IXL - Properties of matrices (Algebra 2 practice)

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or

Learn with an example

question

If

T

=

–

1

–

3

–

6

1

and

U

=

7

–

3

–

2

,

is the following statement true or false?

T

U

=

U

T

true

false

key idea

A matrix product exists if and only if the number of columns in the first matrix equals the number of rows in the second matrix. If

A

has dimensions

m

×

n

and

W

has dimensions

n

×

p

,

then

A

W

has dimensions

m

×

p

.

A

W

=

A

W

(

m

×

n

)

(

n

×

p

)

m

×

p

Two matrices are equal if and only if their corresponding elements are equal.

solution

To see if the statement is true or false, find

T

U

and

U

T

.

Since

T

=

–

1

–

3

–

6

1

and

U

=

7

–

3

–

2

are square matrices with the same number of rows and columns, you can multiply them in either order.

T

U

and

U

T

will also be square matrices with 2 rows and 2 columns.

Since both matrix products exist, you can calculate them.

Calculate

T

U

.

Find the element in the

1

^st

row and

1

^st

column of

T

U

.

Multiply each element in the

1

^st

row of

T

by the corresponding element in the

1

^st

column of

U

.

Then, add the products.

T

U

=

–

1

–

3

–

6

1

7

–

3

–

2

=

–

1

(

7

)

+

(

–

3

)

(

–

3

)

?

=

–

7

+

9

?

=

2

?

Apply a similar process to find the element in the

1

^st

row and

2

^nd

column of

T

U

.

–

1

–

3

–

6

1

7

–

3

–

2

=

2

–

1

(

7

)

+

(

–

3

)

(

–

2

)

?

=

2

–

7

+

6

?

=

2

–

1

?

Repeat this process to calculate the remaining elements of

T

U

.

–

1

–

3

–

6

1

7

–

3

–

2

=

2

–

1

–

45

–

44

Calculate

U

T

.

Find the element in the

1

^st

row and

1

^st

column of

U

T

.

Multiply each element in the

1

^st

row of

U

by the corresponding element in the

1

^st

column of

T

.

Then, add the products.

U

T

=

7

–

3

–

2

–

1

–

3

–

6

1

=

7

(

–

1

)

+

7

(

–

6

)

?

=

–

7

–

42

?

=

–

49

?

Apply a similar process to find the element in the

1

^st

row and

2

^nd

column of

U

T

.

7

–

3

–

2

–

1

–

3

–

6

1

=

–

49

7

(

–

3

)

+

7

(

1

)

?

=

–

49

–

21

+

7

?

=

–

49

–

14

?

Repeat this process to calculate the remaining elements of

U

T

.

7

–

3

–

2

–

1

–

3

–

6

1

=

–

49

–

14

15

7

So,

T

U

=

2

–

1

–

45

–

44

and

U

T

=

–

49

–

14

15

7

.

Compare

T

U

and

U

T

.

Since

T

U

≠

U

T

,

the statement is false. In general, matrices do not satisfy the commutative property of multiplication. It is rare to find two matrices that commute.

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