The last exercise (exercise 4), says that 0(A+B) and (A+B)0 give us 0. But, both final results (the two 0) won't have the same dimensions right? Because it is not commutative, so: 0(A+B) is not equal to (A+B)0 Am I right? I hope my question was clear

It was assumed that all A, B and 0 are nxn, therefore 0(A+B)=(A+B)0.

Hi, everyone. In the second challenge question, which is given as follows: 2) Which of the following expressions are equivalent to A(B+C)? Select all that apply. Selecting i): AB+AC, & ii):A(C+B) will mark the answers right, but selecting " (B+C)A " is wrong? I think it should be right as well, that is, there are overall three answers. My logic is this that, first we should add 'B' and 'C' and then multiply it by A. We did so in A(C+B), since addition is commutative, ( C+B=B+C ). So, therefore A(C+B)=(B+C)A, right? I often do this in my Maths book: (x+a)(x+b) =x(x+b)+a(x+b) =x^2+xb+xa+ab =x^2+(b+a)x+ab Thanks for reading and you time.

We are multiplying Matrices, not scalars. Matrix multiplication is NOT commutative. If A and B are matrices such that AB and BA are defined (can be multiplied) AB≠BA. Check the very first property listed here: https://www.khanacademy.org/math/precalculus/precalc-matrices/properties-of-matrix-multiplication/a/properties-of-matrix-multiplication

I still don't get the whole point in making a matrix full of zeros. Isn't it it redundant? Shouldn't the best and easiest way to multiply a matrix to get 0, be to just use the scalar quantity 0 rather than a matrix full of zeros?

Using the Zero matrix has a lot of use in computing and allows us to compare matrices to algabraic rules

in the following question which is Which of the following expressions are equivalent to I2 (AB) Option AB and (AB) I2 were correct i get why AB is correct, however, i m a bit doubtful about the second option for instance if I 2 is a 2 * 2 matrix and A is 2*3 while B is 3*4 well then AB would be 2*4 so I2 ( AB) would be defined but (AB) I2 wouldnt be possible. BTW i dont know what I2 really means but what i have understood after goin through the lecture is that I 2 means identity matrix with 2 rows and 2 columns

Above all the questions there is a note stating that *For the problems below, let A, B, and C be 2×2 matrices and let O be the 2×2 zero matrix*

what is the union and intersection of two matrices?

Union and intersection are defined on sets, not on matrices.

in Q2 of "check your understanding it says: _Which of the following expressions are equivalent to A(B+C)?_ Why is (B+C)A wrong?

Because it is matrix multipliation and you are multiplying rows with columns. Because of that, changing the order changes which numbers get multiplied. Try it out yourself. Take two 2x2 matrices like: ``` [ 1 2 ] [ 5 6 ] [ 3 4 ] [ 7 8 ] ``` And do the dot product, then swap them and do the dot product.

Main content

Properties of matrix multiplication

Google Classroom

Learn about the properties of matrix multiplication (like the distributive property) and how they relate to real number multiplication.

Properties of matrix multiplication

In this table,

A

B

, and

C

are

n \times n

matrices,

I

is the

n \times n

identity matrix, and

O

is the

n \times n

zero matrix

Property	Example
The commutative property of multiplication $does not hold!$ ‍	$A B \neq B A$ ‍
Associative property of multiplication	$(A B) C = A (B C)$ ‍
Distributive properties	$A (B + C) = A B + A C$ ‍
	$(B + C) A = B A + C A$ ‍
Multiplicative identity property	$I A = A$ ‍ and $A I = A$ ‍
Multiplicative property of zero	$O A = O$ ‍ and $A O = O$ ‍
Dimension property	The product of an $m \times n$ ‍ matrix and an $n \times k$ ‍ matrix is an $m \times k$ ‍ matrix.

Let's take a look at matrix multiplication and explore these properties.

What you should be familiar with before taking this lesson

In matrix multiplication, each entry in the product matrix is the dot product of a row in the first matrix and a column in the second matrix.

If this is new to you, we recommend that you check out our matrix multiplication article.

Here are other relevant articles:

Matrix multiplication is not commutative

One of the biggest differences between real number multiplication and matrix multiplication is that matrix multiplication is not commutative.

