Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Monday, February 28 3.2 Assignment Page 199 Problem 10, 20, 21, 40 Main Idea: You can expand along any row or column. Key Words: Adjoint, Aij Det[A] = SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) Goal: Learn how to expand a determinant. ============================================================== Previous Assignment: Assignment Page 191 Problem 18 through 30. --------------------------------------------------------- Page 191 Problem 18 | 1 3 5 | | 2 1 1 | = 2+9+40-(15+4+12) = 20 | 3 4 2 | 19. Switch two rows changes sign 20. Multiply row by scalar, multiplies determinant by same scalar. 21. Add a multiple of one row to another does not change the determinant 22. Add a multiple of one row to another does not change the determinant 23. Multiply a row by k multiplies determinant by k. 24. Switch two rows changes the sign of the determinant. 25. Determinant = 1 26. Determinant = 1 27. Determinant = k 28. Determinant = k 29. Determinant = -1 30. Determinant = -1 ------------------------------------------------------------------------ Today we do the following. Theorem: If Anxn is any square matrix, then n+1 Det[A] = a Det[A ] - a Det[A ] + ..... + (-1) Deg[A ] 11 11 12 12 1n Where Aij is the n-1 x n-1 matrix obtained from A by deleting row i and column j. Proof. Det[A] = SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) In this sum there are n possible choices for a . One 1 p(1) can choose any element from the top row. The possibilities are a , a , ... a . 11 12 1n We split up the sum into n parts depending on which element was chosen from the top row. Det[A] = SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) with p(1) = 1 + SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) with p(1) = 2 + SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) with p(1) = 3 . . . + SUM sgn(p) a a ... a all p 1 p(1) 2 p(2) n p(n) with p(1) = n ---------------------------------------------------------- Det[A] = SUM sgn(p) a a ... a all p 11 2 p(2) n p(n) with p(1) = 1 + SUM sgn(p) a a ... a all p 12 2 p(2) n p(n) with p(1) = 2 + SUM sgn(p) a a ... a all p 13 2 p(2) n p(n) with p(1) = 3 . . . + SUM sgn(p) a a ... a all p 1n 2 p(2) n p(n) with p(1) = n -------------------------------------------------------- Det[A] = a SUM sgn(p) a ... a 11 all p 2 p(2) n p(n) with p(1) = 1 +a SUM sgn(p) a ... a 12 all p 2 p(2) n p(n) with p(1) = 2 + a SUM sgn(p) a ... a 13 all p 2 p(2) n p(n) with p(1) = 3 . . . + a SUM sgn(p) a ... a 1n all p 2 p(2) n p(n) with p(1) = n These SUMS are very close to Det[A ]. 1j The only difference is that the sgn(p) is based on the total number of times that a bigger precedes a smaller in all n positions of p(1) p(2) p(3) ... p(n) while in the Det[A ] one only counts the number of times that 1j a bigger precedes a smaller in the n-1 positions of p(2) p(3) ... p(n). But if p(1) = k then p(1) precedes exactly k-1 smaller values. We can then write Det[A] = a Det[A ] 11 11 - a Det[A ] 12 12 + a Det[A ] 13 13 . . n-1 . (-1) a Det[A ] 1n 1n -------------------------------------------------- n i+j Theorem: Det[A] = Sum (-1) a Det[A ] j=1 ij ij Proof: This is expansion by the i th row. We first interchange rows i-1 times to move the i th row to the first row. Then we expand along the first row. This gives i-1 Det[A]=(-1) [ a Det[A ] - a Det[A ] + a Det[A ] ... i1 i1 i2 i2 i3 i3 n-1 (-1) a Det[A ] in in This can be written as n i+j-2 Det[A] = SUM (-1) a Det[A ] j=1 ij ij 2 Which can be simplified since (-1) = 1 to n i+j Det[A] = SUM (-1) a Det[A ] j=1 ij ij ----------------------------------------------------------- Definition: Given a matrix A, then | +Det[A ] - Det[A ] +Det[A ] .... | | 11 21 31 | | | | -Det[A ] + Det[A ] -Det[A ] .... | | 12 22 32 | | | Adj(A) = | +Det[A ] - Det[A ] +Det[A ] .... | | 13 23 33 | | | | -Det[A ] + Det[A ] -Det[A ] .... | | 14 24 34 | | . . . | | : : : | Theorem: A Adj(A) = Adj(A) A = Det[A] I. Find the inverse of | 3 1 7 | | 1 2 3 | | 2 3 1 | T check | 3 1 7 | | -7 (-5) -1 | | -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 | -1 | 3 1 7 | | -7 20 -11 | | 1 2 3 | = -1/23 | 5 -11 -2 | | 2 3 1 | | -1 -7 5 | ------------------------------------------------------------ T Theorem: Det[A] = Det[A ] -1 T T -1 Proof: Since (A ) = (A ) If A is invertible, then A is a product of Elementary Row Operation Matrices. Since for Elementary Row Operation T T Matrices Det[E ] = Det[E ], we have Det[A] = Det[A ]. T T If A is not invertible, then neither is A and Det[A ] = Det[A ] = 0. -------------------------------------------------------- Theorem: The determinant of a linear transformation is invariant under change of basis. -1 -1 Proof: Det[ P A P] = Det[P ] Det[A] Det[P] = -1 = Det[P P] Det[A] = Det[A]. --------------------------------------------------- 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. Find the Determinant of differentiation on the space with x -x basis e , e x -x e e x e 1 0 -x e 0 -1 The determinant is -1 here as well. ------------------------------------------------------ Page 190 Problem 10 A = | 1 -2 5 2 | | 0 0 3 0 | | 2 -6 -7 5 | | 5 0 4 4 | Find the determinant of A. | 1 -2 2 | Det[A ] = -3 | | | 2 -6 5 | | 5 0 4 | | 1 -2 2 | Det[A ] = -3 | | |-1 0 -1 | | 5 0 4 | | | Det[A ] = -3*2 | | |-1 -1 | | 5 4 | Det[A] = -3*2*(-4+5) = -6 --------------------------------------------------------- For a triangular matrix, the determinant is the product of the elements on the diagonal.