001 ppt Matrix Algebra Analysis and uses.ppt

Detailed study of Matrix

2. If Aand Bare both m×n matrices then the sumof Aand B,
denoted A+ B, is a matrix obtained by adding corresponding
elements of Aand B. 








310
221
A 








412
403
B 






2
BA
If Aand Bare both m×n matrices then the sumof Aand B,
denoted A+ B, is a matrix obtained by adding corresponding
elementsof Aand B.
add these








310
221
A 








412
403
B 




 

22
BA
add these








310
221
A 








412
403
B 




 

622
BA
add these








310
221
A 








412
403
B 




 

2
622
BA add these








310
221
A 








412
403
B 




 

02
622
BA
add these








310
221
A 








412
403
B 








102
622
BA
add these

3. ABBA  CBACBA  )()(

4. If Ais an m×n matrix and sis a scalar, then we let kA denote the
matrix obtained by multiplying every element ofA by k. This
procedure is called scalar multiplication.

 
k hA kh A
k h A kA hA
k A B kA kB

  
   








310
221
A 



















930
663
331303
232313
3A
PROPERTIES OF SCALAR MULTIPLICATION

5. The m×nzero matrix, denoted 0, is the m×n
matrix whose elements are all zeros.00
0)(
0



A
AA
AA 





00
00  000
2 ×2
1 ×3

6. The multiplication of matrices is easier shown than put
into words. You multiply the rows of the first matrix
with the columns of the second adding products








140
123
A 











13
31
42
B
Find AB
First we multiply across the first row and down the
first column adding products. We put the answer in
the first row, first column of the answer.23 1223  5311223 

7. 








140
123
A 











13
31
42
B Find AB
We multiplied across first row and down first column
so we put the answer in the first row, first column.






5
AB
Now we multiply across the first row and down the second
column and we’ll put the answer in the first row, second
column.43 3243  7113243  






75
AB
Now we multiply across the secondrow and down the first
column and we’ll put the answer in the second row, first
column.20 1420  1311420  







1
75
AB
Now we multiply across the secondrow and down the
second column and we’ll put the answer in the second row,
second column.40 3440 11113440  







111
75
AB
Notice the sizes of Aand Band the size of the product AB.

8. To multiply matrices Aand B
look at their dimensionspnnm 
MUST BE SAME
SIZE OF PRODUCT
If the number of columns of Adoes not
equal the number of rows of Bthen the
product ABis undefined.

9. 










6
BA 










126
BA 








 

2126
BA 











3
2126
BA 











 143
2126
BA 











 4143
2126
BA 













9
4143
2126
BA 













109
4143
2126
BA 













4109
4143
2126
BA Now let’s look at the product BA.











13
31
42
B 








140
123
A BAAB
23
32
across first row as
we go down first
column:60432 
across first row as
we go down
second column:124422 
across first row as
we go down third
column:21412 
across second row
as we go down
first column:30331 
across second row
as we go down
second column:144321 
across second row
as we go down
third column:41311 
across third row
as we go down
first column:90133 
across third row
as we go down
second column:104123 
across third row
as we go down
third column:41113 
Completely different than AB!
Commuter's Beware!

10. 
 
  BCACCBA
ACABCBA
CABBCA


 PROPERTIES OF MATRIX
MULTIPLICATIONBAAB
Is it possible for AB = BA ?,yes it is possible.

11. an nnmatrix with ones on the main diagonal
and zeros elsewhere










100
010
001
3
I
What is AI?
What is IA?












322
510
212
A 










100
010
001
3
I A












322
510
212 A












322
510
212
Multiplying a
matrix by the
identity gives the
matrix back again.

12. 






















10
01
24
13 ?
Let Abe an nnmatrix. If there exists a matrix B
such that AB = BA= I then we call this matrix the
inverseof Aand denoteit A
-1
.










2
3
2
2
1
1 























10
01
24
13
2
3
2
2
1
1
Can we find a matrix to multiply the first matrix by to
get the identity?

13. If Ahas an inverse we say that Ais nonsingular.
If A
-1
does not exist we say Ais singular.
To find the inverse of a matrix we put the matrix A, a
line and then the identity matrix. We then perform row
operations on matrix A to turn it into the identity. We
carry the row operations across and the right hand side
will turn into the inverse.
To find the inverse of a matrix we put the matrix A, a
line and then the identity matrix. We then perform row
operations on matrix A to turn it into the identity. We
carry the row operations across and the right hand side
will turn into the inverse.







72
31
A 





 1210
0131
2r
1+r
2





 1072
0131 





 1210
0131
r
2





 1210
3701
r
1 r
2

14. 







72
31
A 








12
37
1
A Check this answer by multiplying. We should
get the identity matrix if we’ve found the
inverse.







10
01
1
AA

15. We can use A
-1
to solve a system of equations352
13


yx
yx bxA
To see how, we can re-write a
system of equations as matrices.
coefficient
matrix
variable
matrix
constant
matrix





52
31 





y
x 






3
1

16. bx
1
A bx
11 
AAA bxA left multiply both sides
by the inverse of A
This is just the identitybx
1
AI
but the identity times a
matrix just gives us
back the matrix so we
have:
This then gives us a formula
for finding the variable
matrix: MultiplyAinverse
by the constants.

17. 352
13


yx
yx 






52
31
A find the inverse





1052
0131 





 1210
0131
-2r
1+r
2





1210
0131
-r
2







1210
3501
r
1-3r
2























1
4
3
1
12
35
1
bA
This is the
answer to
the system
x
y