In other words, in matrix multiplication, the order in which two matrices are multiplied matters!

See for yourselves!

Let's take a look at a concrete example with the following matrices.

A = [\begin{array}{rr} 3 & 4 \\ 1 & 2 \end{array}]

B = [\begin{array}{rr} 6 & 2 \\ 3 & 2 \end{array}]

1) Find $A B$ ‍ and $B A$ ‍.

$A B =$ ‍	$B A =$ ‍

Finding $A B$ ‍	Finding $B A$ ‍
Let's label the rows of matrix $A$ ‍ and the columns of matrix $B$ ‍ so that it is easier to express $A B$ ‍.	Let's label the rows of matrix $B$ ‍ and the columns of matrix $A$ ‍ so that it is easier to express $B A$ ‍.

We can find $A B$ ‍ as follows:	We can find $B A$ ‍ as follows:
$\begin{aligned} A B & = [\begin{array}{rr} \vec{a_{1}} \cdot \vec{b_{1}} & \vec{a_{1}} \cdot \vec{b_{2}} \\ \vec{a_{2}} \cdot \vec{b_{1}} & \vec{a_{2}} \cdot \vec{b_{2}} \end{array}] \\ = [\begin{array}{rr} 3 \cdot 6 + 4 \cdot 3 & 3 \cdot 2 + 4 \cdot 2 \\ 1 \cdot 6 + 2 \cdot 3 & 1 \cdot 2 + 2 \cdot 2 \end{array}] \\ = [\begin{array}{rr} 30 & 14 \\ 12 & 6 \end{array}] \end{aligned}$ ‍	$\begin{aligned} B A & = [\begin{array}{rr} \vec{b_{1}} \cdot \vec{a_{1}} & \vec{b_{1}} \cdot \vec{a_{2}} \\ \vec{b_{2}} \cdot \vec{a_{1}} & \vec{b_{2}} \cdot \vec{a_{2}} \end{array}] \\ = [\begin{array}{rr} 6 \cdot 3 + 2 \cdot 1 & 6 \cdot 4 + 2 \cdot 2 \\ 3 \cdot 3 + 2 \cdot 1 & 3 \cdot 4 + 2 \cdot 2 \end{array}] \\ = [\begin{array}{rr} 20 & 28 \\ 11 & 16 \end{array}] \end{aligned}$ ‍

Notice that the products are not the same! Since

A B \neq B A

, matrix multiplication is not commutative!

Other than this major difference, however, the properties of matrix multiplication are mostly similar to the properties of real number multiplication.

Associative property of multiplication: $(A B) C = A (B C)$ ‍

This property states that you can change the grouping surrounding matrix multiplication.

For example, you can multiply matrix

A

by matrix

B

, and then multiply the result by matrix

C

, or you can multiply matrix

B

by matrix

C

, and then multiply the result by matrix

A

When using this property, be sure to pay attention to the order in which the matrices are multiplied, since we know that the commutative property does not hold for matrix multiplication!

Distributive properties

We can distribute matrices in much the same way we distribute real numbers.

$A (B + C) = A B + A C$ ‍
$(B + C) A = B A + C A$ ‍

If a matrix

A

is distributed from the left side, be sure that each product in the resulting sum has

A

on the left! Similarly, if a matrix

A

is distributed from the right side, be sure that each product in the resulting sum has

A

on the right!

Let's illustrate these properties using the matrices

A = [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}]

B = [\begin{array}{rr} 3 & 1 \\ 0 & 1 \end{array}]

, and

C = [\begin{array}{rr} 1 & 4 \\ 3 & 2 \end{array}]

$\begin{aligned} A (B + C) & = [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] ([\begin{array}{rr} 3 & 1 \\ 0 & 1 \end{array}] + [\begin{array}{rr} 1 & 4 \\ 3 & 2 \end{array}]) \\ = [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \cdot [\begin{array}{rr} 4 & 5 \\ 3 & 3 \end{array}] \\ = [\begin{array}{rr} 11 & 13 \\ 18 & 21 \end{array}] \\ A B + A C & = [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \cdot [\begin{array}{rr} 3 & 1 \\ 0 & 1 \end{array}] + [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \cdot [\begin{array}{rr} 1 & 4 \\ 3 & 2 \end{array}] \\ = [\begin{array}{rr} 6 & 3 \\ 9 & 5 \end{array}] + [\begin{array}{rr} 5 & 10 \\ 9 & 16 \end{array}] \\ = [\begin{array}{rr} 11 & 13 \\ 18 & 21 \end{array}] \end{aligned}$ ‍

Notice that the two matrices are the same.

