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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Lecture 19. A procedure (an algorithm) for finding the inverse of an invertible matrix. Further investigate the system of linear equation and invertibility of matrices. Determinants: Finding determinants by duplicate column method.

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 19

  2. A procedure (an algorithm) for finding the inverse of an invertible matrix. Further investigate the system of linear equation and invertibility of matrices. Determinants: Finding determinants by duplicate column method. Finding determinants by cofactor expansion along the first row. Finding determinants by cofactor expansion along any row or column. Lecture Outline

  3. Finite sequence of row operation A B Inverse of the row operation in reverse order A B Definition: Row Equivalent Matrices that can be obtained from one another by a finite sequence of elementary row operations are said to berow equivalent. A is row equivalent to B

  4. If Ais annxnmatrix, then the following statements are equivalent, that is, all are true: (a)Ais invertible (b)AX=0has only the trivial solution (the only solution isx1=0, x2=0…xn=0) (c)Ais row equivalent toIn. Finite sequence of row operation A In Theorem 1 If A-1 exists than

  5. a11 a12….… a1n 0 a21 a22….… a2n 0 : : : : : : : : an1 an2….… ann 0 1 0 0….… 0 0 0 1 0….… 0 0 0 0 1….… 0 0 : : : 0 0 0….… 1 0 A= Finite sequence of row operation (b) AX=0has only the trivial solution (the only solution isx1=0, x2=0…xn=0) x1=0, x2=0…xn=0

  6. A In The same sequence of row operation -1 In A Theorem 2 • If Ais annxninvertible matrix, then the sequence of row operations that reducesA to Inreduces Into A-1.

  7. The procedure The same sequence of row operation A | In In | A-1

  8. 1 2 3 2 5 3 1 0 8 A= * –2 * –1 1 2 3 2 5 3 1 0 8 1 0 0 0 1 0 0 0 1 1 2 3 0 1 -3 0 -2 5 1 0 0 -2 1 0 -1 0 1 * 2 Example 1: Find the inverse of

  9. 1 2 3 0 1 -3 0 0 -1 1 0 0 -2 1 0 -5 2 1 * -1 1 2 3 0 1 -3 0 0 1 1 0 0 -2 1 0 5 -2 -1 * 3 1 2 3 0 1 0 0 0 1 1 0 0 13 -5 -3 5 -2 -1 * -3

  10. 1 2 0 0 1 0 0 0 1 -14 6 3 13 -5 -3 5 -2 -1 * -2 1 0 0 0 1 0 0 0 1 -40 16 9 13 -5 -3 5 -2 -1 Thus: -40 16 9 13 -5 -3 5 -2 -1 A-1=

  11. 1 0 0….… 0 0 1 0….… 0 0 0 1….… 0 : : : 0 0 0….… 0 Statement: If the procedure used in this example is attempt on a matrix that is not invertible, then, it will be impossible to reduce the left side to I by row operation. At some point in the computation a row of zeros will occur on the left side, and it can be concluded that the given matrix is not invertible.

  12. 1 2 3 2 5 3 1 0 8 A= x1+2x2+3x3=0 2x1+5x2+3x3=0 x1 +8x3=0 In last example we found that A is invertible. What can we say about the following system of linear equations? (hint: use theorem 1) Answer: The system has only the trivial solution using the second statement in theorem 1 : x1=x2=x3=0

  13. 3 1 5 2 4 1 -4 2 -9 1 0 1 0 1 1 1 1 0 Class Assignment: Exercise set 1.6-5 please submit to TA at the end of the lecture Find the inverse of the following matrices (if the matrices are invertible), and verify (for the invertible matrices) by multiplying the original matrix A= B=

  14. System of Linear Equations and Invertibility

  15. a11 a12….… a1n a21 a22….… a2n : : : an1 an2….… ann x1 x2 : xn B1 B2 : Bn = -1 x1 x2 : xn a11 a12….… a1n a21 a22….… a2n : : : an1 an2….… ann B1 B2 : Bn = Theorem 1 • If Ais an invertiblenxnmatrix, then for eachnx1matrix B,the system of equationAX=Bhas exactly one solution, namelyX= A-1B.

