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![Ch1_9
Submatrix ( 子矩陣 )
Amatrix
215
032
471
−
=A
Row ( 列 ) and Column ( 行 )
[ ] [ ]
3column2column1column2row1row
1
4
5
3
7
2
157432
−
−
−−
157
432
−
−
=A
Aofssubmatrice
25
41
1
3
7
15
32
71
−
=
=
= RQP](https://image.slidesharecdn.com/linearalgebra-131010034008-phpapp02/75/Linear-algebra-9-2048.jpg)
![Ch1_10
Identity Matrices ( 單位矩陣 )
diagonal ( 對角線 ) 上都是 1 ，其餘都是 0 ， I 的下標表示 size
=
=
100
010
001
10
01
32 II
Location
7,4
157
432
2113 =−=
−
−
= aaA
aij 表示在 row i, column j 的元素值
也寫成 location (1,3) = −4
Size and Type
[ ]
matrixcolumnamatrixrowamatrixsquarea
matrix13matrix41matrix3332:Size
2
3
8
5834
853
109
752
542
301
××××
−
−
−
−](https://image.slidesharecdn.com/linearalgebra-131010034008-phpapp02/75/Linear-algebra-10-2048.jpg)
![Ch1_17
Summary
−−−
−
−
=
7012
32863
441284
]:[ BA
A BUse row operations to [A: B] :
.
1100
3010
2001
−
≈≈
−−−
−
−
7012
32863
441284
]:[]:[ XIBA n≈≈i.e.,
Def. [In : X] is called the reduced echelon form ( 簡化梯
式 )
of [A : B].Note. 1. If A is the matrix of coefficients of a system of n equations
in n variables that has a unique solution,
then A is row equivalent to In (A ≈ In).
2. If A ≈ In, then the system has unique solution.
72
32863
441284
21
321
321
−=−−
=−+
=−+
xx
xxx
xxx](https://image.slidesharecdn.com/linearalgebra-131010034008-phpapp02/75/Linear-algebra-17-2048.jpg)
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Linear algebra
- 1. Chapter 1 Systems ofLinear Equations 感謝 大葉大學 資訊工程系 鈴玲老師 提供黃 Linear Algebra
- 2. Ch1_2 Definition • An equationsuch as x+3y=9 is called a linear equation. ( 線性方程式 ) • The graph of this equation is a straight line in the x-y plane. • A pair of values of x and y that satisfy the equation is called a solution. 1.1 Matrices and Systems of Linear Equations system of linear equations ( 線性聯立方程式 )
- 3. Ch1_3 Definition A linear equationin n variables ( 變數 ) x1, x2, x3, …, xn has the form a1 x1 + a2 x2 + a3 x3 + … + an xn = b where the coefficients ( 係數 ) a1, a2, a3, …, an and b are real numbers ( 實數 ). 常見數系的英文名稱： natural number ( 自然數 ), integer ( 整數 ), rational number ( 有理數 ), real number ( 實數 ), complex number ( 複數 ) positive ( 正 ), negative ( 負 )
- 4. Ch1_4 Figure 1.2 No solution( 無解 ) –2x + y = 3 –4x + 2y = 2 Lines are parallel. No point of intersection. No solutions. Solutions for system of linear equations Figure 1.1 Unique solution ( 唯一解 ) x + 3y = 9 –2x + y = –4 Lines intersect at (3, 2) Unique solution: x = 3, y = 2. Figure 1.3 Many solution ( 無限多 解 ) 4x – 2y = 6 6x – 3y = 9 Both equations have the same graph. Any point on the graph is a solution. Many solutions.
- 5. Ch1_5 A linear equationin three variables corresponds to a plane in three-dimensional ( 三維 ) space. Unique solution ※ Systems of three linear equations in three variables:
- 6.
