Solution The Gauss-Jordan method yields the following sequence of matrices: [1001 [1001 [100] A=010→010→010=A 010 010000 Thus,rank 4 rank 4 =2 <3.This shows that Vis a linearly dependent set of vectors.In fact,the EROs used to transform 4 to A can be used to show that [11 0]= [1 0 0]+[0 1 0].This equation also shows that Vis a linearly dependent set of vectors. PROBLEMS Group A Group B Determine if each of the following sets of vectors is linearly 7 Show that the linear system Ax b has a solution if and independent or linearly dependent. only if b can be written as a linear combination of the 1V={101],[121],[222]} columns of A. 2V={[210],[120],[331]} 8 Suppose there is a collection of three or more two- dimensional vectors.Provide an argument showing that the 3V={21],[12]} collection must be linearly dependent. 4V={20,[30]} 9 Show that a set of vectors V(not containing the 0 vector) is linearly dependent if and only if there exists some vector in I that can be written as a nontrivial linear combination 5V= 卧 of other vectors in V. 0 0110 2.5 The Inverse of a Matrix To solve a single linear equation such as 4x=3,we simply multiply both sides of the equation by the multiplicative inverse of 4,which is 4,or.This yields 4(4x)= (4)3,or x=(Of course,this method fails to work for the equation Ox=3,because zero has no multiplicative inverse.)In this section,we develop a generalization of this technique that can be used to solve "square"(number of equations number of un- knowns)linear systems.We begin with some preliminary definitions. DEFINITIO NA square matrix is any matrix that has an equal number of rows and columns. The diagonal elements of a square matrix are those elements ay such that i=j. A square matrix for which all diagonal elements are equal to I and all nondiagonal elements are equal to 0 is called an identity matrix. The m x m identity matrix will be written as Im.Thus, [100] 6=0 3=010 001
Solution The Gauss–Jordan method yields the following sequence of matrices: A → → A Thus, rank A rank A 2 3. This shows that V is a linearly dependent set of vectors. In fact, the EROs used to transform A to A can be used to show that [1 1 0] [1 0 0] [0 1 0]. This equation also shows that V is a linearly dependent set of vectors. PROBLEMS Group A 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 36 CHAPTER 2 Basic Linear Algebra Determine if each of the following sets of vectors is linearly independent or linearly dependent. 1 V {[1 0 1], [1 2 1], [2 2 2]} 2 V {[2 1 0], [1 2 0], [3 3 1]} 3 V {[2 1], [1 2]} 4 V {[2 0], [3 0]} 5 V , , 6 V , , 1 0 1 0 2 1 1 0 0 5 7 9 4 5 6 1 2 3 Group B 7 Show that the linear system Ax b has a solution if and only if b can be written as a linear combination of the columns of A. 8 Suppose there is a collection of three or more twodimensional vectors. Provide an argument showing that the collection must be linearly dependent. 9 Show that a set of vectors V (not containing the 0 vector) is linearly dependent if and only if there exists some vector in V that can be written as a nontrivial linear combination of other vectors in V. 2.5 The Inverse of a Matrix To solve a single linear equation such as 4x 3, we simply multiply both sides of the equation by the multiplicative inverse of 4, which is 41 , or 1 4 . This yields 41 (4x) (41 )3, or x 3 4 . (Of course, this method fails to work for the equation 0x 3, because zero has no multiplicative inverse.) In this section, we develop a generalization of this technique that can be used to solve “square” (number of equations number of unknowns) linear systems. We begin with some preliminary definitions. DEFINITION ■ A square matrix is any matrix that has an equal number of rows and columns. ■ The diagonal elements of a square matrix are those elements aij such that i j. ■ A square matrix for which all diagonal elements are equal to 1 and all nondiagonal elements are equal to 0 is called an identity matrix. ■ The m m identity matrix will be written as Im. Thus, I2 , I3 , 0 0 1 0 1 0 1 0 0 0 1 1 0
