1563 matrix algebra
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Transcript of 1563 matrix algebra
If A and B are both m × n matrices then the sum of A and B, denoted A + B, is a matrix obtained by adding corresponding elements of A and B.
310221
A
412403
B
2BA
If A and B are both m × n matrices then the sum of A and B, denoted A + B, is a matrix obtained by adding corresponding elements of A and B.
add these
310221
A
412403
B
22BA
add these
310221
A
412403
B
622BA
add these
310221
A
412403
B
2622
BA
add these
310221
A
412403
B
02622
BA
add these
310221
A
412403
B
102622
BA
add these
ABBA
CBACBA )()(
If A is an m × n matrix and s is a scalar, then we let kA denote the matrix obtained by multiplying every element of A by k. This procedure is called scalar multiplication.
k hA kh A
k h A kA hA
k A B kA kB
310221
A
930663
331303232313
3A
PROPERTIES OF SCALAR MULTIPLICATION
The m × n zero matrix, denoted 0, is the m × n matrix whose elements are all zeros.
000)(
0
AAA
AA
0000 000
2 × 21 × 3
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
140123
A
133142
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
140123
A
133142
B
Find AB
We multiplied across first row and down first column so we put the answer in the first row, first column.
5AB
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
75AB
Now we multiply across the second row and down the first column and we’ll put the answer in the second row, first column.
20 1420 1311420
1
75AB
Now we multiply across the second row and down the second column and we’ll put the answer in the second row, second column.
40 3440 11113440
11175
AB
Notice the sizes of A and B and the size of the product AB.
To multiply matrices A and B look at their dimensions
pnnm MUST BE SAME
SIZE OF PRODUCT
If the number of columns of A does not equal the number of rows of B then the
product AB is undefined.
6BA
126BA
2126BA
3
2126BA
143
2126BA
41432126
BA
9
41432126
BA
109
41432126
BA
410941432126
BA
Now let’s look at the product BA.
133142
B
140123
A
BAAB
2332can multiply
size of answeracross 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!
BCACCBAACABCBA
CABBCA
PROPERTIES OF MATRIX MULTIPLICATION
BAAB Is it possible for AB = BA ?,yes it is possible.
an n n matrix with ones on the main diagonal and zeros elsewhere
100010001
3I
What is AI?
What is IA?
322510212
A
100010001
3I
A
322510212
A
322510212
Multiplying a matrix by the
identity gives the matrix back again.
1001
2413
?
Let A be an n n matrix. If there exists a matrix B such that AB = BA = I then we call this matrix the inverse of A and denote it A-1.
232211
1001
2413
232211
Can we find a matrix to multiply the first matrix by to get the identity?
If A has an inverse we say that A is nonsingular. If A-1 does not exist we say A is 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
31A
1210
01312r1+r2
1072
0131
12100131
r2
12103701r1 r2
72
31A
12371A
Check this answer by multiplying. We should get the identity matrix if we’ve found the inverse.
10011AA
We can use A-1 to solve a system of equations
35213
yx
yx
bx A
To see how, we can re-write a system of equations as matrices.
coefficient matrix
variable matrix
constant matrix
5231
yx
31
bx 1A
bx 11 AAA
bx A left multiply both sides by the inverse of A
This is just the identity
bx 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: Multiply A inverse by the constants.
35213
yx
yx
5231
A find the inverse
10520131
1210
0131-2r1+r2
12100131
-r2
1210
3501r1-3r2
14
31
12351bA
This is the answer to the system
xy
Your calculator can compute inverses and determinants of matrices. To find out how, refer to the manual or click here to check out the website.
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.
Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au