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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications,

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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

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  1. Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher

  2. Friday, April 4 Chapter 6.2 Page 266 Problem 4,14,46,48 Main Idea: Expand along any row or column. Key Words: Adjoint, Aij Det[A] = SUM sgn(p) a 1 p(1) a 2 p(2) ... a n p(n) all p Goal: Learn how to expand a determinant.

  3. Theorem: If A n x n is any square matrix, then Det[A] = a11 Det[A11] - a12 Det[A12]+...+(-1)n+1 Det[A1n] Where Aij is the n-1 x n-1 matrix obtained by deleting row i and column j from A.

  4. Proof Det [A] = SUM sgn(p) a 1 p(1) a 2 p(2) ... a n p(n) all p There are n possible choices for a 1 p(1) . One can choose any element from the top row. The possibilities are a11, a12, ... a1n.

  5. We split up the sum into n parts depending on which element was chosen from the top row.

  6. Det[A] = SUMsgn(p) a 1 p(1)a 2 p(2)…a n p(n) p (1)=1 + SUM sgn(p) a 1 p(1)a 2 p(2) .... a n p(n) p(1)=2 + SUM sgn(p) a 1 p(1)a 2 p(2) … a n p(n) p(1)=3: + SUM sgn(p) a 1 p(1)a 2 p(2)… a n p(n) p(1)=n

  7. Det[A] = SUM sgn(p) a 11 a 2 p(2)... a n p(n) p(1) = 1 + SUM sgn(p) a 12a 2 p(2)... a n p(n) p(1) = 2 + SUM sgn(p) a 13 a 2 p(2)... a n p(n) p(1) = 3 : + SUM sgn(p) a 1na 2 p(2)... a n p(n) p(1) = n

  8. Det[A] = a11 SUM sgn(p) a 2 p(2)... a n p(n) p(1) = 1 + a12 SUM sgn(p) a 2 p(2)... a n p(n) p(1) = 2 + a13 SUM sgn(p) a2 p(2)... a n p(n) p(1) = 3: + a1n SUM sgn(p) a 2 p(2)... a n p(n) p(1) = n

  9. These SUMS are close to Det[Aij]. The only difference is that the sgn(p) is based on all n positions of p(1) p(2) p(3) ... p(n). In Det [Aij] the sgn is based on n-1 positions of p(2) p(3) ... p(n).

  10. Det[A] = a11 Det[A11] - a12 Det[A12] + a13 Det[A13] . . (-1)n+1 a1n Det[A1n]

  11. n Theorem: Det[A] = Sum (-1) i+jaij Det[Aij] j=1 Proof: This is expansion by the i th row.

  12. We first interchange rows i-1 times to move the ith row to the first row. Then we expand along the first row. This gives | ai1 Det[Ai1] | | - ai2 Det[Ai2] | Det[A]=(-1)i-1 | + ai3 Det[Ai3] | | : | | (-1)n+1 ain Det[Ain] |

  13. This can be written as Det[A] = SUM (-1) i+j-2 aij Det[Aij] Which can be simplified since (-1)2 = 1 to Det[A] = SUM (-1)i+j aij Det[Aij]

  14. Definition: Given a matrix A, then |+Det [A11] - Det [A21] +Det [A31] ....| |-Det [A12] + Det [A22] -Det [A32] ....| Adj(A) =|+Det [A13] - Det [A23] +Det [A33] ....| |-Det [A14] + Det [A24] -Det [A34] ....| | . . . | | . . . | | . . . |

  15. Theorem: A Adj(A) = Adj(A) A = Det[A] I

  16. Find the inverse of | 3 1 7 | | 1 2 3 | | 2 3 1 |

  17. check |3 1 7| | -7 (-5) -1| T |-7 20 -11| | -23 0 0| Adj|1 2 3| =|(-20)-11 (7)| = |5 -11 -2 | | 0 -23 0| |2 3 1| | -11 (2) 5 | | -1 -7 5| | 0 0 -23 |

  18. | 3 1 7 | -1 | -7 20 -11 | | 1 2 3 | = -1/23 | 5 -11 -2 | | 2 3 1 | | -1 -7 5 |

  19. Theorem: Det[A] = Det[AT] Proof: Since (A-1)T = (AT) -1 If A is invertible, then A is a product of Elementary Row Operation Matrices. For Elementary Row Operation Matrices Det [ET] = Det [E], we have Det [A] = Det[AT]. If A is not invertible, then neither is AT and Det [A] = Det[AT] = 0.

  20. Theorem: The determinant of a linear transformation is invariant under change of basis. Proof: Det [ P -1A P] = Det[P-1] Det [A] Det [P] = Det[ P-1 P] Det [A] = Det [A].

  21. Find the Determinant of differentiation on the space with basis Sinh[x] Cosh[x]. Sinh[x] Cosh[x] Sinh[x] 0 1 Cosh[x] 1 0 Det[ derivation ] = -1.

  22. Find the Determinant of differentiation on the space with basis e x, e -x. ex e -x e x 1 0 e -x 0 -1 Det [ derivation ] = -1.

  23. Page 263 Example 8. A = | 1 0 1 2 | | 9 1 3 0 | | 9 2 2 0 | | 5 0 0 3 | Find the determinant of A.

  24. Add -2 Row 2 to row 3 | 1 0 1 2 | | 9 1 3 0 | |-9 0-4 0 | | 5 0 0 3 |

  25. Expand along second column | 1 1 2 | +1 | | |-9 -4 0 | | 5 0 3 |

  26. Add 4 Row 1 to Row 2 | 1 1 2 | +1 | | |-5 0 8 | | 5 0 3 |

  27. Expand along column 2 | | (-1) | | |-5 8 | | 5 3 | Evaluate the final 2 x 2 determinant. -(-15 - 40) = 55.

  28. Page 265 Problem 5. | 1 1 1 1 1 | | 1 2 2 2 2 | A = | 1 1 3 3 3 | | 1 1 1 4 4 | | 1 1 1 1 5 | Find the determinant of A.

  29. Eliminate the first column | 1 1 1 1 1 | | 0 1 1 1 1 | A = | 0 0 2 2 2 | | 0 0 0 3 3 | | 0 0 0 0 4 | The determinant of a triangular matrix is the product of the elements on the diagonal. Det [A] = 24.

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