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# LINEAR ALGEBRA MATH 304

Texas A&M

GPA 3.6

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This 64 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 304 at Texas A&M University taught by Yaroslav Vorobets in Fall. Since its upload, it has received 32 views. For similar materials see /class/226002/math-304-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15

MATH 304 Linear Algebra Lecture 8 Inverse matrix continued Elementary matrices Transpose of a matrix Inverse matrix Definition Let A be an ngtltn matrix The inverse of A is an ngtltn matrix denoted A l such that AA 1 A lA If A 1 exists then the matrix A is called invertible Otherwise A is called singular lnverting diagonal matrices Theorem A diagonal matrix D diagd1 7 dquot is invertible if and only if all diagonal entries are nonzero d y 0 for 1 S i S n If D is invertible then D 1 diagdf1d1 Inverting 2 by2 matrices Definition The determinant of a 2x2 matrix A lt3 b is detAad bc C d Theorem A matrix A Z is invertible if and only if detA y 0 If detA y 0 then abili 1 d b Cd iad bc C a39 Fundamental results on inverse matrices Theorem 1 Given a square matrix A the following are equivalent i A is invertible ii x 0 is the only solution of the matrix equation Ax 0 iii the row echelon form of A has no zero rows iv the reduced row echelon form of A is the identity matrix Theorem 2 Suppose that a sequence of elementary row operations converts a matrix A into the identity matrix Then the same sequence of operations converts the identity matrix into the inverse matrix A l Theorem 3 For any ngtltn matrices A and B BA l ltgt AB l Row echelon form of a square matrix 3 3 3 D invertible case noninvertible case 3 2 0 Example A 1 0 1 2 3 0 To Check Whether A is invertible we convert it to row echelon form Interchange the lst row with the 2nd row 1 0 1 3 2 0 2 3 0 Add 3 times the lst row to the 2nd row 1 0 1 0 2 3 2 3 0 Add 2 times the lst row to the 3rd row 1 0 1 0 2 3 0 3 2 Multiply the 2nd row by 12 1 0 1 0 1 15 0 3 2 Add 3 times the 2nd row to the 3rd row 1 0 1 0 1 15 0 0 25 Multiply the 3rd row by 25 0 1 0 15 0 0 We already know that the matrix A is invertible Let39s proceed towards reduced row echelon form Add 32 times the 3rd row to the 2nd row 1 0 1 0 1 0 0 0 1 Add 1 times the 3rd row to the lst row 100 010 001 To obtain A l we need to apply the following sequence of elementary row operations to the identity matrix interchange the 1st row with the 2nd row add 3 times the 1st row to the 2nd row add 2 times the 1st row to the 3rd row multiply the 2nd row by 12 add 3 times the 2nd row to the 3rd row multiply the 3rd row by 25 add 32 times the 3rd row to the 2nd row add 1 times the 3rd row to the 1st row A convenient way to compute the inverse matrix A 1 is to merge the matrices A and into one 3x6 matrix A l and apply elementary row operations to this new matrix 3 20 100 A 101 010 2 30 001 3 20100 A 101010 2 30001 3 20100 1 1010 0001 00 2 Interchange the lst row with the 2nd row 1 0 1 0 1 0 3 2 0 1 0 0 2 3 0 0 0 1 Add 3 times the lst row to the 2nd row 1 0 1 0 1 0 0 2 3 1 3 0 2 3 0 0 0 1 Add 2 times the lst row to the 3rd row 1 0 1 0 1 0 0 2 3 1 3 0 0 3 2 0 2 1 Multiply the 2nd row by 12 1 0 1 0 1 0 0 1 15 05 15 0 0 3 2 0 2 1 Add 3 times the 2nd row to the 3rd row 1 0 1 0 1 0 0 1 15 05 15 0 0 0 25 15 25 1 Multiply the 3rd row by 25 1 0 1 0 1 0 0 1 15 05 15 0 0 0 1 06 1 04 Add 32 times the 3rd row to the 2nd row 1 0 1 0 1 0 0 1 0 04 0 06 0 0 1 06 1 04 Add 1 times the 3rd row to the lst row 1 0 0 06 0 04 0 1 0 04 0 06 0 0 1 06 1 04 25 35 25 001 35 25 35 That is 001 010 100 25 3 5 25 0 0 1 35 25 35 010 203 312 001 010 