Let's also find

(B + C) A

and

B A + C A

$\begin{aligned} (B + C) A & = ([\begin{array}{rr} 3 & 1 \\ 0 & 1 \end{array}] + [\begin{array}{rr} 1 & 4 \\ 3 & 2 \end{array}]) [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \\ = [\begin{array}{rr} 4 & 5 \\ 3 & 3 \end{array}] \cdot [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \\ = [\begin{array}{rr} 23 & 14 \\ 15 & 9 \end{array}] \end{aligned}$ ‍

$\begin{aligned} B A + C A & = [\begin{array}{rr} 3 & 1 \\ 0 & 1 \end{array}] \cdot [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] + [\begin{array}{rr} 1 & 4 \\ 3 & 2 \end{array}] \cdot [\begin{array}{rr} 2 & 1 \\ 3 & 2 \end{array}] \\ = [\begin{array}{rr} 9 & 5 \\ 3 & 2 \end{array}] + [\begin{array}{rr} 14 & 9 \\ 12 & 7 \end{array}] \\ = [\begin{array}{rr} 23 & 14 \\ 15 & 9 \end{array}] \end{aligned}$ ‍

Again, notice that the two matrices are equivalent.

These examples also show us that

A (B + C) \neq (B + C) A

and that

A B + A C \neq B A + C A

. This reminds us that the commutative property of matrix multiplication is not true. We must take extra care with the ordering of matrix expressions!

Multiplicative identity property

The

n \times n

identity matrix, denoted

I_{n}

, is a matrix with

n

rows and

n

columns. The entries on the diagonal from the upper left to the bottom right are all

1

's, and all other entries are

0

For example:

I_{2} = [\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}] I_{3} = [\begin{array}{rr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] I_{4} = [\begin{array}{rr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]

The multiplicative identity property states that the product of any

n \times n

matrix

A

and

I_{n}

is always

A

, regardless of the order in which the multiplication was performed. In other words,

A \cdot I = I \cdot A = A

The role that the

n \times n

identity matrix plays in matrix multiplication is similar to the role that the number

1

plays in the real number system. If

a

is a real number, then we know that

a \cdot 1 = a

and

1 \cdot a = a

Multiplicative property of zero

A zero matrix is a matrix in which all of the entries are

0

. For example, the

3 \times 3

zero matrix is

O_{3 \times 3} = [\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

A zero matrix is indicated by

O

, and a subscript can be added to indicate the dimensions of the matrix if necessary.

The multiplicative property of zero states that the product of any

n \times n

matrix and the

n \times n

zero matrix is the

n \times n

zero matrix. In other words,

A \cdot O = O \cdot A = O

The role that the

n \times n

zero matrix plays in matrix multiplication is similar to the role that the number

0

plays in the real number system. If

a

is a real number, then we know that

a \cdot 0 = 0

and

0 \cdot a = 0

The dimension property

One property that is unique to matrices is the dimension property. This property has two parts:

The product of two matrices will be defined if the number of columns in the first matrix is equal to the number of rows in the second matrix.
If the product is defined, the resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.

For example, if

A

is a

3 \times 2

matrix and if

B

is a

2 \times 4

matrix, the dimension property tells us:

The product $A B$ ‍ is defined.
$A B$ ‍ will be a $3 \times 4$ ‍ matrix.

Check your understanding

Now that you are familiar with matrix multiplication and its properties, let's see if you can use them to determine equivalent matrix expressions.

For the problems below, let

A

B

, and

C

2 \times 2

matrices and let

O

be the

2 \times 2

zero matrix.

2) Which of the following expressions are equivalent to $A (B + C)$ ‍?

