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3.2 Determinants; Mtx InversesPowerPoint Presentation

3.2 Determinants; Mtx Inverses

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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(A1A2…Ak) = detA1detA2…detAk
- also: det (Ak) = (det A)k

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)

- <== 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

- Find all values of b for which A will have an inverse.

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)

- 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

- 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 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

- 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 (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

- 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

Linear Equations

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

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

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

Examples

- Solve the following using Cramer’s rule.

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...

- 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

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)

- 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

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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