- 7. Ch1_7 How to solvea system of linear equations? Gauss-Jordan elimination. ( 高斯 - 喬登消去法 ) 1.2 節會介紹
- 8. Ch1_8 Definition • A matrix( 矩陣 ) is a rectangular array of numbers. • The numbers in the array are called the elements ( 元素 ) of the matrix. Matrices −= − = − − = 1298 520 653 C 38 50 17 B 157 432 A 注意矩陣左右兩邊是中括號不是直線，直線表示的是行列式。
- 9. Ch1_9 Submatrix ( 子矩陣) Amatrix 215 032 471 − =A Row ( 列 ) and Column ( 行 ) [ ] [ ] 3column2column1column2row1row 1 4 5 3 7 2 157432 − − −− 157 432 − − =A Aofssubmatrice 25 41 1 3 7 15 32 71 − = = = RQP
- 10. Ch1_10 Identity Matrices (單位矩陣 ) diagonal ( 對角線 ) 上都是 1 ，其餘都是 0 ， I 的下標表示 size = = 100 010 001 10 01 32 II Location 7,4 157 432 2113 =−= − − = aaA aij 表示在 row i, column j 的元素值 也寫成 location (1,3) = −4 Size and Type [ ] matrixcolumnamatrixrowamatrixsquarea matrix13matrix41matrix3332:Size 2 3 8 5834 853 109 752 542 301 ×××× − − − −
- 11. Ch1_11 matrix of coefficientand augmented matrix 62 332 2 321 321 321 −=−− =++ =++ xxx xxx xxx Relations between system of linear equations and matrices 係數矩陣 擴大矩陣 隨堂作業： 5(f) tcoefficienofmatrix 211 132 111 −− matrixaugmented 6211 3132 2111 −−−
- 12. Ch1_12 給定聯立方程式後， 不會改變解的一些轉換 Elementary Transformation 1. Interchangetwo equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a multiple of one equation to another equation. 將左邊的轉換對應到矩陣上 Elementary Row Operation ( 基本列運算 ) 1. Interchange two rows of a matrix. ( 兩列交換 ) 2. Multiply the elements of a row by a nonzero constant. ( 某列的元素同乘一非零常數 ) 3. Add a multiple of the elements of one row to the corresponding elements of another row. ( 將一個列的倍數加進另一列裡 ) Elementary Row Operations of Matrices
- 13. Ch1_13 Example 1 Solving thefollowing system of linear equation. 62 332 2 321 321 321 −=−− =++ =++ xxx xxx xxx −−− 6211 3132 2111 62 332 2 321 321 321 −=−− =++ =++ xxx xxx xxx Solution Equation Method Initial system: Analogous Matrix Method Augmented matrix: 1 2 32 321 −=− =++ xx xxx Eq2+(–2)Eq1 Eq3+(–1)Eq1 −−− −− 8320 1110 2111 R2+(–2)R1 R3+(–1)R1 ≈ 符號表示 row equivalent 832 32 −=−− xx
- 14. Ch1_14 105 1 32 3 32 31 −=− −=− =+ x xx xx −− −− 01500 1110 3201 2 1 32 3 32 31 = −=− =+ x xx xx −− 2100 1110 3201 2 1 1 3 2 1 = = −= x x x − 2100 1010 1001 Eq1+(–1)Eq2 Eq3+(2)Eq2 (–1/5)Eq3 Eq1+(–2)Eq3 Eq2+Eq3 The solutionis .2,1,1 321 ==−= xxx The solution is .2,1,1 321 ==−= xxx 832 1 2 32 32 321 −=−− −=− =++ xx xx xxx −−− −− 8320 1110 2111 R1+(–1)R2 R3+(2)R2 ≈ (–1/5)R3 ≈ R1+(–2)R3 R2+R3 ≈ 隨堂作業： 7(d)
- 15. Ch1_15 Example 2 Solving thefollowing system of linear equation. 833 1852 1242 321 321 321 −=−+− =+− =+− xxx xxx xxx Solution ( 請先自行練習 ) −−− − − 8331 18512 12421 −− − 4110 6330 12421 R2 3 1 ≈ −− − 4110 2110 12421 −− 6200 2110 8201 −− 3100 2110 8201 R1R3 2)R1(R2 + −+ ≈ R2)1(R3 (2)R2R1 −+ + ≈ R3 2 1 ≈ 3100 1010 2001 R3R2 2)R3(R1 + −+ ≈ . 3 1 2 solution 3 2 1 = = = x x x
