Worksheet on Addition of Matrices

Worksheet on Addition of Matrices | Matrix Addition Worksheets with Answers PDF

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Worksheet on Addition of Matrices consists of all problems related to properties of addition, Addition of Two Matrices, Addition of more than two matrices, etc. Solve all the Matrix Addition Problems with Solutions on your own to know how to add matrices.

Various tricks and tips are also included in the Adding Matrices Worksheet for easy solving of matrix addition questions. Quickly go through the entire article and gain high knowledge of matrix addition problems. Complete matrices information is given in the 10th Grade Math articles.

Also, find:

Adding Matrices Worksheet with Answers PDF | Sum of Matrices Worksheet

Check out the below problems to learn the addition of the matrices concepts deeply. Different addition of matrix problems along with their answers is given below.

Question 1. Find the sum of A and B where \( A =\left[
\begin{matrix}
4&6 \cr
-10&14 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
8&12 \cr
4&-22 \cr
\end{matrix}
\right]
\)

Solution:

Given matrices are \( A =\left[
\begin{matrix}
4&6 \cr
-10&14 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
8&12 \cr
4&-22 \cr
\end{matrix}
\right]
\)
The two matrices are of the same order of 2 × 2. So, it is possible to perform an addition operation on the given matrices.
Now, Add the elements of the first matrix with the respective elements of the second matrix.
A + B = \( \left[
\begin{matrix}
4&6 \cr
-10&14 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
8&12 \cr
4&-22 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
4 + 8&6 + 12 \cr
-10 + 4&14 – 22 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
14&18 \cr
-6&-8 \cr
\end{matrix}
\right]
\)

Therefore, the sum of the given matrices A and B is \( \left[
\begin{matrix}
14&18 \cr
-6&-8 \cr
\end{matrix}
\right]
\)

Question 2. Add the matrices \( A =\left[
\begin{matrix}
3&4 \cr
9&10 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
4&0&5 \cr
6&59&4 \cr
\end{matrix}
\right]
\)

Solution:

Given matrices are \( A =\left[
\begin{matrix}
3&4 \cr
9&10 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
4&0&5 \cr
6&59&4 \cr
\end{matrix}
\right]
\)
The matrix A order is 2 × 2. And the matrix B order is 2 × 3.
We can’t perform an addition operation on the matrices with different dimensions.

Therefore, addition operation is not possible for given matrices.

Question 3. Write the elements of the sum matrix Z = X + Y explicitly by addition of matrices X and Y of dimension 1 × 2 whose elements are given as: x11 = 5, x12 = 15 and y11 = -16, y12 = -22.

Solution:

Given that the sum matrix Z = X + Y explicitly by addition of matrices X and Y of dimension 1 × 2.
We can perform the addition operation for X and Y as the X and Y order is 1 × 2. The output will also have the same order of 1 × 2 by adding the corresponding elements.
z11 = x11 + y11 = 5 + (-16) = -11.
z12 = x12 + y12 = 15 + (-22) = -7.

Therefore, addition operation is not possible for given matrices.

Question 4. Determine the element of the second row and third column of the matrix M + N using the addition of matrices definition if m23 = -26 is an element of M and n23 = 40 is an element in N.

Solution:

Given that m23 = -26 is an element of M and n23 = 40 is an element in N.
We need to add m23 and n23 to find the element of the second row and third column of the matrix M + N.
m23 + n23 = -26 + 40 = 14.

Therefore, the element in the second row and third column of M + N is 14.

Question 5. Find X + Y when \( X =\left[
\begin{matrix}
4&6&8 \cr
10&12&14 \cr
16&10&22 \cr
\end{matrix}
\right]
\) and \( Y =\left[
\begin{matrix}
6&-4&-6 \cr
10&8&6 \cr
2&6&4 \cr
\end{matrix}
\right]
\)

Solution:

Given matrices are \( X =\left[
\begin{matrix}
4&6&8 \cr
10&12&14 \cr
16&10&22 \cr
\end{matrix}
\right]
\) and \( Y =\left[
\begin{matrix}
6&-4&-6 \cr
10&8&6 \cr
2&6&4 \cr
\end{matrix}
\right]
\)
The two matrices are of the same order of 3 × 3. So, it is possible to perform an addition operation on the given matrices.
Now, Add the elements of the first matrix with the respective elements of the second matrix.
X + Y = \( \left[
\begin{matrix}
4&6&8 \cr
10&12&14 \cr
16&10&22 \cr
\end{matrix}
\right]
\)  + \( \left[
\begin{matrix}
6&-4&-6 \cr
10&8&6 \cr
2&6&4 \cr
\end{matrix}
\right]
\)
= \( \left[
\begin{matrix}
4 + 6&6 – 4&8 – 6 \cr
10 + 10&12 + 8&14 + 6 \cr
16 + 2&10 + 6&22 + 4 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
10&2&2 \cr
20&20&20 \cr
18&16&26 \cr
\end{matrix}
\right]
\)

Therefore, X + Y = \( \left[
\begin{matrix}
10&2&2 \cr
20&20&20 \cr
18&16&26 \cr
\end{matrix}
\right]
\)

Question 6. If \( A =\left[
\begin{matrix}
-2&4&-6 \cr
-4&2&8 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
0&-2&4 \cr
6&0&2 \cr
\end{matrix}
\right]
\), then find the sum of A and B.