If the multiplications I 4 and Al are defined,it is easy to show that Im A=Al=A. Thus,just as the number 1 serves as the unit element for multiplication of real numbers, m serves as the unit element for multiplication of matrices. Recall that is the multiplicative inverse of 4.This is because 4()=()4=1.This motivates the following definition of the inverse of a matrix. DEFINITION■ For a given m X m matrix A,the m X m matrix B is the inverse of A if BA AB=I 16) (It can be shown that if BA=I or 4B =Im then the other quantity will also equal In)■ Some square matrices do not have inverses.If there does exist an m X m matrix B that satisfies Equation(16),then we write B=A-.For example,if 「20-1] A= 31 -101 the reader can verify that 20 -1[10 12 [100 3 1 2 -5 1 -7 010 -10 1 10 2 001 and 10 120 -1 [100 -7 1 2 0 10 10 2-10 001 Thus, 1 0 1 10 2 To see why we are interested in the concept of a matrix inverse,suppose we want to solve a linear system Ax =b that has m equations and m unknowns.Suppose that 4- exists.Multiplying both sides of Ax b by 4,we see that any solution of Ax =b must also satisfy 4(Ax)=Ab.Using the associative law and the definition of a matrix in- verse,we obtain (4-A)x=Ab or Imx=A厂lb or x=A-b This shows that knowing 4 enables us to find the unique solution to a square linear sys- tem.This is the analog of solving 4x=3 by multiplying both sides of the equation by 4 The Gauss-Jordan method may be used to find 4-(or to show that 4-does not ex- ist).To illustrate how we can use the Gauss-Jordan method to invert a matrix,suppose we want to find for
If the multiplications Im A and AIm are defined, it is easy to show that Im A AIm A. Thus, just as the number 1 serves as the unit element for multiplication of real numbers, Im serves as the unit element for multiplication of matrices. Recall that 1 4 is the multiplicative inverse of 4. This is because 4( 1 4 ) ( 1 4 )4 1. This motivates the following definition of the inverse of a matrix. DEFINITION ■ For a given m m matrix A, the m m matrix B is the inverse of A if BA AB Im (16) (It can be shown that if BA Im or AB Im, then the other quantity will also equal Im.) ■ Some square matrices do not have inverses. If there does exist an m m matrix B that satisfies Equation (16), then we write B A1 . For example, if A the reader can verify that and Thus, A1 To see why we are interested in the concept of a matrix inverse, suppose we want to solve a linear system Ax b that has m equations and m unknowns. Suppose that A1 exists. Multiplying both sides of Ax b by A1 , we see that any solution of Ax b must also satisfy A1 (Ax) A1 b. Using the associative law and the definition of a matrix inverse, we obtain (A1 A)x A1 b or Imx A1 b or Imx A1 b This shows that knowing A1 enables us to find the unique solution to a square linear system. This is the analog of solving 4x 3 by multiplying both sides of the equation by 41 . The Gauss–Jordan method may be used to find A1 (or to show that A1 does not exist). To illustrate how we can use the Gauss–Jordan method to invert a matrix, suppose we want to find A1 for A 5 3 2 1 1 7 2 0 1 0 1 5 1 0 0 1 0 1 0 1 0 0 1 2 1 0 1 0 2 3 1 1 7 2 0 1 0 1 5 1 0 0 1 0 1 0 1 0 0 1 7 2 0 1 0 1 5 1 1 2 1 0 1 0 2 3 1 1 2 1 0 1 0 2 3 1 2.5 The Inverse of a Matrix 37
This requires that we find a matrix a b =A- d that satisfies GT 2-63 (70 From Equation(17),we obtain the following pair of simultaneous equations that must be satisfied by a,b,c,and d: G9-母B8-9 Thus,to find (the first column of),we can apply the Gauss-Jordan method to the augmented matrix [2511 130 Once EROs have transformed to12, [ will have been transformed into the first column of4.To determine [ (the second column of 4),we apply EROs to the augmented matrix When has been transformed into 12, [☒ will have been transformed into the second column of4.Thus,to find each column of 4,we must perform a sequence of EROs that transform 「251 L13
This requires that we find a matrix A1 that satisfies (17) From Equation (17), we obtain the following pair of simultaneous equations that must be satisfied by a, b, c, and d: ; Thus, to find (the first column of A1 ), we can apply the Gauss–Jordan method to the augmented matrix Once EROs have transformed to I2, will have been transformed into the first column of A1 . To determine (the second column of A1 ), we apply EROs to the augmented matrix When has been transformed into I2, will have been transformed into the second column of A1 . Thus, to find each column of A1 , we must perform a sequence of EROs that transform 5 3 2 1 0 1 5 3 2 1 0 1 5 3 2 1 b d 1 0 5 3 2 1 1 0 5 3 2 1 a c 0 1 b d 5 3 2 1 1 0 a c 5 3 2 1 0 1 1 0 b d a c 5 3 2 1 b d a c 38 CHAPTER 2 Basic Linear Algebra