- 16. Ch1_16 Example 3 Solve thesystem 72 32863 441284 21 321 321 −=−− =−+ =−+ xx xxx xxx −−− − − 7012 32863 441284 −−− − − 7012 32863 11321 − − − 15630 1100 11321 − − − 1100 5210 11321 − − 1100 5210 1101 . 1100 3010 2001 − .1,3,2issolutionThe 321 −=== xxx R2 3 1 ≈ R12R3 3)R1(R2 + −+ ≈ R3R2 ↔ ≈ − − − 1100 15630 11321 R1 4 1 ≈ 2)R2(R1 −+ ≈ 2R3R2 1)R3(R1 + −+ ≈ Solution ( 請先自行練習 ) 隨堂作業： 10(d)(f)
- 17. Ch1_17 Summary −−− − − = 7012 32863 441284 ]:[ BA A BUserow operations to [A: B] : . 1100 3010 2001 − ≈≈ −−− − − 7012 32863 441284 ]:[]:[ XIBA n≈≈i.e., Def. [In : X] is called the reduced echelon form ( 簡化梯 式 ) of [A : B].Note. 1. If A is the matrix of coefficients of a system of n equations in n variables that has a unique solution, then A is row equivalent to In (A ≈ In). 2. If A ≈ In, then the system has unique solution. 72 32863 441284 21 321 321 −=−− =−+ =−+ xx xxx xxx
- 18. Ch1_18 Example 4 ManySystems Solving the following three systems of linear equation, all of which have the same matrix of coefficients. 3321 2321 1321 42 for42 3 bxxx bxxx bxxx =−+− =+− =+− in turn 4 3 3 , 2 1 0 , 11 11 8 3 2 1 − − = b b b Solution −−−− − − 4211421 3111412 308311 . 2 1 2 , 1 3 0 , 2 1 1 3 2 1 3 2 1 3 2 1 = = −= = = = = −= = x x x x x x x x x −−− −−− − 123110 315210 308311 −−− 212100 315210 013101 − − 212100 131010 201001 R2+(–2)R1 R3+R1 ≈ 1)R2(R3 R2R1 −+ + ≈ R32R2 R3)1(R1 + −+ ≈ 隨堂作業： 13(b) The solutions to the three systems are
- 19. Ch1_19 Homework Exercise 1.1 ： 1,2, 4, 5, 6, 7, 10, 13
- 20. Ch1_20 1-2 Gauss-Jordan Elimination Definition Amatrix is in reduced echelon form ( 簡化梯式 ) if 1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix. 2. The first nonzero element of each other row is 1. This element is called a leading 1. 3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row. 4. All other elements in a column that contains a leading 1 are zero.
- 21. Ch1_21 Examples for reducedechelon form 10000 04300 03021 9100 3010 7001 3100 0000 4021 000 210 801 () ()() () 利用一連串的 elementary row operations ，可讓 任何矩陣變成 reduced echelon form The reduced echelon form of a matrix is unique. 隨堂作業： 2(b)(d) (h)
- 22. Ch1_22 Gauss-Jordan Elimination System oflinear equations ⇒ augmented matrix ⇒ reduced echelon from ⇒ solution
- 23. Ch1_23 −− − − −+ ≈ 41200 22200 43111 4)R1(R3 − − −+ + ≈ 61000 11100 52011 R2)2(R3 R2R1 Example 1 Use themethod of Gauss-Jordan elimination to find reduced echelon form of the following matrix. − − − 1211244 129333 22200 Solution − − − ↔ ≈ 1211244 22200 129333 R2R1 pivot ( 樞軸，未來的 leading 1) − − − ≈ 1211244 22200 43111 R1 3 1 pivot − − + −+ ≈ 61000 50100 170011 R3R2 2)R3(R1 The matrix is the reduced echelon form of the given matrix. ≈ −− − − 41200 11100 43111 R2 2 1
