Lecture 9: 3.4 The Geometry of Linear Systems

Transcription

1 Lectre 9: 3.4 The Geometry of Linear Systems Wei-Ta Ch 200/0/5

2 Dot Prodct Form of a Linear System Recall that a linear eqation has the form a x +a 2 x 2 + +a n x n = b (a,a 2,, a n not all zero) The corresponding homogeneos eqation is a x +a 2 x 2 + +a n x n = 0 (a,a 2,, a n not all zero) These eqations can be rewritten in ector form by letting a = (a,a 2,,a n ) and x=(x,x 2,,x n ) Two eqations can be written as 2

3 Dot Prodct Form of a Linear System It reeals that each soltion ector x of a homogeneos eqation is orthogonal to the coefficient ector a. Consider the homogeneos system a a a 2 x m x x a a 2 a 22 x x m2 2 2 x a a... a If we denote the sccessie row ectors of the coefficient matrix by r, r 2,, r m, then we can write this system as n 2n x mn n x x n n 3

4 Theorem If A is an m n matrix, then the soltion set of the homogeneos linear system Ax=0 consists of all ectors in R n that are orthogonal to eery row ector of A. Example: the general soltion of is x =-3r-4s-2t, x 2 =r, x 3 =-2s, x 4 =s, x 5 =t, x 6 =0 Vector form: x = (-3r-4s-2t, r, -2s, s, t, 0) 4

5 Theorem According to Theorem 3.4.3, the ector x mst be orthogonal to each of the row ectors r = (,3,-2,0,2,0) r 2 = (2, 6, -5, -2, 4, -3) r 3 = (0,0,5,0,0,5) r 4 = (2,6,0,8,4,8) Verify that r.x = (-3r-4s-2t)+3(r)+(-2)(-2s)+0(s)+2(t)+0(0) = 0 5

6 The Relationship Between Ax=0 and Ax=b Compare the soltions of the corresponding linear systems Homogeneos system: x =-3r-4s-2t, x 2 =r, x 3 =-2s, x 4 =s, x 5 =t, x 6 =0 Nonhomogeneos system: x =-3r-4s-2t, x 2 =r, x 3 =-2s, x 4 =s, x 5 =t, x 6 =/3 6

7 The Relationship Between Ax=0 and Ax=b We can rewrite them in ector form: Homogeneos system: x = (-3r-4s-2t, r, -2s, s, t, 0) Nonhomogeneos system: x = (-3r-4s-2t, r, -2s, s, t, /3) By splitting the ectors on the right apart and collecting terms with like parameters, Homogeneos system: (x,x 2,x 3,x 4,x 5 ) = r(-3,,0,0,0) + s(-4,0,- 2,,0,0) + t(-2,0,0,0,,0) Nonhomogeneos system: (x,x 2,x 3,x 4,x 5 ) = r(-3,,0,0,0) + s(-4,0,- 2,,0,0) + t(-2,0,0,0,,0) + (0,0,0,0,0,/3) Each soltion of the nonhomogeneos system can be obtained by adding (0,0,0,0,0,/3) to the corresponding soltion of the homogeneos system. 7

8 Theorem The general soltion of a consistent linear system Ax=b can be obtained by adding any specific soltion of Ax=b to the general soltion of Ax=0. Proof: Let x 0 be any specific soltion of Ax=b, Let W denote the soltion set of Ax=0, and let x 0 +W denote the set of all ectors that reslt by adding x 0 to each ector in W. Show that if x is a ector in x 0 +W, then x is a soltion of Ax=b, and conersely, that eery soltion of Ax=b is in the set x 0 +W. 8

9 Theorem Assme that x is a ector in x 0 +W. This implies that x is expressible in the form x=x 0 +w, where Ax 0 =b and Aw=0. Ths, Ax = A(x 0 +w) = Ax 0 + Aw = b + 0 = b which shows that x is a soltion of Ax=b. Conersely, let x be any soltion of Ax=b. To show that x is in the set x 0 +W we mst show that x is expressible in the form: x = x 0 +w, where w is in W (Aw = 0). We can do this by taking w = x-x 0. It is in W since Aw = A(x-x 0 ) = Ax Ax 0 = b b = 0. 9

10 Geometric Interpretation of Theorem We interpret ector addition as translation, then the theorem states that if x 0 is any specific soltion of Ax=b, then the entire soltion set of Ax=b can be obtained by translating the soltion set of Ax=0 by the ector x 0. Ax = b x 0 Ax = 0 0 0

11 Lectre 9: 3.5 Cross Prodct Wei-Ta Ch 200/0/5

12 Definition If = (, 2, 3 ) and =(, 2, 3 ) are ectors in 3-space, then the cross prodct is the ector defined by = ( , 3 3, 2 2 ) Or, in determinant notation Remark: For the matrix to find the first component of, delete the first colmn and take the determinant, 2

