Download 2.2 Matrix Multiplication - La Jolla Country Day School

2.3 Matrix Multiplication
Background Example
• We return to our Sweatshirt store example:
– We wish to find the value of the inventory by size
– Smalls - $10, Med - $11, Large - $12, XL - $13
...s.........m........l........xl
100 200 300 400


200 300 100 200


100 250 100 200


100 40
20 100 


• How would we set up the mult. to do this?
Matrix Multiplication Algorithm
• The mtx mult algorithm is defined to do just that:
...s.........m........l........xl
100 200 300 40010 10(100) 11(200) 12(300)13(400)

  

200 300 100 20011 10(200)11(300) 12(100) 13(200)

  

100 250 100 20012 10(100) 11(250)12(100) 13(250) 

  

100 40
  

20 100 

13  10(100) 11(40) 12(20) 13(100) 
• (i,j): multiply entries in row i of first mtx by the
corresponding entries in col j of second mtx, and then add
terms.
- Note that this is the dot product of row i and col j
A few things to note
• Given the way that the algorithm is defined, what must
be true about the dimensions of the matrices in order for
multiplication to work?
– # of entries in each row of first matrix must equal # of
entries in each column of second matrix
– (i.e. number of columns of first matrix must equal the
number of rows of second matrix)
– So if multiplying matrices of following order: (a x b)
x (c x d), b = c
• Note: order of solution mtx is:
axd
Example
• Given matrices A and B, find the results of the
following if possible:
– A2
1 1
2 
– B2

, B   
A

– AB
2

3
3


 
– BA
Example#2
• Given the matrices A and B, find AB and BA:
1 1 0 
1 0 0 




A  2 3 1 , B  0 1 0 




2 1 1
0 0 1 
•
B is the identity matrix, I3, since AB = BA = A
• The identity is always a square matrix with 1’s on the
main diagonal and 0’s elsewhere.
•
I is only the identity for a matrix of the same size.
Properties of Mtx Multiplication
1
2
3
4
5
6
IA = A, BI = B
A(BC) = (AB)C
A(B+C) = AB + AC, A(B-C) = AB - AC
(B+C)A = BA + CA, (B - C)A = BA - CA
k(AB) = (kA)B = A(kB)
(AB)T = BTAT
- Note: commutativity does not hold (AB ≠ BA in
most cases).
- Therefore the order of the factors in a product of
matrices makes a difference!
Helpful Notation
• To help us prove the properties, it is useful to understand
the following notation for matrix multiplication.
• A = [aij] is m x n, B = [bij] is n x p
b1j 
 
• ith row of A is [ ai1 ai2 …… ain]
b2 j 

• jth column of B is: ---------------------> 
...
 
b 
 nj 
• ij entry of prod mtx is dot prod of rown i of A and col j
of B: a b  a b ...  a b 
ab
i1 1 j
i2 2 j
in nj

k 1
ik kj
Proving Property 3 (also in book)
• Property 3: A(B + C) = AB + AC
• A is m x n, B is n x p, C is n x p
• B + C = [bij + cij]
n
n
n
n
k 1
k 1
k 1
k 1
A(B  C)   aik (bkj  ckj )  (aik bkj  aik ckj )   aik bkj   aik ckj
• This is just the (i,j) entry of AB + the (i,j) entry of AC
• Therefore A(B + C) = AB + AC
•
Matrix form of linear system
• Try to write the following system as a single matrix
equation: 2x  3x  x  6
1
2
3
x1  2x2  x3  4
x 
2 3 1  1  6

x2   
1 2 1  4
x 3 
AX  B
• A is coefficient mtx, X is solution mtx, B is
constant mtx
Precursor to Theorem 2
• If X1 is a sol’n to AX=B and X0 is a sol’n to the related
homogeneous system AX = 0, then:
– X1 + X0 is a solution to AX = B:
A(X1  X0 )  AX1  AX0  B  0  B
• Theorem 2 is a converse of this.
Theorem 2
• If X1 is a sol’n to AX=B, then every solution, X2, to
AX=B is of the form:
– X2 = X1 + X0 where X0 is a solution to the associated
homogeneous system AX=0.
Proof of Theorem 2
• If X2 and X1 are both sol’ns to AX=B,
– So AX1 = B and AX2 = B:
– say X0 = X2 - X1 so X2 = X0+X1
– AX0 = A(X2 - X1) = AX2 - AX1= B - B = 0
– Therefore, X0 is a solution to the associated
homogeous system AX = 0. •
Example
• Express every solution to the following system as the
sum of a specific solution plus a solution to the
associated homogeneous system.
x  2y  4z  10
2x  y  2z  5
x  y  2z  7
Solution
1

0

0
x   4  4 0
0 0 4 

  
    
1 2 3  so...X  y  2t  3 3 t 2 
  
    

x   t  0 1 
0 0 0 
• If we take t=0, we get a specific solution, X1
• Therefore, if t≠0, the solution, X2 = X1 + X0 where X0 is
a solution to the associated homogeneous stm: AX = 0
0 
• So,  
gives all sol’ns to assoc. hom. System
t 2  X0
• (Show)  
1 
Example
1 2
1 1 2 0 
1 




A  0 1 2,...,B  1 3 2 1 




1 1
2 
0 2
0 1 
• Find row 3 and column 2 of AB.
• In many situations, it is helpful to write a matrix as a
column of rows or as a row of columns:
R 
 1 
A  R2 ,...B  C1 C2 C3 C4 
 
R3 
Simplifying Notation
• So then we can use the following notation in mtx mult:
R1 
R1 X 
 


R2 
R2 X 
AX   X  

... 
 ... 




R
R
X
 m 
 m 
A further simplification
• We can partition a matrix into smaller blocks:
1

0
A  
1

0
0
1
0 0
0 0
2
2
3 1
1 2
0 

0  I2 O23 

 
1 X Y 
2 3 
3 



4

B  0
2


2
1 
 W 
1   
Z 

1 
3

Continuing the Example
I2
AB  
X
2
3 


 4
O23 W   W
1 
  
 

Y Z  XW  YZ  6 12 


10
4


• We need to make sure that the way we partition the
matrices allows us to multiply matrices which match up
appropriately by dimension. (show)
Another Block Multiplication Ex.
• Go through ex. 11 with finding powers of mtx:
1 1 0 

 X
A  1 1
0  

 Y
1 1 2
• Find A8
O

Z 
Example continued
X OX
2

A  
Y Z Y
O  X 2
 
Z  
YX  ZY
0  X 2
 
Z 2 
 
0
• This was convenient since we have a 0 matrix
2

X
A 4  (A2 ) 2  

 0
0 X 2

2
Z 

 0
0  X 4
  
2
Z 
 
 0
0 

4
Z 

X 4
A  (A )  

 0
0 X 4

4
Z 

 0
0  X 8
  
4
Z 
 
 0
0 

8
Z 

8
4 2
• This was convenient since we had a diagonal matrix
0 

Z 2 

Finally...
Z8  2 8  256
X 2  2X
so...X 4  (2X) 2  4X 2  4(2 X)  8X
so...X 8  (8X) 2  64X 2  64(2X)  128X
128X
A  
 0
8
1 1 0 
X 0 
0 


 128
 1281 1 0 
256
0 2 


0 0 2 
Other topics in homework
• Trace: sum of elements on main diagonal of square mtx
• Idempotent: square mtx,P, is idempotent if P2 = P