  16. We can write the this system as AX=B x1 x2 x3 5 3 17 1 2 3 2 5 3 1 0 8 A= X= B= Example: consider the system of linear equations: x1+2x2+3x3=5 2x1+5x2+3x3=3 x1 +8x3=17

  17. -40 16 9 13 -5 -3 5 -2 -1 A-1= By theorem 1 the solution of the system is: -40 16 9 13 -5 -3 5 -2 -1 5 3 17 1 -1 2 X=A-1B= = In example 1 we found that the inverse of A-1is : or: x1=1, x2=-1, x3=2

  18. Many problems in engineering and science involve systems of n linear equation with n unknown (that is square matrix). The method is particularly useful when it is necessary to solve a series of systems: AX=B1, AX=B2…….. AX=Bk, Why it is useful to find the inverse of a matrix? In this case each has the same square matrix A and the solutions are: X=A-1B1, X=A-1B2……. X=A-1Bk

  19. a11 a12….… a1n a21 a22….… a2n : : : an1 an2….… ann c1 c2 : cn v1 v2 : vn = -1 c1 c2 : cn a11 a12….… a1n a21 a22….… a2n : : : an1 an2….… ann v1 v2 : vn = Electronic Circuitry AC=V voltage current

  20. Theorem 3 • If Ais annxnmatrix, then the following statements are equivalent: • (a)Ais invertible • (b)AX=0has only the trivial solution (the only solution isx1=0, x2=0…xn=0) • (c)Ais row equivalent toIn. • (d)AX=Bis consistent for everynx1matrixB.

  21. Class Assignment: Exercise set 1.7-3 please submit to TA at the end of the lecture Solve the following system using the method learn today in class x1+2x2+2x3=-1 x1+3x2+ x3=4 x1+3x2+ 2x3=3

  22. 2 1 -3 1 0 5 4 3 0 0 1 2 0 0 0 3 5 1 4 1 0 0 2 -1 0 0 1 1 0 0 0 7 B= Hint: Consider the associated homogeneous systems 2x1 +x2 -3x3 -2x4 =0 5x2+4x3 +3x4=0 x3 +2x4 =0 3x4 =0 5x1 +x2 +4x3 +x4 =0 2x3 -x4=0 x3 + x4 =0 7x4 =0 Without using pencil and paper, determine whether the following matrices are invertible A=

  23. Why study determinants? • They have important applications to system of linear equations and can be used to produce formula for the inverse of an invertible matrix. • The determinant of a square matrixAis denoted bydet(A) or |A|. IfAis 1x1matrix A=[a11], then:det(A)= a11 For exampleA=[-7] det (A)=det(-7)=-7

  24. If Ais a 2x2 matrix a11 a12 a21 a22 A= a11 a12 a21 a22 det(A)= = det(A)= a11a22 -a12a21 Determinant of a 2x2 matrix Then we define a11 a12 a21 a22

  25. 5 4 3 2 5 4 3 2 A= det(A)= = det(A)= 5X2-4X3=-2 Example 1 Then we define 5 4 3 2

  26. If Ais a 3x3 matrix a11 a12 a13 a21 a22 a23 a31 a32 a33 A= det(A)= • 5 -3 • 1 0 2 • 3 -1 2 a11 a12 a13 a21 a22 a23 a31 a32 a33 0 2 -1 2 a22 a23 a32 a33 a21 a23 a31 a33 a21 a22 a31 a32 = 1 = a11 -a12 +a13 Example 2 1 0 3 -1 1 2 3 2 det(A)= -3 -5 Determinant of a 3x3 matrix Then we define =1[0*2 - (-1*2)] –5 [1*2-2*3] - 3[-1*1-0*3]=25

