by the property define by scalar multiplication the determinant of a n-dimension matrix will be the scalar^n times, why is it not times the scalar

It sounds like you know the rule, but aren't sure why it works that way? Well, let's look at an example. Here is a 2 x 2 matrix, so of course n = 2 `┌ ┐` ` 3 2 ` ` -1 5 ` `└ ┘` We multiply by scalar c `┌ ┐` ` 3c 2c ` ` -1c 5c ` `└ ┘` Now we take the determinant which will be [3c∙5c - 2c∙(-1c)] That will be 15c²- (- 2c²) = 15c²+ 2c² = 17c² The operation of multiplying the elements to produce the terms of the determinant effectively squares the constant c, and the last operation of subtraction does not affect the exponent of the c Likewise, if we use a 3 x 3 matrix, three elements would be multiplied in each diagonal direction to get the determinant, and of course, n = 3 this time. If each element has been multiplied by constant k, then the product result would be a number multiplied by k³ so the last step of subtraction again leaves us with a multiple of k³ , or kⁿ I hope that helps

Is there such a thing as Ac=cA? Where A is a matrix and c is a scalar. Is Ac even a valid thing?

As long as there is a scalar times a matrix, the scalar is applied to every value in the matrix, regardless of which comes first. So yes, Ac=cA is correct. Just watch out for matrix x matrix, which is not commutative.

could you prove the properties? not examples. Thank you

The proofs as such are not required as you should be able to figure them yourself out yourself using basic arithmetic skills.

"c" is whatever scalar you are using for a certain problem. It's a variable that is usually given to you, and if it isn't, you'll be given all the other information you need to work backwards.

what if you multiply a matrix by a complex number?

Then you will have a matrix whose entries are complex numbers. You can add, multiply, and row-reduce such matrices in the usual way.

Main content

Precalculus

Course: Precalculus > Unit 7

Lesson 5: Properties of matrix addition & scalar multiplication

Properties of matrix scalar multiplication

Google Classroom

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

In the table below,

A

and

B

are matrices of equal dimensions,

c

and

d

are scalars, and

O

is a zero matrix.

Property	Example
Associative property of multiplication	$(c d) A = c (d A)$ ‍
Distributive properties	$c (A + B) = c A + c B$ ‍
	$(c + d) A = c A + d A$ ‍
Multiplicative identity property	$1 A = A$ ‍
Multiplicative properties of zero	$0 \cdot A = O$ ‍
	$c \cdot O = O$ ‍
Closure property of multiplication	$c A$ ‍ is a matrix of the same dimensions as $A$ ‍.

This article explores these properties.

Matrices and scalar multiplication

A matrix is a rectangular arrangement of numbers into rows and columns.

When we work with matrices, we refer to real numbers as scalars.

The term scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each entry in the matrix is multiplied by the given scalar.

\begin{aligned} 2 \cdot [\begin{array}{rr} 5 & 2 \\ 3 & 1 \end{array}] & = [\begin{array}{ll} 2 \cdot 5 & 2 \cdot 2 \\ 2 \cdot 3 & 2 \cdot 1 \end{array}] \\ = [\begin{array}{rr} 10 & 4 \\ 6 & 2 \end{array}] \end{aligned}

If any of this is new to you, you should check out the following articles before you proceed:

Dimensions considerations

Notice that a scalar times a

2 \times 2

matrix is another

2 \times 2

matrix. In general, a scalar multiple of a matrix will be another matrix of the same dimension. This is what is meant by the closure property of scalar multiplication!

Matrix scalar multiplication & real number multiplication

Because scalar multiplication relies heavily on real number multiplication, many of the multiplication properties that we know to be true with real numbers are also true in scalar multiplication.

Let's take a look at each property individually.

Associative property of multiplication: $(c d) A = c (d A)$ ‍

This property states that if a matrix is multiplied by two scalars, you can multiply the scalars together first, and then multiply by the matrix. Or you can multiply the matrix by one scalar, and then the resulting matrix by the other.

The following example illustrates this property for

c = 2

d = 3

, and

A = [\begin{array}{rr} 5 & 4 \\ 8 & 1 \end{array}]

In each column we simplified one side of the identity into a single matrix. Notice that these two matrices are equal because of the associative property of multiplication for real numbers. For example,

(2 \cdot 3) \cdot 5 = 2 \cdot (3 \cdot 5)

This shows that the original expressions must be equivalent as well!

Distributive properties:

$c (A + B) = c A + c B$ ‍

This property states that a scalar can be distributed over matrix addition.

