Transcript 2.3 Matrix Inverses - San Diego Private School | La Jolla

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,
1
1
det (A ) 
det A
• Proof: ==> given A invertible, AA-1=I
• det (AA-1)=detI=1=detAdetA-1
• and
1
det A 
det A
1
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.
1
2
6b 


A  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 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.
det A if A is orthogonal.
• I = AA-1 = AAT
• 1 = det I = det A det AT = (det A)2
• So det A = ± 1
Find
Adjoint
• Adjoint of a (2x2) is just the right part of inverse:
a b
d b
, ...,adj(A)  

A  
c d 
c a 
• Recall that:
1
A 
adj(A)
det A
1
• 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:
–
1
A 
adj(A)
det A
1
For a (2x2)
a b
A  

c d 
d c

[Cij (A)] 
b a 
d b
T

[Cij (A)]  
c a 
Example
• Find the adjoint of A:
1 1 2 
1 1 2




A  3 1 0  adj(A)  3 1
6 




0 1 1 
3 1 4 
1
A 
adj( A)
det (A)
(det A)I  A(adj(A))
1
• So we could find det(A)
For (nxn)
• A(adjA) = (detA)I
a11

A(adj( A))  a21

a31
a12
a22
a32
for any (nxn): ex. (3 x 3)
a13 C11

a23 C12

a33 C13
C21
C22
C23
C31 

C32 

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,
1
A 
adj(A)
det A
1
• 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...
1
X
(adj(A))B
det A
x1 
C11 (A) C21 (A)
 

...
x2 
1 C12 ( A)
 

...
... det A  ...


...
xn 

C1n ( A)
... Cn1 ( A)b1 
 
...
... b2 
 
...
... ...
... Cnn (A)

bn 

Finding determinants is easier
1
xi 
b1C1i (A)  b2C2i (A) ...  bnCni (A)

det A
• 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:
det A
xi 
i
det 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
• Solve the following using Cramer’s rule.
5x  y  z  7
2x  y  2z  6
3x  2z  7
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 •