13 Example Find, where =(,2,-2) and =(3,0,) Soltion 3

14 Theorems Theorem 3.5. (Relationships Inoling Cross Prodct and Dot Prodct) If, and w are ectors in 3-space, then ( ) = 0 ( is orthogonal to ) ( ) = 0 ( is orthogonal to ) 2 = 2 2 ( ) 2 (Lagrange s identity) ( w) = ( w) ( ) w prodct) (relationship between cross & dot ( ) w = ( w) ( w) prodct) (relationship between cross & dot 4

15 Proof of Theorem 3.5.(a) Let =(, 2, 3 ) and =(, 2, 3 ) Example: =(, 2, -2) and =(3, 0, ) 5

16 Proof of Theorem 3.5.(c) ( 23 32, 3 3, 2 2) 6

17 Theorems Theorem (Properties of Cross Prodct) If, and w are any ectors in 3-space and k is any scalar, then = - ( ) ( + w) = + w ( + ) w = w + w k( ) = (k) = (k) 0 = 0 = 0 = 0 Proof of (a) Interchanging and interchanges the rows of the three determinants and hence changes the sign of each component in the cross prodct. Ths = - ( ). 7

18 Standard Unit Vectors The ectors x i = (,0,0), j = (0,,0), k = (0,0,) hae length and lie along the coordinate axes. They are called the standard nit ectors in 3-space. Eery ector = (, 2, 3 ) in 3-space is expressible in terms of i, j, k since we can write = (, 2, 3 ) = (,0,0) + 2 (0,,0) + 3 (0,0,) = i + 2 j + 3 k z k=(0,0,) i=(,0,0) j=(0,,0) y For example, (2, -3, 4) = 2i 3j +4k Note that i i = 0, j j = 0, k k = 0 i j = k, j k = i, k i = j j i = -k, k j = -i, i k = -j 8

19 9 Cross Prodct A cross prodct can be represented symbolically in the form of 33 determinant: Example: if =(,2,-2) and =(3,0,) k j i k j i

20 Cross Prodct It s not tre in general that For example: Right-hand rle If the fingers of the right hand are cpped so they point in the direction of rotation, then the thmb indicates the direction of 20

21 Geometric Interpretation of Cross Prodct From Lagrange s identity, we hae Since so, it follows that sin 2

22 Geometric Interpretation of Cross Prodct From Lagrange s identity in Theorem 3.5. If θdenotes the angle between and, then Since, it follows that, ths 22

23 Geometric Interpretation of Cross Prodct sin is the altitde ( 頂垂線 ) of the parallelogram determined by and. Ths, the area A of this parallelogram is gien by This reslt is een correct if and are collinear, since we hae when 23

24 Area of a Parallelogram Theorem (Area of a Parallelogram) If and are ectors in 3-space, then is eqal to the area of the parallelogram determined by and. Example Find the area of the triangle determined by the point (2,2,0), (-,0,2), and (0,4,3). 24

25 25 Triple Prodct Definition If, and w are ectors in 3-space, then ( w) is called the scalar triple prodct ( 純量三乘積 ) of, and w ) ( w w w w

26 Example = 3i 2j 5k, = i + 4j 4k, w = 3j + 2k 26

27 Triple Prodct Remarks: The symbol ( ) w make no sense becase we cannot form the cross prodct of a scalar and a ector. ( w) = w ( ) = (w ), since the determinants that represent these prodcts can be obtained from one another by two row interchanges. 27

28 Theorem The absolte ale of the determinant det 2 is eqal to the area of the parallelogram in 2-space determined by the ectors = (, 2 ), and = (, 2 ), The absolte ale of the determinant 2 3 det 2 3 w w2 w3 is eqal to the olme of the parallelepiped in 3-space determined by the ectors = (, 2, 3 ), = (, 2, 3 ), and w = (w, w 2, w 3 ), 28

29 Proof of Theorem 3.5.4(a) View and as ectors in the xy-plane of an xyzcoordinate system. Express =(, 2,0) and =(, 2,0) It follows from Theorem and the fact that that the area A of the parallelogram determined by and is 29

30 Proof of Theorem 3.5.4(b) The area of the base is The height h of the parallelepiped is the length of the orthogonal projection of on The olme V of the parallelepiped is 30

31 Remark 3

32 Remark We can conclde that where + or reslts depending on whether makes an acte or an obtse angle with 32

33 Theorem If the ectors = (, 2, 3 ), = (, 2, 3 ), and w = (w, w 2, w 3 ) hae the same initial point, then they lie in the same plane if and only if ( w ) w w w

34 Exercises Sec. 3.: 4, 22, 38 (Tre and False) Sec. 3.2:, 9, 33, 35 (Tre and False) Sec. 3.3: 7, 23, 38, 44, 47 (Tre and False) Sec. 3.4: 0, 25, 27, 29 (Tre and False) Sec. 3.5: 8, 9, 32, 4 (Tre and False) 34