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3.2 Determinants; Mtx Inverses

3.2 Determinants; Mtx Inverses. Theorem 1- Product Theorem. If A and B are (n x n), then det(AB)=det A det B (come back to prove later) Show true for 2 x 2 of random variables. Extension. Using induction, we could show that: det(A 1 A 2 …A k ) = detA 1 detA 2 …detA k

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3.2 Determinants; Mtx Inverses

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  1. 3.2 Determinants; Mtx Inverses

  2. Theorem 1- Product Theorem • If A and B are (n x n), then det(AB)=det A det B • (come back to prove later) • Show true for 2 x 2 of random variables

  3. Extension • Using induction, we could show that: • det(A1A2…Ak) = detA1detA2…detAk • also: det (Ak) = (det A)k

  4. Theorem 2 • An (n x n) matrix A is invertible iff det A ≠ 0. If it is invertible, • Proof: ==> given A invertible, AA-1=I • det (AA-1)=detI=1=detAdetA-1 • and

  5. Proof (continued) • <== Given det A ≠ 0 • A can clearly be taken to reduced row ech form w/ no row of zeros (call it R = Ek…E2E1A) (otherwise the determinant would be 0) • (det R = det Ek … det E2 det E1 det A≠0) • Since R has no row of zeros, R=I, and A is clearly invertible.

  6. Example • Find all values of b for which A will have an inverse.

  7. Theorem 3 • If A is any square matrix, det AT = det A • Proof: For E of type I or type II, ET = E (show ex) • For E of type III, ET is also of type III, and det E = 1 = det ET by theorem 2 in 3.1 (which says that if we add a multiple of a row to another row, we do not change the determinant). • So det E = det ET for all E • Given A is any square matrix: • If A is not invertible, neither is AT (since the row operations to reduce A which would take A to a row of zeros could be used as column ops on AT to get a column of zeros) so det A = 0 = det AT

  8. Theorem 3 (continued) • If A is invertible, then A = Ek…E2E1 and AT= E1TE2T…EkT • So det AT = det E1T det E2T …det EkT = detE1detE2…detEk = det A 

  9. Examples • If det A =3, det B =-2 find det (A-1B4AT) • A square matrix is orthogonal is A-1 = AT. Find det A if A is orthogonal. • I = AA-1 = AAT • 1 = det I = det A det AT = (det A)2 • So det A = ± 1

  10. Adjoint • Adjoint of a (2x2) is just the right part of inverse: • Recall that: • Now we will show that it is also true for larger square matrices.

  11. Adjoint--definition • If A is square, the cofactor matrix of A , [Cij(A)], is the matrix whose (I,j) entry is the (i,j) cofactor of A. • The adjoint of A, adj(A), is the transpose of the cofactor matrix: • adj(A) = [Cij(A)]T • Now we need to show that this will allow our definition of an inverse to hold true for all square matrices:

  12. For a (2x2)

  13. Example • Find the adjoint of A: • So we could find det(A)

  14. For (nxn) • A(adjA) = (detA)I for any (nxn): ex. (3 x 3) • we have 0’s off diag since they are like determinants of matrices with two identical rows (like prop 5 of last chapter)

  15. Theorem 4: Adjoint Formula • If A is any square matrix, then • A(adj(A)) = (det A)I = (adj(A))A • If det A ≠ 0, • Good theory, but not a great way to find A-1

  16. Example • Use thm 4 to find the values of c which make A invertible: c ≠ 0

  17. Linear Equations • Recall that if AX = B, and if A is invertible (det A ≠ 0), then • X = A-1B So...

  18. Finding determinants is easier • the right part is just the det of a matrix formed by replacing column i with the B column matrix

  19. Theorem 5: Cramer’s Rule If A is an invertible (n x n) matrix, the solution to the system AX = B of n equations in n variables is: Where Ai is the matrix obtained by replacing column i of A with the column matrix B. This is not very practical for large matrices, and it does not give a solution when A is not invertible

  20. Examples • Solve the following using Cramer’s rule.

  21. Proof of Theorem 1 • for A,B (nxn): det AB = det A detB • det E = -1 if E is type 1. • = u if E is type 2 (and u is multiplied by one row of I) • = 1 if E is of type 3 • If E is applied to A, we get EA • det (EA) = -det(A) if E is of type I • = udet(A) if E is of type II • = det (A) if E is of type III • So det (EA) = det E det A

  22. Continued... • So if we apply more elementary matrices: • det(E2E1A) = det E2(det(E1A)) = det E2 det E1 det A • This could continue and we get the following: • Lemma 1: If E1, E2, …, Ek are (n x n) elementary matrices, • and A is (n x n), then: • det(Ek …E2 E1A) = det Ek … det E2 det E1 det A

  23. Continued • Lemma 2: If A is a noninvertible square matrix, then det A =0 • Proof: A is not invertible ==> when we put A into reduced row echelon form, the resulting matrix, R will have a row of zeros. • det R = 0 • det R = det (Ek…E2E1A) = det Ek … det E2 det E1 det A= 0 • since det E’s never 0, det A = 0

  24. Proving Theorem 1 (finally) • Show that det AB = det A det B • Proof : Case 1: A is not invertible: • Then det A = 0 • If AB were invertible, then AB(AB)-1 = I • so A(BB-1A) = I, which would mean that A is • invertible, but it is not, so AB is not invertible. • Therefore, det AB = 0 = det A det B

  25. Continued.. • Case 2: A is invertible: • A is a product of elementary matrices so • det A = det(Ek…E2E1) = det Ek …det E2 det E1 (by L 1) • det(AB) = det (Ek … E2 E1 B) • = det Ek … det E2 det E1 det B (by L 1) • = det A det B 

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