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(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,

det(A1 ) 1

det A

• Proof: ==> given A invertible, AA-1=I

• det (AA-1)=detI=1=detAdetA-1

• and det A 1 1

det A

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.

A1 2 6b

0 b 2 2

0 4b 12 b 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 A T 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= E1

TE2T…Ek

T

• So det AT = det E1T det E2

T …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:

Aa b

c d

, ...,adj(A)

d b c a

• Recall that: A 1

1

det Aadj(A)

• 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:– A 1

1

det Aadj(A)

For a (2x2)

Aa b

c d

[Cij(A)] d c b a

[Cij(A)]T d b c a

Example• Find the adjoint of A:

A1 1 2

3 1 0

0 1 1

adj(A) 1 1 2

3 1 6

3 1 4

A 1 1

det(A)adj(A)

(det A)I A(adj(A))

• So we could find det(A)

For (nxn)

• A(adjA) = (detA)I for any (nxn): ex. (3 x 3)

A(adj(A)) a11 a12 a13

a21 a22 a23

a31 a32 a33

C11 C21 C31

C12 C22 C32

C13 C23 C33

det A 0 0

0 det A 0

0 0 det A

• 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,

A 1 1

det Aadj(A)

• Good theory, but not a great way to find A-1

Example

• Use thm 4 to find the values of c which make A invertible:

0 c c 1 2 1

c c c

c ≠ 0

Linear Equations

• Recall that if AX = B, and if A is invertible (det A ≠ 0), then

X = A-1B So...

X 1

det A(adj(A))B

x1

x2

...

xn

1

det A

C11(A) C21(A) ... Cn1(A)

C12(A) ... ... ...

... ... ... ...

C1n(A) ... ... Cnn(A)

b1

b2

...

bn

Finding determinants is easier

• the right part is just the det of a matrix formed by replacing column i with the B column matrix

xi 1

det Ab1C1i(A) b2C2i(A) ... bnCni(A)

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:

xi det Aidet A

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

5x y z 7

2x y 2z 6

3x 2z 7

• 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 EEkk … det E … det E22 det E det E11 det A= 0 det A= 0

since det E’s never 0, 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 sodet 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 