# Properties of Scalar Multiplication of a Matrix with Proofs | Multiplying a Matrix by a Scalar Properties

There are different properties for Scalar Multiplication of Matrices like commutative, associative, Multiplicative Identity, and Multiplicative Property of Zero. We have given Scalar Multiplication of Matrix properties and their proofs in this article.

Check out every property and learn to solve the problems related to them. Â Refer to the entire 10th Grade Math article and be well versed with the article and score good grades in your exams.

## Scalar Multiplication of Matrices Properties | Properties of Scalar Multiplication of a Matrix

There are different properties that are applicable to the Multiplication of Matrices by a scalar. They are

• Commutative Property of Multiplication of a Scalar Matrix
• Associative Property of Multiplication of a Matrix by a Scalar
• Multiplicative Identity Property
• Multiplicative Property of Zero
• Distributive Property of Multiplication
• Closure Property of Multiplication

### Commutative Property of Multiplication

The Scalar Multiplication of a Matrix is said to be commutative when the result of the matrix multiplication is equal to the multiplication of its interchanged position. If A is a matrix and k is a scalar, then the multiplication of k and A and multiplication of A and k must be equal.
kA = Ak

Proof: Let A = [aij]m Ã— n where m is the number of rows and n is the number of columns of a matrix and k is a scalar.
Now, Multiply k and A. k * A = kA which is also equal to k[aij]m Ã— n = kaij.
Then, Multiply A and k. A * k = Ak = [aij]m Ã— nk which is also equal to Ak
Since there is no change in the order of the matrix and Ak is equal to kA, the Scalar Multiplication of a Matrix obeys the Commutative Property of Multiplication.
From the definition of Commutative Property of Multiplication, Ak = aijk = kaij = kA.

Hence, kA and Ak are commutative.

### Associative Property of Multiplication

The Associative Property of Multiplication of a Matrix states that when any two real numbers are multiplied with the matrix, then multiplying one real number with the matrix and again multiplying with the other number does not affect the result. If A is the matrices and k and c are the scalars, then (ck)A = k(cA).

Proof: Let A = [aij]m Ã— n, where m is the number of rows and n is the number of columns of a matrix, and c and k are scalars.
Firstly, find the (ck)A.
Now, substitute the A = [aij]m Ã— n in (ck)A.
(ck)A = (ck) (aij) = ckaij (by using the definition of scalar multiplication of matrices)
Now, find k(cA)
From the definition of scalar multiplication of matrices,
k(cA) = k[caij],
= [k(caij)]
Again from the definition of scalar multiplication of matrices,
= [(kc) aij],
= (kc) [aij]
= (kc)A

Therefore, k(cA) = (kc)A are Associative.

### Multiplicative Identity Property

The Multiplicative Identity Property of Matrix defines that when we multiply a matrix by 1, then the resultant matrix is the same as the given matrix. The order and elements of the matrix will not the same by multiplying 1 with the given matrix. If A is a matrix with the order of m Ã— n and the number 1, then 1. A = A.

Proof: Let A = [aij]m Ã— n, where m is the number of rows and n is the number of columns of a matrix, and 1 is the scalar.
Now, multiply the matrix A with the number 1. substitute the A = [aij]m Ã— n in 1A.
1A = 1[aij]
= [1 âˆ™ aij]
= [aij]
= A

Therefore, 1A = A shows Multiplicative Identity Property.

### Multiplicative Property of Zero

The Multiplicative Property of Zero of Matrix defines that when we multiply a matrix by 0, then the resultant matrix becomes zero or null matrix. The order and elements of the matrix will be zero by multiplying 0 with the given matrix. If A is a matrix with the order of m Ã— n and the number 0, then 0. A = 0.

Proof: Let A = [aij]m Ã— n, where m is the number of rows and n is the number of columns of a matrix, and 0 is the scalar.
Now, multiply the matrix A with the number 0. substitute the A = [aij]m Ã— n in 0.A.
0A = 0[aij]
= [0 âˆ™ aij]
= [0]
= 0

Therefore, 0A = 0 shows Multiplicative Property of Zero.

### Distributive Property of Multiplication

The distributive property of multiplication of matrices defines that when a number is multiplied by the sum of two matrices, the first number can be distributed to both of those matrices and multiplied by each of them separately, then adding the two matrices together for the same result as multiplying the first number by the sum of the matrices.
(i) k(A + B) = kA + kB
(ii) (k + c)A = kA + cA

Proof:Â Let A = [aij]m Ã— nÂ and B = [bij]m Ã— n where m is the number of rows and n is the number of columns of a matrix and c and k are the scalars.
Now, substitute the A = [aij]m Ã— nÂ and B = [bij]m Ã— n in k(A + B).
k(A + B) = k([aij] + [bij])
From the definition of addition of matrices
= k[aij + bij],
From the definition of scalar multiplication of matrices
= [k(aij + bij)],
= [kaij + kbij]
= [kaij] + [kbij]
= k[aij] + k[bij]
= kA + kB

Therefore, k(A + B) = kA + kB obeys Distributive Property of Multiplication.