100 010 203 312 25 35 25 0 0 1 35 25 35 Why does it work 1 0 0 31 32 33 31 a2 a3 0 2 0 b1 b2 b3 2b1 2b2 2133 0 0 1 C1 C2 C3 C1 C2 C3 1 0 0 31 32 33 a1 a2 33 3 1 0 b1 b2 b3 b13a1 b23a2 b33as 0 0 1 C1 C2 C3 C1 C2 C3 1 0 0 a1 22 23 a1 a2 33 0 0 1 b1 b2 b3 C1 C2 C3 0 1 0 C1 C2 C3 b1 b2 b3 Proposition Any elementary row operation can be simulated as left multiplication by a certain matrix Elementary matrices E r row i 1 To obtain the matrix EA from A multiply the ith row by r To obtain the matrix AE from A multiply the ith column by r Elementary matrices row i row j C w j 0 on 1 To obtain the matrix EA from A add r times the ith row to the jth row To obtain the matrix AE from A add r times the jth column to the ith column Elementary matrices 0 1 row i l row j o quot 1 To obtain the matrix EA from A interchange the ith row with the jth row To obtain AE from A interchange the ith column with the jth column Why does it work Assume that a square matrix A can be converted to the identity matrix by a sequence of elementary row operations Then EkEk1 E2E1A I where E E2 7 Ek are elementary matrices corresponding to those operations Applying the same sequence of operations to the identity matrix we obtain the matrix B EkEk1E2E1 EkEk1E2E1 Thus BA which implies that B A l Transpose of a matrix Definition Given a matrix A the transpose of A denoted AT is the matrix whose rows are columns of A and whose columns are rows of A That is if A aij then AT by where bJ aJi T 1 4 Examples E 2 5 3 6 T 7 T 8 789 4 7 9 7 0 Properties of transposes o ATTA o ABTATBT o rATrAT ABTBTAT A1A2AkTAkTA2TA1T A71TAT71 Definition A square matrix A is said to be symmetric if AT A For example any diagonal matrix is symmetric Proposition For any square matrix A the matrices B AAT and C A AT are symmetric Proof BT AATT ATTAT AAT B CT AATTATATTATA C MATH 304 Linear Algebra Lecture 7 Inverse matrix continued Diagonal matrices Definition A square matrix is called diagonal if all non diagonal entries are zeros 7 0 0 Example 0 1 0 denoted diag712 0 0 2 Theorem Let A diags1 52 5quot B diagt1t2t Then A B diag51 t1752 t2 75 tn rA diagrs1 r52 rsn AB diag51t1 52t2 sntn Identity matrix Definition The identity matrix or unit matrix is a diagonal matrix with all diagonal entries equal to 1 100 10 I11 I2lt0 1 I3 1 0 00 1 1 0 0 0 10 lngeneral 001 Theorem Let A be an arbitrary mgtlt n matrix Then A Al7 A Inverse matrix Definition Let A be an ngtltn matrix The inverse of A is an ngtltn matrix denoted A l such that AA 1 A lA If A 1 exists then the matrix A is called invertible Otherwise A is called singular Let A and B be ngtltn matrices If A is invertible then we can divide B by A left division A lB right division BA l Basic properties of inverse matrices o The inverse matrix if it exists is unique 0 If A is invertible so is A4 and A 1 1 A o If ngtltn matrices A and B are invertible so is AB and ABr1 B lA l o If ngtltn matrices A17A2 Ak are invertible so lS A1142 Ak and A1142 Ak71 A21 lnverting diagonal matrices Theorem A diagonal matrix D diagd1 7 dquot is invertible if and only if all diagonal entries are nonzero d y 0 for 1 S i S n If D is invertible then D 1 diagdf1d1 lnverting diagonal matrices Theorem A diagonal matrix D diagd17 7 dquot is invertible if and only if all diagonal entries are nonzero d 0 for 1 S i S n If D is invertible then D 1 diagdf177d1 Proof If all d 0 then Clearly diagd17 7dquot diagdf17 7d1 diag1771 diagdf17 7 d77 1diagd17 7 dquot diag1771 Now suppose that d 0 for some i Then for any ngtltn matrix B the ith row of the matrix DB is a zero row Hence DB y Inverting 2 by2 matrices Definition The determinant of a 2x2 matrix A lt3 b is detAad bc C d Theorem A matrix