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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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theorem 1 product theorem
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
Extension
  • Using induction, we could show that:
    • det(A1A2…Ak) = detA1detA2…detAk
    • also: det (Ak) = (det A)k
theorem 2
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
proof continued
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.
example
Example
  • Find all values of b for which A will have an inverse.
theorem 3
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
theorem 3 continued
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 
examples
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
adjoint
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.
adjoint definition
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:
example1
Example
  • Find the adjoint of A:
  • So we could find det(A)
for nxn
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)
theorem 4 adjoint formula
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
example2
Example
  • Use thm 4 to find the values of c which make A invertible:

c ≠ 0

linear equations
Linear Equations
  • Recall that if AX = B, and if A is invertible (det A ≠ 0), then
  • X = A-1B So...
finding determinants is easier
Finding determinants is easier
  • the right part is just the det of a matrix formed by replacing column i with the B column matrix
theorem 5 cramer s rule
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

examples1
Examples
  • Solve the following using Cramer’s rule.
proof of theorem 1
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
continued
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
continued1
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
proving theorem 1 finally
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
continued2
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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