Practice Questions on Subtraction of Matrices

Subtraction of Matrices is one of the operations that are performed between two matrices. It is similar to addition of matrices. This article provides practice questions on subtraction of matrices along with solved examples and concepts related to it. This article serves as a one stop solution for practicing problems related to subtraction of matrices to ace the exam.

Suppose, A and B are two matrices of the same dimension m x n, then the subtraction of two matrices is given by matrix C of the dimension m x n

C = A – B, where

c_ij = a_ij – b_ij

for all i = 1, 2, 3, …,m and j = 1, 2, 3…,n.

Commutative Property of Subtraction of matrices

A – B ≠ B – A

Associative Property of Subtraction of matrices

(A – B) – C ≠ A – (B – C)

Q1. Given matrices A and B where

A = [Tex]\begin{bmatrix} 1 &3 \\ 5 &7 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 2 &4 \\ 6 &8 \end{bmatrix}[/Tex]

find the matrix C = A – B.

Solution:

To find the subtraction of matrix A and B is C

c₁₁ = a₁₁ – b₁₁ = 1 – 2 = -1

c₁₂ = a₁₂ – b₁₂ = 3 – 4 = -1

C₂₁ = a₂₁ – b₂₁ = 5 – 6 = -1

c₂₂ = a₂₂ – b₂₂ = 7 – 8 = -1

The resulting matrix C is

C = [Tex]\begin{bmatrix} -1 &-1 \\ -1 &-1 \end{bmatrix}[/Tex]

Q2. Let A = [Tex]\begin{bmatrix} 0 &-1 \\ 3 &2 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 5 &3 \\ -3 &4 \end{bmatrix}[/Tex]

Compute the elements of matrix C = A – B.

Solution:

To find the elements of matrix C

c₁₁ = a₁₁ – b₁₁ = 0 – 5 = -5

c₁₂ = a₁₂ – b₁₂ = -1 – 3 = -4

c₂₁ = a₂₁ – b₂₁ = 3 + 3 = 6

c₂₂ = a₂₂ – b₂₂ = 2 – 4 = -2

The resulting matrix C is

C = [Tex]\begin{bmatrix} -5 &-4 \\ 6 &-2 \end{bmatrix}[/Tex]

Q3. Verify the commutative property for matrices

A = [Tex]\begin{bmatrix} 1 &3 \\ 2 &4 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 4 &3 \\ 2 &1 \end{bmatrix}[/Tex]

Solution:

To verify the commutative property for matrices follow these steps

First we calculate A – B

c₁₁ = a₁₁ – b₁₁ = 1 – 4 = -3

c₁₂ = a₁₂ – b₁₂ = 3 – 3 = 0

c₂₁ = a₂₁ – b₂₁ = 2 – 2 = 0

c₂₂ = a₂₂ – b₂₂ = 4 – 1 = 3

The resulting matrix C1 is

C1 = [Tex]\begin{bmatrix} -3 &0 \\ 0 &3 \end{bmatrix}[/Tex]

Now, we calculate B – A

c₁₁ = a₁₁ – b₁₁ = 4 – 1 = 3

c₁₂ = a₁₂ – b₁₂ = 3 – 3 = 0

c₂₁ = a₂₁ – b₂₁ = 2 – 2 = 0

c₂₂ = a₂₂ – b₂₂ = 1 – 4 = -3

The resulting matrix C is

C2 = [Tex]\begin{bmatrix} 3 &0 \\ 0 &-3 \end{bmatrix}[/Tex]

So, C1 and C2 are not equal so, it follow the commutative property.

Q4. Given Matrices are

A = [Tex]\begin{bmatrix} -1 &0 \\ 4 &5 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 3 &1 \\ -2 &6 \end{bmatrix}[/Tex]

Show that A + B ≠ B + A.

