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
Select 2x2 mtx of variables and multiply to show that could not possibly get I

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:

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
In 1, add col. 2 to dol 3 to get another 0
In 2, factor out common factor across row, swap rows, or add multiples etc.

20. Examples Solve the following using Cramer’s rule.
Either use shortcut, or row reduce to create zeros and then factor out common factor.

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
Column reduce, then factor out factors in col. 2 and 3

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
Column reduce, then factor out factors in col. 2 and 3

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
Column reduce, then factor out factors in col. 2 and 3

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
Column reduce, then factor out factors in col. 2 and 3

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 
Column reduce, then factor out factors in col. 2 and 3