Here's an example where

c = 2

A = [\begin{array}{rr} 5 & 2 \\ 3 & 1 \end{array}]

, and

B = [\begin{array}{rr} 3 & 4 \\ 2 & 6 \end{array}]

If we compare the last matrix in each column, we see that these are equivalent because of the distributive property for real numbers. For example,

2 (5 + 3) = 2 \cdot 5 + 2 \cdot 3

Thus the original two expressions must be equivalent as well!

$(c + d) A = c A + d A$ ‍

This property states that a matrix can be distributed over scalar addition.

Here's an example where

c = 2

d = 3

, and

A = [\begin{array}{rr} 6 & 9 \\ 7 & 4 \end{array}]

Once again, we see that the last matrix in each column are equivalent because of the distributive property for real numbers, making the original expressions equivalent as desired!

Multiplicative identity property: $1 A = A$ ‍

This property says that when you multiply any matrix

A

by the scalar

1

, the result is simply the original matrix

A

So, for example, if

A = [\begin{array}{rr} 2 & 5 \\ 1 & 7 \end{array}]

, then we have:

\begin{aligned} 1 [\begin{array}{rr} 2 & 5 \\ 1 & 7 \end{array}] & = [\begin{array}{rr} 1 \cdot 2 & 1 \cdot 5 \\ 1 \cdot 1 & 1 \cdot 7 \end{array}] \\ = [\begin{array}{rr} 2 & 5 \\ 1 & 7 \end{array}] \end{aligned}

Notice that because

1 \cdot a = a

for any real number

a

, the scalar

1

will always be the multiplicative identity in scalar multiplication!

Multiplicative properties of zero:

$0 \cdot A = O$ ‍

This property states that in scalar multiplication,

0

times any

m \times n

matrix

A

is the

m \times n

zero matrix.

This is true because of the multiplicative properties of zero in the real number system. If

a

is a real number, we know

0 \cdot a = 0

. The following example illustrates this.

\begin{aligned} 0 [\begin{array}{rr} 3 & 8 \\ 6 & 7 \end{array}] & = [\begin{array}{rr} 0 \cdot 3 & 0 \cdot 8 \\ 0 \cdot 6 & 0 \cdot 7 \end{array}] \\ = [\begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array}] \end{aligned}

$c \cdot O = O$ ‍

This property states that any scalar times a zero matrix is the same zero matrix.

Again, this property is true because of the multiplicative properties of zero in the real number system. Here's an example where

c = 3

and

O

is the

2 \times 2

zero matrix.

\begin{aligned} 3 [\begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array}] & = [\begin{array}{rr} 3 \cdot 0 & 3 \cdot 0 \\ 3 \cdot 0 & 3 \cdot 0 \end{array}] \\ = [\begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array}] \end{aligned}

Check your understanding

Now that you are familiar with all of the scalar multiplication properties, let's see if you can use them to determine equivalent matrix expressions.

For the problems below, let

A

and

B

2 \times 2

matrices and let

c

and

d

be scalars.

1) Which of the following are equivalent to $c (1 A + B)$ ‍?

[I need help!]

2) Which of the following are equivalent to $(c d) A + 0 A$ ‍?

[I need help!]

