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