A Z is invertible if and only if detA y 0 If detA y 0 then abili 1 d b Cd iad bc C a39 Theorem A matrix A Z is invertible if and only if detA y 0 If detA y 0 then a b 1 i 1 d b C d i ad bc C a 39 Proof Let B d 7b Then C a adibc 0 ABBA 0 adibc ad 7 bc2 In the case detA 31 0 we have A 1 detA 1B In the case detA 0 the matrix A is not invertible as otherwise AB O gt A lAB A 10 gt B O gt A 0 but the zero matrix is singular 4x3y5 Problem Solve a system 3X 2y 1 This system is equivalent to a matrix equation 4 3 x i 5 3 2 y i1 39 4 3 Let A 3 2 We have detA ily O Hence A is invertible Let39s multiply both sides of the matrix equation by A 1 from the left i if i 3 i i 3H i 31 at 13 System of n linear equations in n variables a11X1 a12x2 39 39 39 a1an b1 a21X1 a22x2 39 39 39 a2an b2 gt bl anlxl l an2X2 l anan bn where 811 812 317 X1 131 A El21322 aquot 7 x X2 7 b anl an2 ann Xn bn Theorem If the matrix A is invertible then the system has a unique solution which is x A lb Problem Solve the matrix equation XA B X 4 2 5 2 where Alt1 1 Blt3 0 Since B is a 2x2 matrix it follows that XA and X are also 2x2 matrices XABX ltgt X XAB ltgt X AB ltgt XB A 1 provided that A is an invertible matrix 3 2 Alt 1 0 3 2 o Alt1 0 det A 3 0 2 1 2 l A lt2 M gt a zgt lt2 gt as 16gt lt2 gt Fundamental results on inverse matrices Theorem 1 Given a square matrix A the following are equivalent i A is invertible ii x 0 is the only solution of the matrix equation Ax 0 iii the row echelon form of A has no zero rows iv the reduced row echelon form of A is the identity matrix Theorem 2 Suppose that a sequence of elementary row operations converts a matrix A into the identity matrix Then the same sequence of operations converts the identity matrix into the inverse matrix A l Theorem 3 For any ngtltn matrices A and B BA l ltgt AB l Row echelon form of a square matrix 3 3 3 D invertible case noninvertible case MATH 304 Linear Algebra Lecture 4 Row echelon form GaussJordan reduction System of linear equations a11X1 a12x2 39 39 39 a1an b1 a21X1 a22x2 39 39 39 a2an b2 am1X1 am2X2 aman bm Coefficient matrix m x n and column vector of the right hand sides m x 1 311 312 31 b1 821 822 327 132 am 3mg am7 b 3 System of linear equations a11X1 a12x2 39 39 39 a1an b1 a21X1 a22x2 39 39 39 a2an b2 am1X1 am2X2 aman bm Augmented m x n 1 matrix 311 312 31 b1 821 822 327 132 Solution of a system of linear equations splits into two parts A elimination and B back substitution Both parts can be done by applying a finite number of elementary operations Since the elementary operations preserve the standard form of linear equations we can trace the solution process by looking on the augmented matrix In terms of the augmented matrix the elementary operations are elementary row operations Elementary operations for systems of linear equations 1 to multiply an equation by a nonzero scalar 2 to add an equation multiplied by a scalar to another equation 3 to interchange two equations Elementary row operations 1 to multiply a row by some r y 0 2 to add a scalar multiple of a row to another row 3 to interchange two rows Remark The rows are added and multiplied by scalars as vectors namely row vectors Vector algebra Let a a17a27a and b b17b27b be n dimensional vectors and r E R be a scalar Vectorsum a b a1 b17a2 b27ab Scalar multiple ra ra1 rag ran Zero vector 0 0 0 7 0 Negative ofa vector b b17 b27 7 bn Vector difference a ba ba1 b17a2 b27an bn Elementary row operations Augmented matrix 311 312 31 31 V1 821 822 327 132 V2 am1 3mg am7 bm vm where v an a2 am b is a row vector Elementary row operations Operation 1 to multiply the ith row by r y 0 V1 V1 V gt I V Vm Vm Elementary row operations Operation 