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jk
Posted 8 years ago. Direct link to jk's post “by the property define by...”
by the property define by scalar multiplication the determinant of a n-dimension matrix will be the scalar^n times, why is it not times the scalar
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- doctorfoxphd
  Posted 7 years ago. Direct link to doctorfoxphd's post “It sounds like you know t...”
  It sounds like you know the rule, but aren't sure why it works that way?
  Well, let's look at an example. Here is a 2 x 2 matrix, so of course n = 2
  ┌ ┐
  3 2
  -1 5
  └ ┘
  We multiply by scalar c
  ┌ ┐
  3c 2c
  -1c 5c
  └ ┘
  Now we take the determinant which will be [3c∙5c - 2c∙(-1c)]
  That will be 15c²- (- 2c²)
  = 15c²+ 2c²
  = 17c²
  The operation of multiplying the elements to produce the terms of the determinant effectively squares the constant c, and the last operation of subtraction does not affect the exponent of the c
  Likewise, if we use a 3 x 3 matrix, three elements would be multiplied in each diagonal direction to get the determinant, and of course, n = 3 this time. If each element has been multiplied by constant k, then the product result would be a number multiplied by k³ so the last step of subtraction again leaves us with a multiple of k³ , or kⁿ
  I hope that helps
  Button navigates to signup page
  (7 votes)
lol
Posted 8 months ago. Direct link to lol's post “Is there such a thing as ...”
Is there such a thing as Ac=cA? Where A is a matrix and c is a scalar. Is Ac even a valid thing?
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(2 votes)
Answer
- Angelica Chen
  Posted 8 months ago. Direct link to Angelica Chen's post “As long as there is a sca...”
  As long as there is a scalar times a matrix, the scalar is applied to every value in the matrix, regardless of which comes first. So yes, Ac=cA is correct. Just watch out for matrix x matrix, which is not commutative.
  Button navigates to signup page
  (4 votes)
g1y
Posted 7 years ago. Direct link to g1y's post “Just making sure, you can...”
Just making sure, you can't add a vector to a scalar, right?
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(3 votes)
Answer
DanDanJ1107
Posted 6 years ago. Direct link to DanDanJ1107's post “could you prove the prope...”
could you prove the properties? not examples. Thank you
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(1 vote)
Answer
- cossine
  Posted 6 years ago. Direct link to cossine's post “The proofs as such are no...”
  The proofs as such are not required as you should be able to figure them yourself out yourself using basic arithmetic skills.
  Button navigates to signup page
  (4 votes)
bella m
Posted 4 years ago. Direct link to bella m's post “how do you find c?”
how do you find c?
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(1 vote)
Answer
- Lawrence
  Posted 4 years ago. Direct link to Lawrence's post “"c" is whatever scalar yo...”
  "c" is whatever scalar you are using for a certain problem. It's a variable that is usually given to you, and if it isn't, you'll be given all the other information you need to work backwards.
  Comment on Lawrence's post “"c" is whatever scalar yo...”
  (4 votes)
Jeff
Posted 4 years ago. Direct link to Jeff's post “should one remember all t...”
should one remember all the property names??
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(2 votes)
Answer
- Polina Vitić
  Posted 4 years ago. Direct link to Polina Vitić's post “Yes, I think it is helpfu...”
  Yes, I think it is helpful to know these well, but most important is knowing how to apply the properties to matrix multiplication.
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  (1 vote)
SilverFlameShadows
Posted 2 years ago. Direct link to SilverFlameShadows's post “So, if I had A(xB)C is th...”
So, if I had A(xB)C is that equivalent to x(ABC) if x is my scalar and A, B, C are all square matrices?
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(2 votes)
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- loumast17
  Posted 2 years ago. Direct link to loumast17's post “Yes, the matrices have to...”
  Yes, the matrices have to stay in order though.
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  (1 vote)
7795225861
Posted 4 months ago. Direct link to 7795225861's post “what if you multiply a ma...”
what if you multiply a matrix by a complex number?
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(1 vote)
Answer
- kubleeka
  Posted 4 months ago. Direct link to kubleeka's post “Then you will have a matr...”
  Then you will have a matrix whose entries are complex numbers. You can add, multiply, and row-reduce such matrices in the usual way.
  Button navigates to signup page
  (2 votes)
choonklog
Posted 7 years ago. Direct link to choonklog's post “Why isn't commutative pro...”
Why isn't commutative property written in the above table? Is that impossible?
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(1 vote)
Answer
- nfsrrocker333
  Posted 6 years ago. Direct link to nfsrrocker333's post “The commutative property ...”
  The commutative property over multiplication is only possible over scalar multiplication and not problems in which two matrices are multiplied together. This is because the two inner dimensions have to be the same. What I mean by inner dimensions being the same is if one matrix has dimensions a x b and another matrix has dimensions c x d, where c and b hold the same value. Notice how the two dimensions in the middle are the same. In some cases, though, the commutative property could work, like if both matrices have dimensions 2 x 2.
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treetroop123
Posted 5 months ago. Direct link to treetroop123's post “confusing”
confusing
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- Wingfeather
  Posted 4 months ago. Direct link to Wingfeather's post “in the previous lessons o...”
  in the previous lessons on matrices, there's a video explaining this
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  (1 vote)

Precalculus

Course: Precalculus > Unit 7

Matrices and scalar multiplication

Dimensions considerations

Matrix scalar multiplication & real number multiplication

Associative property of multiplication: (cd)A=c(dA)‍

Distributive properties:

c(A+B)=cA+cB‍

(c+d)A=cA+dA‍

Multiplicative identity property: 1A=A‍

Multiplicative properties of zero:

0⋅A=O‍

c⋅O=O‍

Check your understanding

Want to join the conversation?

Associative property of multiplication: $(c d) A = c (d A)$ ‍

$c (A + B) = c A + c B$ ‍

$(c + d) A = c A + d A$ ‍

Multiplicative identity property: $1 A = A$ ‍

$0 \cdot A = O$ ‍

$c \cdot O = O$ ‍