Solution:

First we calculate A – B

c₁₁ = a₁₁ – b₁₁ = -1 – 3 = -4

c₁₂ = a₁₂ – b₁₂ = 0 – 1 = -1

?₂₁ = ?₂₁ – ?₂₁ = 4 + 2 = 6

?₂₂ = ?₂₂ – ?₂₂ = 5 – 6 = -1

The resulting matrix C1 is

C1 = [Tex]\begin{bmatrix} -4 &-1 \\ 6 &-1 \end{bmatrix}[/Tex]

Now, we calculate B – A

c₁₁ = b₁₁ – a₁₁ = 3 + 1 = 4

c₁₂ = b₁₂ – a₁₂ = 1 – 0 = 1

?₂₁ = b₂₁ – a₂₁ = -2 – 5 = -7

?₂₂ = b₂₂ – a₂₂ = 6 – 5 = 1

The resulting matrix C2 is

C2 = [Tex]\begin{bmatrix} 4 &1 \\ -7 &1 \end{bmatrix}[/Tex]

So, C1 and C2 are not equal. Therefore it follows the commutative property.

Q5. Given Matrices are

A = [Tex]\begin{bmatrix} 2 &1 \\ 0 &-1 \end{bmatrix}[/Tex], B = [Tex]\begin{bmatrix} 3 &5 \\ 6 &7 \end{bmatrix}[/Tex] and C = [Tex]\begin{bmatrix} 4 &8 \\ 2 &3 \end{bmatrix}[/Tex]

Verify the associative property (A – B) – C ≠ A – (B – C).

Solution:

First we calculate A – B

e₁₁ = a₁₁ – b₁₁ = 2 – 3 = -1

e₁₂ = a₁₂ – b₁₂ = 1 – 5 = -4

e₂₁ = ?₂₁ – ?₂₁ = 0 – 6 = -6

e₂₂ = ?₂₂ – ?₂₂ = -1 – 7 = -8

The resulting matrix E1 is

E1 = [Tex]\begin{bmatrix} -1 &-4 \\ -6 &-8 \end{bmatrix}[/Tex]

(A – B) – C

d₁₁ = e₁₁ – c₁₁ = -1 -4 = -5

d₁₂ = e₁₂ – c₁₂ = -4 -8 = -12

d₂₁ = e₂₁ – c₂₁ = -6 – 2 = -8

d₂₂ = e₂₂ – c₂₂ = -8 – 3 = -11

So, (A – B) – C = [Tex]\begin{bmatrix} -5 &-12 \\ -8 &-11 \end{bmatrix}[/Tex]

Now, we calculate B – C

e₁₁ = b₁₁ – c₁₁ = 3 – 4 = -1

e₁₂ = b₁₂ – c₁₂ = 5 – 8 = -3

e₂₁ = b₂₁ – c₂₁ = 6 – 2 = 4

e₂₂ = b₂₂ – c₂₂ = 7 – 3 = 4

The resulting matrix E2 is

E2 = [Tex]\begin{bmatrix} -1 &-3 \\ 4 &4 \end{bmatrix}[/Tex]

A – (B – C)

d₁₁ = a₁₁ – e₁₁ = 2 + 1 = 3

d₁₂ = a₁₂ – e₁₂ = 1 + 3 = 4

d₂₁ = a₂₁ – e₂₁ = 0 – 4 = -4

d₂₂ = a₂₂ – e₂₂ = -1 – 4 = -5

So, A – (B – C) = [Tex]\begin{bmatrix} 3 &4 \\ 8 &9 \end{bmatrix}[/Tex]

So, (A – B) – C ≠ A – (B – C).

Q6. If A = [Tex]\begin{bmatrix} 1 &4 \\ 5 &7 \end{bmatrix}[/Tex], B = [Tex]\begin{bmatrix} 0 &2 \\ 3 &6 \end{bmatrix}[/Tex] and C = [Tex]\begin{bmatrix} 2 &1 \\ 4 &5 \end{bmatrix}[/Tex],

Prove that matrix addition is associative by calculating both (A – B) – C and A – (B – C).