into 12.This suggests that we can find 4 by applying EROs to the 2 X 4 matrix =9109 When 「2 has been transformed to 12, [8 will have been transformed into the first column of and [☒ will have been transformed into the second column of A.Thus,as A is transformed into 12,12 is transformed into A.The computations to determine A-follow. Step 1 Multiply row 1 of 2 by This yields Step 2 Replace row 2 of 4'by -1(row 1 of 4'2)+row 2 of 4'12.This yields s-0引 0 Step 3 Multiply row 2 of 4"I2 by 2.This yields 4s=0引 Step 4 Replace row 1 of 4I by -(row 2 of 4T)+row 1 of 4"I2.This yields 01 Because A has been transformed into 12,12 will have been transformed into 4-.Hence, -[ The reader should verify that 44-1=A-14 =12. A Matrix May Not Have an Inverse Some matrices do not have inverses.To illustrate,let 4-6到m-[5 (18)
into I2. This suggests that we can find A1 by applying EROs to the 2 4 matrix AI2 When has been transformed to I2, will have been transformed into the first column of A1 , and will have been transformed into the second column of A1 . Thus, as A is transformed into I2, I2 is transformed into A1 . The computations to determine A1 follow. Step 1 Multiply row 1 of AI2 by 1 2 . This yields A I 2 Step 2 Replace row 2 of A I 2 by 1(row 1 of A I 2) row 2 of A I 2. This yields A I 2 Step 3 Multiply row 2 of A I 2 by 2. This yields A I2 Step 4 Replace row 1 of A I 2 by 5 2 (row 2 of A I 2 ) row 1 of A I 2 . This yields Because A has been transformed into I2, I2 will have been transformed into A1 . Hence, A1 The reader should verify that AA1 A1 A I2. A Matrix May Not Have an Inverse Some matrices do not have inverses. To illustrate, let A and A1 (18) f h e g 2 4 1 2 5 2 3 1 5 2 3 1 0 1 1 0 0 2 1 2 1 5 2 1 1 0 0 1 1 2 1 2 5 2 1 2 1 0 0 1 1 2 0 5 2 3 1 1 0 1 1 0 5 3 2 1 0 1 1 0 5 3 2 1 2.5 The Inverse of a Matrix 39
To find we must solve the following pair of simultaneous equations: 阳-周 (18.1) 2明- (18.2) When we try to solve(18.1)by the Gauss-Jordan method,we find that is transformed into [121] L00-2 This indicates that(18.1)has no solution,and cannot exist. Observe that (18.1)fails to have a solution,because the Gauss-Jordan method trans- forms A into a matrix with a row of zeros on the bottom.This can only happen if rank A 2.If m X m matrix A has rank A m,then A will not exist. The Gauss-Jordan Method for Inverting an m x m Matrix A Step 1 Write down the m x 2m matrix Alm. Step 1 Use EROs to transform Al into I B.This will be possible only if rank A m. In this case,B =A.If rank A m,then A has no inverse. Using Matrix Inverses to Solve Linear Systems As previously stated,matrix inverses can be used to solve a linear system Ax b in which the number of variables and equations are equal.Simply multiply both sides of Ax=b by 4-to obtain the solution x =4b.For example,to solve 2x1+5x2=7 19 x+3x2=4 write the matrix representation of (19): (20) Let 4- We found in the previous illustration that
To find A1 we must solve the following pair of simultaneous equations: (18.1) (18.2) When we try to solve (18.1) by the Gauss–Jordan method, we find that is transformed into This indicates that (18.1) has no solution, and A1 cannot exist. Observe that (18.1) fails to have a solution, because the Gauss–Jordan method transforms A into a matrix with a row of zeros on the bottom. This can only happen if rank A 2. If m m matrix A has rank A m, then A1 will not exist. The Gauss–Jordan Method for Inverting an m m Matrix A Step 1 Write down the m 2m matrix AIm. Step 1 Use EROs to transform AIm into ImB. This will be possible only if rank A m. In this case, B A1 . If rank A m, then A has no inverse. Using Matrix Inverses to Solve Linear Systems As previously stated, matrix inverses can be used to solve a linear system Ax b in which the number of variables and equations are equal. Simply multiply both sides of Ax b by A1 to obtain the solution x A1 b. For example, to solve 2x1 5x2 7 (19) x1 3x2 4 write the matrix representation of (19): (20) Let A We found in the previous illustration that A1 5 2 3 1 5 3 2 1 7 4 x1 x2 5 3 2 1 1 2 2 0 1 0 1 0 2 4 1 2 0 1 f h 2 4 1 2 1 0 e g 2 4 1 2 40 CHAPTER 2 Basic Linear Algebra