3) Which of the following expressions are equivalent to $I_{2} (A B)$ ‍?

4) Which of the following expressions are equivalent to $O (A + B)$ ‍?

Want to join the conversation?

Sort by:

Pedro Santos
Posted 8 years ago. Direct link to Pedro Santos's post “The last exercise (exerci...”
The last exercise (exercise 4), says that 0(A+B) and (A+B)0 give us 0.
But, both final results (the two 0) won't have the same dimensions right?
Because it is not commutative, so: 0(A+B) is not equal to (A+B)0

Am I right? I hope my question was clear
Button navigates to signup pageButton navigates to signup page
(32 votes)
Answer
- Stefen
  Posted 8 years ago. Direct link to Stefen's post “It was assumed that all A...”
  It was assumed that all A, B and 0 are nxn, therefore 0(A+B)=(A+B)0.
  Comment on Stefen's post “It was assumed that all A...”
  (46 votes)
bluefirefighter9019
Posted 8 years ago. Direct link to bluefirefighter9019's post “Hi, everyone. In the sec...”
Hi, everyone.

In the second challenge question, which is given as follows:

2) Which of the following expressions are equivalent to A(B+C)?
Select all that apply.

Selecting i): AB+AC, & ii):A(C+B) will mark the answers right, but selecting " (B+C)A " is wrong? I think it should be right as well, that is, there are overall three answers.

My logic is this that, first we should add 'B' and 'C' and then multiply it by A. We did so in A(C+B), since addition is commutative, ( C+B=B+C ).

So, therefore A(C+B)=(B+C)A, right? I often do this in my Maths book:

(x+a)(x+b)
=x(x+b)+a(x+b)
=x^2+xb+xa+ab
=x^2+(b+a)x+ab

Thanks for reading and you time.
Button navigates to signup pageComment on bluefirefighter9019's post “Hi, everyone. In the sec...”
(6 votes)
Answer
- Stefen
  Posted 8 years ago. Direct link to Stefen's post “We are multiplying Matric...”
  We are multiplying Matrices, not scalars.
  Matrix multiplication is NOT commutative. If A and B are matrices such that AB and BA are defined (can be multiplied) AB≠BA.
  
  Check the very first property listed here:
  https://www.khanacademy.org/math/precalculus/precalc-matrices/properties-of-matrix-multiplication/a/properties-of-matrix-multiplication
  Button navigates to signup page
  (20 votes)
Alex
Posted 7 years ago. Direct link to Alex's post “I still don't get the who...”
I still don't get the whole point in making a matrix full of zeros. Isn't it it redundant? Shouldn't the best and easiest way to multiply a matrix to get 0, be to just use the scalar quantity 0 rather than a matrix full of zeros?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- GaryEdwin
  Posted 6 years ago. Direct link to GaryEdwin's post “Using the Zero matrix has...”
  Using the Zero matrix has a lot of use in computing and allows us to compare matrices to algabraic rules
  Button navigates to signup page
  (2 votes)
elinamazorliade
Posted a year ago. Direct link to elinamazorliade's post “can a 4x5 multiply a 6x2”
can a 4x5 multiply a 6x2
Button navigates to signup pageComment on elinamazorliade's post “can a 4x5 multiply a 6x2”
(4 votes)
Answer
- Bilal Mohammed Ali
  Posted 10 months ago. Direct link to Bilal Mohammed Ali's post “No, and the reason for th...”
  No, and the reason for that is you are going to have an undefined matrix where you can't perform multiplication, such as Matrix 4 x 5 by Matrix 6 x 2 is undefined. So, how can we have a defined matrix? Well, a matrix is defined when the number of columns of the first matrix is equal to the number of rows of the second matrix. For example, let A, B be two matrices. Find out C = AB.
  