Solution:

First we calculate A – B

e₁₁ = a₁₁ – b₁₁ = 1 – 0 = 1

e₁₂ = a₁₂ – b₁₂ = 4 – 2 = 2

e₂₁ = a₂₁ – b₂₁ = 5 – 3 = 2

e₂₂ = a₂₂ – b₂₂ = 7 – 6 = 1

The resulting matrix E1 is

E1 = [Tex]\begin{bmatrix} 1 &2 \\ 2 &1 \end{bmatrix}[/Tex]

(A – B) – C

d₁₁ = e₁₁ – c₁₁ = 1 – 2 = -1

d₁₂ = e₁₂ – c₁₂ = 2 – 1 = 1

d₂₁ = e₂₁ – c₂₁ = 2 – 4 = -2

d₂₂ = e₂₂ – c₂₂ = 1 – 5 = -4

So, (A – B) – C = [Tex]\begin{bmatrix} -1 &1 \\ -2 &-4 \end{bmatrix}[/Tex]

Now, we calculate B – C

e₁₁ = b₁₁ – c₁₁ = 0 – 2 = -2

e₁₂ = b₁₂ – c₁₂ = 2 – 1 = 1

e₂₁ = b₂₁ – c₂₁ = 3 – 4 = -1

e₂₂ = b₂₂ – c₂₂ = 6 – 5 = 1

The resulting matrix E2 is

E2 = [Tex]\begin{bmatrix} -2 &1 \\ -1 &1 \end{bmatrix}[/Tex]

A – (B – C)

d₁₁ = a₁₁ – e₁₁ = 1 + 2 = 3

d₁₂ = a₁₂ – e₁₂ = 4 – 1 = 3

d₂₁ = a₂₁ – e₂₁ = 5 – 7 = -2

d₂₂ = a₂₂ – e₂₂ = 7 – 11 = -4

So, A – (B – C) = [Tex]\begin{bmatrix} 3 &3 \\ -2 &-4 \end{bmatrix}[/Tex]

So, (A – B) – C ≠ A – (B – C).

Q7. Given matrix A = [Tex]\begin{bmatrix} 6 &8 \\ 7 &9 \end{bmatrix}[/Tex] and the zero matrix 0 = [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex], show that A – 0 = A.

Solution:

A – 0 = [Tex]\begin{bmatrix} 6 &8 \\ 7 &9 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex]

c₁₁ = a₁₁ – 0₁₁ = 6 – 0 = 6

c₁₂ = a₁₂ – 0₁₂ = 8 – 0 = 8

c₂₁ = a₂₁ – 0₂₁ = 7 – 0 = 7

c₂₂ = a₂₂ – 0₂₂ = 9 – 0 = 9

So, Subtraction of matrix A and 0 is [Tex]\begin{bmatrix} 6 &8 \\ 7 &9 \end{bmatrix}[/Tex]

So, it is equal to A matrix.

Q8. Given matrix A = [Tex]\begin{bmatrix} -3 &5 \\ 7 &2 \end{bmatrix}[/Tex], find the result of A – zero matrix.

Solution:

A – 0 = [Tex]\begin{bmatrix} -3 &5 \\ 7 &2 \end{bmatrix}[/Tex] – [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex]

c₁₁ = a₁₁ – 0₁₁ = -3 – 0 = -3

c₁₂ = a₁₂ – 0₁₂ = 5 – 0 = 5

c₂₁ = a₂₁ – 0₂₁ = 7 – 0 = 7

c₂₂ = a₂₂ – 0₂₂ = 2 – 0 = 2

So, Subtraction of matrix A and 0 is [Tex]\begin{bmatrix} -3 &5 \\ 7 &2 \end{bmatrix}[/Tex]

Q9. Given matrix A = [Tex]\begin{bmatrix} 2 &-3 \\ 4 &-5 \end{bmatrix}[/Tex], find the additive inverse −A and verify that A + (−A) = 0.

Solution:

Additive inverse -A = [Tex]\begin{bmatrix} -2 &3 \\ -4 &5 \end{bmatrix}[/Tex]

So, A + (-A) = [Tex]\begin{bmatrix} 2 &-3 \\ 4 &-5 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} -2 &3 \\ -4 &5 \end{bmatrix}[/Tex]

= [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex]

So, A + (-A) = 0.