### Closure Property of Multiplication of a Matrix

Closure Property of Matrix Scalar Multiplication defines that after multiplying a matrix by a scalar, then the order of the matrix remains the same. If A =[aij] and k is a scalar, then multiplying the A =[aij] by k does not affect the order of the matrix.

Also, check:

### Scalar Multiplication of a Matrix and its Properties Examples

Every property of a Scalar Multiplication of a Matrix is clearly explained using different solved examples. Practice every problem given here to understand deeply the properties of Multiplication of a Matrix by a Scalar.

Example 1. Find the Commutative Property of Multiplication of a Scalar Matrix where k = 2 and $$A =\left[ \begin{matrix} 1&3 \cr 3&7 \cr \end{matrix} \right]$$

Solution: Given that scalar k = 2 and matrix $$A =\left[ \begin{matrix} 1&3 \cr 3&7 \cr \end{matrix} \right]$$
Now, find kA = 2 $$\left[ \begin{matrix} 1&3 \cr 3&7 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 2 * 1&2 * 3 \cr 2 * 3&2 * 7 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 2&6 \cr 6&14 \cr \end{matrix} \right]$$
Now, find Ak = $$\left[ \begin{matrix} 1&3 \cr 3&7 \cr \end{matrix} \right]$$ 2
= $$\left[ \begin{matrix} 1 * 2&3 * 2 \cr 3 * 2&7 * 2 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 2&6 \cr 6&14 \cr \end{matrix} \right]$$
kA = $$\left[ \begin{matrix} 2&6 \cr 6&14 \cr \end{matrix} \right]$$ = Ak

Therefore, kA = Ak.

Example 2. Find the Associative Property of Multiplication of a Scalar Matrix where k = 2, c = 3, and $$A =\left[ \begin{matrix} 4&2 \cr 2&5 \cr \end{matrix} \right]$$

Solution: Given that k = 2, c = 3, and $$A =\left[ \begin{matrix} 4&2 \cr 2&5 \cr \end{matrix} \right]$$
Now, find (ck)A = (2 * 3) $$\left[ \begin{matrix} 4&2 \cr 2&5 \cr \end{matrix} \right]$$
= (6) $$\left[ \begin{matrix} 4&2 \cr 2&5 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 6 * 4&6 * 2 \cr 6 * 2&6 * 5 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 24&12 \cr 12&30 \cr \end{matrix} \right]$$
Now, find k(cA).
k(cA) = 2 $$3 \left[ \begin{matrix} 4&2 \cr 2&5 \cr \end{matrix} \right]$$
= 2 $$\left[ \begin{matrix} 3 * 4&3 * 2 \cr 3 * 2&3 * 5 \cr \end{matrix} \right]$$
= 2 $$\left[ \begin{matrix} 12&6 \cr 6&15 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 2 * 12&2 * 6 \cr 2 * 6&2 * 15 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 24&12 \cr 12&30 \cr \end{matrix} \right]$$
(ck)A = $$\left[ \begin{matrix} 24&12 \cr 12&30 \cr \end{matrix} \right]$$ = k(cA)

Therefore, k(cA) = (kc)A.

Example 3. Find the Multiplicative Identity Property of Multiplication of a Scalar Matrix where scalar is the 1 and $$A =\left[ \begin{matrix} 12&4 \cr 6&4 \cr \end{matrix} \right]$$

Solution:Â Given that the number is 1 and matrix is $$A =\left[ \begin{matrix} 12&4 \cr 6&4 \cr \end{matrix} \right]$$
Now, multiply the matrix A with 1.
1A = (1) $$\left[ \begin{matrix} 12&4 \cr 6&4 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 1 * 12&1 * 4 \cr 1 * 6&1 * 4 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 12&4 \cr 6&4 \cr \end{matrix} \right]$$
1A = $$\left[ \begin{matrix} 12&4 \cr 6&4 \cr \end{matrix} \right]$$ = A.

Therefore, 1A = A.

