3 2 determinants mtx inverses
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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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3.2 Determinants; Mtx Inverses

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3 2 determinants mtx inverses

3.2 Determinants; Mtx Inverses


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:


For a 2x2

For a (2x2)


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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