
1 0 0 0 1 0 0 0a21a11 1 0 0 (01) 1 0 0
a31a11 a32(1)a22
(1)1 0 (01) (11) 1 0
a41a11 a42(1)a22
(1) a43(2)a33
(2)1 (11) (-11) (-12) 1
1 0 0 00 1 0 00 0 1 01 -1 -05 1
=
=
=[U] =
=[L]
Now check if [L][U] = [A]
Tutorial Example 1 (WITHOUT row interchange)
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 4
(i) Determinant
1 2 4 8 1200 1 2 3 -0200 0 2 9 1000 0 0 05 060
det (A) = (1)(1)(2)(05) = 1
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)0(1)(1)(2)(05) = 1
For triangular matrix
det [A] = product of diagonal terms aii
Triangular obtained from [A] WITHOUT any row interchange
Tutorial Example 1 (WITHOUT row interchange)
Alternatively
24 Jul 2007 Week 2 Page 5
(ii) LU decomposition of [A]
a11 a12 a13 a14 1 2 4 8
0 a22(1) a23
(1) a24(1)
0 1 2 3
0 0 a33(2) a34
(2)0 0 2 9
0 0 0 a44(3)
0 0 0 05
1 0 0 0 1 0 0 0a21a11 1 0 0 (01) 1 0 0
a31a11 a32(1)a22
(1)1 0 (01) (11) 1 0
a41a11 a42(1)a22
(1) a43(2)a33
(2)1 (11) (-11) (-12) 1
1 0 0 00 1 0 00 0 1 01 -1 -05 1
=
=
=[U] =
=[L]
Now check if [L][U] = [A]
Tutorial Example 1 (WITHOUT row interchange)
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 5
(ii) LU decomposition of [A]
a11 a12 a13 a14 1 2 4 8
0 a22(1) a23
(1) a24(1)
0 1 2 3
0 0 a33(2) a34
(2)0 0 2 9
0 0 0 a44(3)
0 0 0 05
1 0 0 0 1 0 0 0a21a11 1 0 0 (01) 1 0 0
a31a11 a32(1)a22
(1)1 0 (01) (11) 1 0
a41a11 a42(1)a22
(1) a43(2)a33
(2)1 (11) (-11) (-12) 1
1 0 0 00 1 0 00 0 1 01 -1 -05 1
=
=
=[U] =
=[L]
Now check if [L][U] = [A]
Tutorial Example 1 (WITHOUT row interchange)
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 6
Tutorial Example 1 (WITH row interchange)
[A]x = b1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
ORIGINAL
[A] [b]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
[A]x = b
[A] [b]
INTERCHANGE ROWS 2-4
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 7
Tutorial Example 1 (WITH row interchange)
Forward elimination
[A]
1 2 4 8 120
1 1 1 1 150
0 1 4 12 080
0 1 2 3 -020
1 2 4 8 120
0 -1 -3 -7 030
0 1 4 12 080
0 1 2 3 -020
R2(1) = R2(0) - (11)R1(0)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 -1 -4 010
R3(2) = R3(1) - (1-1)R2(1)
R4(2) = R4(1) - (1-1)R2(1)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
R4(3) = R4(2) - (-11)R3(2)
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 8
(i) Determinant
det (A) = (-1)pa11a22(1)a33
(2)a44(3)
= (-1)1(1)(-1)(1)(1) = 1
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Equals det (A) previously NOT det (A)
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 9
Alternatively
det (A) = (1)(-1)(1)(1) = -1 For triangular matrix
det [ ] = product of diagonal terms aii
Triangular obtained from [A]
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 -1 -3 -7 0300 0 1 5 1100 0 0 1 120
Hence det (A) = (1)p(-1) = 1
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 10
Tutorial Example 1 (WITH row interchange)
1 2 4 8 1200 1 2 3 -0200 1 4 12 0801 1 1 1 150
1 2 4 8 120
1 1 1 1 150
0 1 2 3 -020
0 1 4 12 080
How many row interchange p =
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 11
Tutorial Example 2
Refer exercise in Chapter 1 Question 3
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 12
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2
24 Jul 2007 Week 2 Page 13
Tutorial Example 2