Example 4. Find the Multiplicative Property of Zero of a Scalar Matrix where scalar is the 0 and $$A =\left[ \begin{matrix} 25&15 \cr 53&48 \cr \end{matrix} \right]$$

Solution:Â Given that the number is 0 and matrix is $$A =\left[ \begin{matrix} 25&15 \cr 53&48 \cr \end{matrix} \right]$$
Now, multiply the matrix A with 0.
0A = (0) $$\left[ \begin{matrix} 25&15 \cr 53&48 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 0 * 25&0 * 15 \cr 0 * 53&0 * 48 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 0&0 \cr 0&0 \cr \end{matrix} \right]$$
0A = $$\left[ \begin{matrix} 0&0 \cr 0&0 \cr \end{matrix} \right]$$ = 0.

Therefore, 0A = 0.

Example 5. Find the Distributive Property of of a Scalar Matrix where scalars k = 3, c = 5, $$A =\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$, and $$B =\left[ \begin{matrix} 9&11 \cr 7&12 \cr \end{matrix} \right]$$. Find
(i) k(A + B) = kA + kB
(ii) (k + c)A = kA + cA

Solution: Given that k = 3, c = 5, $$A =\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$, and $$B =\left[ \begin{matrix} 9&11 \cr 7&12 \cr \end{matrix} \right]$$
(i) k(A + B) = kA + kB
Firstly, find k(A + B) = 3($$\left[ \begin{matrix} 1 + 9&2 + 11 \cr 3 + 7&4 + 12 \cr \end{matrix} \right]$$)
= 3 $$\left[ \begin{matrix} 10&13 \cr 10&16 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 * 10&3 * 13 \cr 3 * 10&3 * 16 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 30&39 \cr 30&48 \cr \end{matrix} \right]$$
Now, find kA = 3 $$\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 * 1&3 * 2 \cr 3 * 3&3 * 4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3&6 \cr 9&12 \cr \end{matrix} \right]$$
Now, find kB = 3 $$\left[ \begin{matrix} 9&11 \cr 7&12 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 * 9&3 * 11 \cr 3 * 7&3 * 12 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 27&33 \cr 21&36 \cr \end{matrix} \right]$$
kA + kB = $$\left[ \begin{matrix} 3&6 \cr 9&12 \cr \end{matrix} \right]$$ + $$\left[ \begin{matrix} 27&33 \cr 21&36 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 + 27&6 + 33 \cr 9 + 21&12 + 36 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 30&39 \cr 30&48 \cr \end{matrix} \right]$$
k(A + B) = $$\left[ \begin{matrix} 30&39 \cr 30&48 \cr \end{matrix} \right]$$ = kA + kB

Therefore, (A + B) = kA + kB

(ii) (k + c)A = (3 + 5) $$\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$ = 8 $$\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} 8 * 1&8 * 2 \cr 8 * 3&8 * 4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8&16 \cr 24&32 \cr \end{matrix} \right]$$
Now, find kA.
kA = 3 $$\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 * 1&3 * 2 \cr 3 * 3&3 * 4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3&6 \cr 9&12 \cr \end{matrix} \right]$$
Now, find cA = 5 $$\left[ \begin{matrix} 1&2 \cr 3&4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 5 * 1&5 * 2 \cr 5 * 3&5 * 4 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 5&10 \cr 15&20 \cr \end{matrix} \right]$$
kA + cA = $$\left[ \begin{matrix} 3&6 \cr 9&12 \cr \end{matrix} \right]$$ + $$\left[ \begin{matrix} 5&10 \cr 15&20 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 + 5&6 + 10 \cr 9 + 15&12 + 20 \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8&16 \cr 24&32 \cr \end{matrix} \right]$$
(k + c)A = $$\left[ \begin{matrix} 8&16 \cr 24&32 \cr \end{matrix} \right]$$ = kA + cA

Therefore, (k + c)A = kA + cA.

### FAQs on Scalar Matrix Multiplication Properties

1. What are the Properties of Scalar Matrix Multiplication?

The Properties of Scalar Matrix Multiplication are Commutative, Associative, Multiplicative Identity Property, Multiplicative Property of Zero, Distributive, and Closure Property of Multiplication.

2. What is the Multiplication of a Matrix by a Scalar?

Multiplying the matrix by a real number is known as the Multiplication of a Matrix by a Scalar.

3. What is the Distributive Property of Multiplication of a matrix by a scalar?

• k(A + B) = kA + kB
• (k + c)A = kA + cA are the Distributive Property of the Multiplication of a matrix by a scalar.

4. What is the Closure Property of Multiplication of a Matrix?

The Closure Property of Multiplication of a Matrix states that the order of the matrix remains the same after multiplying it with the scalar.

### Conclusion

Scalar Multiplication of a Matrix Properties are explained along with the proofs and examples. We have given the complete details of the scalar matrix and their related concepts on our website for free. Go through all the articles and start learning immediately.

Scroll to Top
Scroll to Top