  Solution:
  
  A=
  | 4 0 9|,
  
  B=
  |-1 14 |
  | 2 -2 |
  |-8 5 |
  
  Now, Matrix A= 1 x 3 and B= 3 x 2, which means that we can go ahead a apply multiplication to AB to get C since Matrix C is defined by having the number of Columns in A equal to the number of Rows in B.
  Button navigates to signup page
  (1 vote)
Alizay Hayder
Posted 7 years ago. Direct link to Alizay Hayder's post “in the following question...”
in the following question which is
Which of the following expressions are equivalent to I2 (AB)
Option
AB and (AB) I2 were correct i get why AB is correct, however, i m a bit doubtful about the second option for instance if I 2 is a 2 * 2 matrix and A is 2*3 while B is 3*4 well then AB would be 2*4 so I2 ( AB) would be defined but (AB) I2 wouldnt be possible. BTW i dont know what I2 really means but what i have understood after goin through the lecture is that I 2 means identity matrix with 2 rows and 2 columns
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Sajjad Bin Samad
  Posted 6 years ago. Direct link to Sajjad Bin Samad's post “Above all the questions t...”
  Above all the questions there is a note stating that For the problems below, let A, B, and C be 2×2 matrices and let O be the 2×2 zero matrix
  Button navigates to signup page
  (2 votes)
shagullreader
Posted 3 years ago. Direct link to shagullreader's post “what is the union and int...”
what is the union and intersection of two matrices?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “Union and intersection ar...”
  Union and intersection are defined on sets, not on matrices.
  Button navigates to signup page
  (2 votes)
Icedlatte
Posted 8 years ago. Direct link to Icedlatte's post “in Q2 of "check your unde...”
in Q2 of "check your understanding it says:
Which of the following expressions are equivalent to A(B+C)?

Why is (B+C)A wrong?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Robert Stone
  Posted 8 years ago. Direct link to Robert Stone's post “Because it is matrix mult...”
  Because it is matrix multipliation and you are multiplying rows with columns. Because of that, changing the order changes which numbers get multiplied. Try it out yourself. Take two 2x2 matrices like:
  [ 1 2 ] [ 5 6 ] [ 3 4 ] [ 7 8 ]
  
  And do the dot product, then swap them and do the dot product.
  Button navigates to signup page
  (4 votes)
Cameron Milinkovic
Posted 6 years ago. Direct link to Cameron Milinkovic's post “In question 2(d), is (B +...”
In question 2(d), is (B + C)A wrong because it would end up being BA + CA? I'm used to being able to switch around the order of scalars. The side on which it A is matters even when the terms its multiplying into are in parentheses?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- ajlee2006
  Posted 5 years ago. Direct link to ajlee2006's post “Yes, it will become `BA +...”
  Yes, it will become BA + CA. Remember, matrix multiplication is not commutative.
  Button navigates to signup page
  (1 vote)
John He
Posted 7 years ago. Direct link to John He's post “I remember when a real nu...”
I remember when a real number times its inverse,will get 1.How to find inverse of a matrix?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Akshat Sanghvi
  Posted 6 years ago. Direct link to Akshat Sanghvi's post “Hello! This is Akshat of ...”
  Hello! This is Akshat of Class 9! Check this link - https://www.khanacademy.org/math/precalculus/precalc-matrices/intro-to-matrix-inverses/v/inverse-matrix-part-1
  Button navigates to signup page
  (1 vote)
Max Duan
Posted 8 years ago. Direct link to Max Duan's post “All of the zero matrices ...”
All of the zero matrices and identity matrices I've seen here are n times n matrices. Is there any m times n zero matrix or m times n identity matrix ?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- kiwimaniac2014
  Posted 8 years ago. Direct link to kiwimaniac2014's post “An identity matrix would ...”
  An identity matrix would seem like it would have to be square. That is the only way to always have 1's on a diagonal- which is absolutely essential. However, a zero matrix could me mxn. Say you have O which is a 3x2 matrix, and multiply it times A, a 2x3 matrix. That is defined, and would give you a 3x3 O matrix.
  Button navigates to signup page
  (1 vote)

Precalculus

Course: Precalculus > Unit 7

Properties of matrix multiplication

What you should be familiar with before taking this lesson

Matrix multiplication is not commutative

See for yourselves!

Associative property of multiplication: (AB)C=A(BC)‍

Distributive properties

Multiplicative identity property

Multiplicative property of zero

The dimension property

Check your understanding

Want to join the conversation?

Associative property of multiplication: $(A B) C = A (B C)$ ‍