  27. From the previous slide: = det(A)= a11a22a33+a12a23a31+a13a21a32- a13a22a31 - a11a23a32 -a12a21a33 a11 a12 a13 a21 a22 a23 a31 a32 a33 a11 a12 a13 a21 a22 a23 a31 a32 a33 det(A)= a22 a23 a32 a33 a21 a23 a31 a33 a21 a22 a31 a32 = a11 -a12 +a13 • 5 -3 • 1 0 2 • 3 -1 2 det(A)= Example 3: Duplicate Column Method – for 3x3 a11 a12 a21 a22 a31 a32 • 5 • 1 0 • 3 -1 =0+30+3-0-(-2)-10=25

  28. a11 a12 a13 A= a21 a22 a23 For 3x3 matrix a31 a32 a33 a22 a23 a32 a33 Minor and Cofactor of a Matrix Entry • Definition: • IfAis a square matrix, then the minor of entryaijis denoted byMijand is defined to be the determinant of the sub-matrix that remains after the ith row and jth column are deleted fromA. The number(-1)i+jMijis denoted byCijand is called the cofactor of entryaij. M11 = C11 =(-1)1+1 M11 = M11

  29. 1 2 3 2 5 3 1 0 8 A= The minor of a11 is: 1 2 1 0 5 3 0 8 M23= M11= = 1X0 - 2X1=-2 = 5X8 - 3X0=40 The cofactor of a11 is: C11=(-1)i+jMij=(-1)2M11=M11=40 The minor of a23 is: The cofactor of a23 is: C23=(-1)i+jMij=(-1)5M23 =-M23 =2 Example 4

  30. a11 a12 a13 a21 a22 a23 a31 a32 a33 a21 a23 a31 a33 a21 a22 a31 a32 C12= - C13= a11 a12 a13 a21 a22 a23 a31 a32 a33 a22 a23 a32 a33 C11= = a11C11 +a12C12 +a13C13 Finding determinant using the cofactor And for nxn matrix: = a11C11 +a12C12 +…..+a1nC1n

  31. 4 0 1 5 2 -2 1 1 3 3 0 1 -1 2 -2 0 1 3 3 4 1 -1 5 -2 0 1 3 3 4 0 -1 5 2 0 1 1 1 3 -1 0 0 4 5 1 2 0 2 1 -3 1 -2 3 A= det(A)=(1) - (0) + 2 - (-3) Example: evaluate det(A) for: det(A) = a11C11 +a12C12 + a13C13 +a14C14 = (1)(35)-0+(2)(62)-(-3)(13)=198

  32. The determinant of nxn matrix A can be computed by multiplying the entries in any row (or column) by their cofactors and adding the resulting products; that is, for each 1 < i < n and 1 < j < n, Theorem 1 det(A)=a1jC1j +a2jC2j +…..+anjCnj Cofactor expansion along the jth column det(A)=ai1Ci1 +ai2Ci2 +…..+ainCin Cofactor expansion along the ith row

  33. 5 • 1 0 • 3 -1 -3 2 2 det(A)= 1 0 3 -1 1 5 3 -1 1 5 1 0 -3* (-1)4 det(A)= +2*(-1)5 +2*(-1)6 Example 5: evaluate By a cofactor along the third column det(A)=a13C13 +a23C23+a33C33 = det(A)= -3(-1-0)+2(-1)5(-1-15)+2(0-5)=25

  34. 1 0 0….… 0 0 1 0….… 0 0 0 1….… 0 : : : 0 0 0….… 0 Theorem 2 • If A is a square matrix with a row column of zeros, then det(A)=0 Proof • Evaluate det(A) by cofactor expansion along the row or column of zeros.

  35. Let -1 2 3 4 1 -6 -3 5 2 A= Class Assignment: Exercise set 2.1-2 please submit to TA at the end of the lecture • Find det(A) using • (a) the duplicate column method. • (b) cofactor expansion along the first row. • (c) cofactor expansion along the third row. • (d) cofactor expansion along the first column. • (e) cofactor expansion along the second column.

  36. Solution a)

  37. b)

  38. c)

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