- 24. Ch1_24 Example 2 Solve, ifpossible, the system of equations 753 742 9333 321 321 321 =−− =+− =+− xxx xxx xxx Solution ( 請先自行練習 ) ( ) −− − −≈ −− − − 7153 7412 3111 7153 7412 9333 R1 3 1 −−− − −+ −+ ≈ 2420 1210 3111 3)R1(R3 2)R1(R2 + + ≈ 0000 1210 4301 R2R3 R2R1 12 43 12 43 32 31 32 31 +−= +−= ⇒ =+ =+ ⇒ xx xx xx xx The general solution to the system is .parameter)a(callednumberrealiswhere, 12 43 3 2 1 rrx rx rx = +−= +−= 隨堂作業： 5(c)
- 25. Ch1_25 Example 3 Solve thesystem of equations 446942 14232 58101242 4321 4321 4321 =−+− −=+−+− =−+− xxxx xxxx xxxx Solution( 自行練習 ) ( ) −− −−− −−≈ −− −−− −− 446942 142321 295621 446942 142321 58101242 R1 2 1 −− − −−≈ −+ + 144300 153300 295621 2)R1(R3 R1R2 ( ) −− − −−≈ 144300 51100 295621 R2 3 1 − −−≈ + −+ 11000 51100 11021 3R2R3 6)R2(R1 −−≈ + −+ 11000 60100 20021 R3R2 1)R3(R1 .somefor, 1 6 22 1 6 22 4 3 2 1 4 3 21 r x x rx rx x x xx = = = −= ⇒ = = −=− ⇒ 變數個數 > 方程式個數 many sol.
- 26. Ch1_26 Example 4 Solve thesystem of equations 432 636242 232 54321 54321 54321 =+−+−− =++−+ =++−+ xxxxx xxxxx xxxxx Solution( 自行練習 ) −≈ −−− − − + −+ 642000 210000 213121 431121 636242 213121 R1R3 2)R1(R2 − ≈ ↔ 210000 642000 213121 R3R2 ( ) −≈ 210000 321000 213121 2R 2 1 −−− ≈ −+ 210000 321000 750121 3)R2(R1 − −≈ −+ + 210000 101000 300121 2)R3(R2 5R3R1 .andsomefor, 2 ,1,, 32 2 1 32 5 432 1 5 4 321 sr x xsxrx srx x x xxx = −=== ++−= ⇒ = −= ++−= ⇒
- 27. Ch1_27 Example 5 This exampleillustrates a system that has no solution. Let us try to solve the system 122 32 4522 32 32 321 321 321 =+ −=−+ =+− =+− xx xxx xxx xxx Solution( 自行練習 ) ≈ − −− − ↔ 1220 2100 6330 3211 R3R2 ( ) ≈ − −− − 1220 2100 2110 3211 R2 3 1 ≈ − −− −+ + 5400 2100 2110 1101 2)R2(R4 R2R1 ≈ − − −+ + −+ 13000 2100 4010 3001 4)R3R(R4 R3R2 1)R3(R1 ( ) ≈ − − 1000 2100 4010 3001 R4 13 1 The system has no solution. ≈ −− − − −− − − −+ −+ 1220 6330 2100 3211 1220 3121 4522 3211 1)R1(R3 2)R1(R2 0x1+0x2+0x3=1 隨堂作業： 5(d)
- 28. Ch1_28 Homogeneous System oflinear Equations Definition A system of linear equations is said to be homogeneous ( 齊 次 ) if all the constant terms ( 等號右邊的常數項 ) are zeros. Example: =+−− =−+ 0632 052 321 321 xxx xxx Observe that is a solution.0,0,0 321 === xxx Theorem 1.1 A system of homogeneous linear equations in n variables always has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called the trivial solution.
- 29. Ch1_29 Homogeneous System oflinear Equations Theorem 1.2 A system of homogeneous linear equations that has more variables than equations has many solutions. Note. 除 trivial solution 外，可能還有其他解。 隨堂作業： 8(e) − ≈≈ −− − 0410 0301 0632 0521 =+−− =−+ 0632 052 321 321 xxx xxx The system has other nontrivial solutions. rxrxrx ==−=∴ 321 ,4,3 Example:
- 30. Ch1_30 Homework Exercise 1.2: 2, 5,8, 14 (Section 1.3 跳過 )