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Understanding Determinants in Linear Algebra Classes

Learn how to expand determinants using rows or columns and understand key concepts in linear algebra theory. Explore proofs and theorems to enhance your knowledge.

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Understanding Determinants in Linear Algebra Classes

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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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