Q10. Given matrix A = [Tex]\begin{bmatrix} 3 &4 \\ -2 &1 \end{bmatrix}[/Tex], find the additive inverse −A and verify that A + (−A) = 0.

Solution:

Additive inverse -A = [Tex]\begin{bmatrix} -3 &-4 \\ 2 &-1 \end{bmatrix}[/Tex]

So, A + (-A) = [Tex]\begin{bmatrix} 3 &4 \\ -2 &1 \end{bmatrix}[/Tex] + [Tex]\begin{bmatrix} -3 &-4 \\ 2 &-1 \end{bmatrix}[/Tex]

= [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex]

So, A + (-A) = 0.

Q1. Given matrices A = [Tex]\begin{bmatrix} 2 &5 \\ 7 &1 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 3 &6 \\ 4 &2 \end{bmatrix}[/Tex], find the matrix C = A – B.

Q2. Given matrices A = [Tex]\begin{bmatrix} -2 &8 \\ 3 &-5 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 4 &-7 \\ -1 &6 \end{bmatrix}[/Tex], find the matrix C = A – B.

Q3. Verify the commutative property for matrices A = [Tex]\begin{bmatrix} 2 &3 \\ 4 &5 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 5 &4 \\ 3 &2 \end{bmatrix}[/Tex]

Q4. Given Matrices are A = [Tex]\begin{bmatrix} -3 &1 \\ 2 &6 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix} 4 &-2 \\ 5 &3 \end{bmatrix}[/Tex]

Show that A + B ≠ B + A.

Q5. Given Matrices are

A = [Tex]\begin{bmatrix} 1 &-2 \\ 3 &4 \end{bmatrix}[/Tex], B = [Tex]\begin{bmatrix} 2 &3 \\ 4 &5 \end{bmatrix}[/Tex] and C = [Tex]\begin{bmatrix} -1 &0 \\ 2 &3 \end{bmatrix}[/Tex]

verify the associative property (A – B) – C ≠ A – (B – C).

Q6. If A =[Tex]\begin{bmatrix} 0 &1 \\ -1 &2 \end{bmatrix}[/Tex], B =[Tex]\begin{bmatrix} 2 &0 \\ 1 &3 \end{bmatrix}[/Tex] and C =[Tex]\begin{bmatrix} 1 &-1 \\ 3 &4 \end{bmatrix}[/Tex]

Prove that matrix subtraction is associative by calculating both (A – B) – C and A – (B – C).

Q7. Given matrix A =[Tex]\begin{bmatrix} 4 &5 \\ 6 &7 \end{bmatrix}[/Tex] and the zero matrix 0 = [Tex]\begin{bmatrix} 0 &0 \\ 0 &0 \end{bmatrix}[/Tex] , show that A – 0 = A.

Q8. Given matrix A =[Tex]\begin{bmatrix} -1 &3 \\ 2 &4 \end{bmatrix}[/Tex], find the result of A – zero matrix.

Q9. Given matrix A = [Tex]\begin{bmatrix} 1 &-4 \\ 3 &2 \end{bmatrix}[/Tex], find the additive inverse −A and verify that A + (−A) = 0.

Q10. Given matrix A = [Tex]\begin{bmatrix} 2 &-3 \\ -4 &1 \end{bmatrix}[/Tex], find the additive inverse −A and verify that A + (−A) = 0.

When we can perform Subtraction of matrices?

To perform subtraction of matrices, the dimension of the matrices need to be same.

What are the properties of Matrix subtraction?

The properties of matrix subtraction are following:

• Closure property
• Commutative Property
• Associative Property

What is Zero matrix?

Zero Matrix is a matrix in which all the elements are zero.

What do you mean by Identity Matrix?

Identity matrix is a matrix in which the elements on the main diagonal is 1 and else where is 0.

Do Subtraction of matrices follow the commutative property?

Yes, Addition of matrices follows the commutative property.