2 to add the ith row multiplied by r to the jth row V1 V1 Vi Vi gt 5 VJ VJ39 I V Elementary row operations Operation 3 to interchange the ith row with the jth row V1 V1 V VJ39 gt Row echelon form Definition Leading entry of a matrix is the first nonzero entry In a row The goal of the Gaussian elimination is to convert the augmented matrix into row echelon form 0 leading entries shift to the right as we go from the first row to the last one 0 each leading entry is equal to 1 1 71 3 0 2 1 4 0 73 7 2 0 1 1 72 0 0 6 1 3 4 0 0 0 0 0 1 2 3 1 74 2 1 0 0 0 0 0 0 1 9 71 2 1 0 0 0 0 0 0 0 1 1 1 73 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Row echelon form General matrix in row echelon form leading entries are boxed all equal to 1 all the entries below the staircase line are zero each step of the staircase has height 1 each circle marks a free variable Strict triangular form is a particular case of row echelon form that can occur for systems of n equations in n variables The original system of linear equations is consistent if there is no leading entry in the rightmost column of the augmented matrix in row echelon form lnconsistent system The goal of the GaussJordan reduction is to convert the augmented matrix into reduced row echelon form 0 all entries below the staircase line are zero 0 each boxed entry is 1 the other entries in its column are zero 0 each circle marks a free variable Example X y 2 1 102 2X y z3 2 1 13 xyz6 1116 Row echelon form also strict triangular x y 2 1 0 2 y z 1 0 1 1 z 2 0 0 2 Reduced row echelon form x 7 3 0 0 3 y 1 0 0 1 z i 2 0 0 2 Another example xy 2z1 11 21 y z3 01 13 X4y 3z1 14 31 Row echelon form xy 2z1 1 21 y z3 0 13 01 00 0 Reduced row echelon form x z0 0 10 y z0 0 10 01 00 0 Yet another example X y 22 y z X 4y 32 14 Row echelon form xy 2z1 1 21 y z 3 0 13 00 0000 Reduced row echelon form x z 2 0 1 2 y z 3 0 13 00 0000 X12X23X34X410 X22X33X46 1 2 3 4 10 0 1 2 3 6 The matrix is already in row echelon form To convert it into reduced row echelon form add 2 times the 2nd row to the 1st row 0 1 2 2 X3 and X4 are 0 2 3 6 free variables X1 X3 2X4 2 X1X32X4 2 X22X33X46 X2 2X3 3X46 New example Augmented matrix lt System of linear equations X12X23X34X4 X22X33X46 General solution X1t25 2 X2 2t 3s6 X3t tseR X4 Example with a parameter y 32 0 xy 220 36R X 2y 32 0 The system is homogeneous all right hand sides are zeros Therefore it is consistent X y z 0 is a solution 0 1 3 0 Augmented matrix 1 1 2 0 1 2 a 0 Since the 1st row cannot serve as a pivotal one we interchange it with the 2nd row Now we can start the elimination First subtract the 1st row from the 3rd row 1 1 2 0 1 1 2 0 0 1 3 0 gt 0 1 3 0 1 2 a 0 0 1 32 0 The 2nd row is our new pivotal row Subtract the 2nd row from the 3rd row 1 1 2 0 1 1 2 0 0 1 3 0 gt 0 1 3 0 0 1 32 0 0 0 a 1 0 At this point row reduction splits into two cases Case 1 a y 1 In this case multiply the 3rd row by a 1 1 1 1 2 0 1 2 0 0 1 3 0 gt 0 3 0 0 0 a 1 0 0 0 0 The matrix is converted into row echelon form We proceed towards reduced row echelon form Subtract 3 times the 3rd row from the 2nd row Add 2 times the 3rd row to the 1st row 11 20 1100 0100 gt0100 0010 0010 Finally subtract the 2nd row from the 1st row 1100 000 00gt000 10 000 Thus X y z 0 is the only solution 01 00 Case 2 a 1 In this case the matrix is already in row echelon form 1 20 0 30 00 00 To get reduced row echelon form subtract the 2nd row from the 1st row 0 0 1 3 0 gt 0 3 0 0 0 0 0 0 0 0 0 z is a free variable X 520 2 X52 y3z0 y 3z System of linear equations y 32 0 X y 22 0 X 2y 32 0 Solution If a y 1 then X7y72 000 if a 1 then X7y72 5t 3t t t E R

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