Solution:

Given matrices are \( A =\left[
\begin{matrix}
-2&4&-6 \cr
-4&2&8 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
0&-2&4 \cr
6&0&2 \cr
\end{matrix}
\right]
\)
The two matrices are of the same order of 2 × 3. So, it is possible to perform an addition operation on the given matrices.
Now, Add the elements of the first matrix with the respective elements of the second matrix.
A + B = \( \left[
\begin{matrix}
-2&4&-6 \cr
-4&2&8 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
0&-2&4 \cr
6&0&2 \cr
\end{matrix}
\right]
\)
=\( \left[
\begin{matrix}
-2 + 0&4 – 2&-6 + 4 \cr
-4 + 6&2 + 0&8 + 2 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
-2&2&-2 \cr
2&2&10 \cr
\end{matrix}
\right]
\)

Therefore, the sum of the given matrices A and B is \( \left[
\begin{matrix}
-2&2&-2 \cr
2&2&10 \cr
\end{matrix}
\right]
\)

Question 7. If \( \left[
\begin{matrix}
4&6 \cr
-10&8 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
-4&2 \cr
x&6 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
0&8 \cr
-6&14 \cr
\end{matrix}
\right]
\), find the value of x.

Solution:

Given matrices are \( \left[
\begin{matrix}
4&6 \cr
-10&8 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
-4&2 \cr
x&6 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
0&8 \cr
-6&14 \cr
\end{matrix}
\right]
\)
Now, add the elements of the first two matrices and compare the elements of it with the output matrix elements.
Firstly, add \( \left[
\begin{matrix}
4&6 \cr
-10&8 \cr
\end{matrix}
\right]
\) and \( \left[
\begin{matrix}
-4&2 \cr
x&6 \cr
\end{matrix}
\right]
\)
\( \left[
\begin{matrix}
4&6 \cr
-10&8 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
-4&2 \cr
x&6 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
4 – 4&6 + 2 \cr
-10 + x&8 + 6 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
0&8 \cr
-10 + x&14 \cr
\end{matrix}
\right]
\)
Compare \( \left[
\begin{matrix}
0&8 \cr
-10 + x&14 \cr
\end{matrix}
\right]
\) with \( \left[
\begin{matrix}
0&8 \cr
-6&14 \cr
\end{matrix}
\right]
\)
So, -10 + x = -6; x = -6 + 10; x = 4.

Therefore, the value of x is 4.

Question 8. If \( A =\left[
\begin{matrix}
2&8 \cr
4&6 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
-8&-2 \cr
-6&-4 \cr
\end{matrix}
\right]
\), compute A + B.

Solution:

Given matrices are \( A =\left[
\begin{matrix}
2&8 \cr
4&6 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
-8&-2 \cr
-6&-4 \cr
\end{matrix}
\right]
\)
The two matrices are of the same order of 2 × 2. So, it is possible to perform an addition operation on the given matrices.
Now, Add the elements of the first matrix with the respective elements of the second matrix.
A + B = \( \left[
\begin{matrix}
2&8 \cr
4&6 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
-8&-2 \cr
-6&-4 \cr
\end{matrix}
\right]
\)
=\( \left[
\begin{matrix}
2 – 8&8 – 2 \cr
4 – 6&6 – 4 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
-6&6 \cr
-2&2 \cr
\end{matrix}
\right]
\)

Therefore, the sum of the given matrices A and B is \( \left[
\begin{matrix}
-6&6 \cr
-2&2 \cr
\end{matrix}
\right]
\)

Question 9. If \( \left[
\begin{matrix}
10&-9 \cr
4&8 \cr
\end{matrix}
\right]
\) + A = \( \left[
\begin{matrix}
1&0 \cr
0&1 \cr
\end{matrix}
\right]
\), find the matrix A.

Solution:

Given that \( \left[
\begin{matrix}
10&-9 \cr
4&8 \cr
\end{matrix}
\right]
\) + A = \( \left[
\begin{matrix}
1&0 \cr
0&1 \cr
\end{matrix}
\right]
\)
So, A = \( \left[
\begin{matrix}
1&0 \cr
0&1 \cr
\end{matrix}
\right]
\) – \( \left[
\begin{matrix}
10&-9 \cr
4&8 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
1 – 10&0 + 9 \cr
0 – 4&1 – 8 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
-9&9 \cr
-4&-7 \cr
\end{matrix}
\right]
\)

Therefore, the matrix is \( A = \left[
\begin{matrix}
-9&9 \cr
-4&-7 \cr
\end{matrix}
\right]
\)

Question 10. Given \( M = \left[
\begin{matrix}
2&6 \cr
4&8 \cr
\end{matrix}
\right]
\), find a matrix N such that \( M + N = \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\)

Solution:

Given that \( M = \left[
\begin{matrix}
2&6 \cr
4&8 \cr
\end{matrix}
\right]
\) and \( M + N = \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\)
Substitute M in \( M + N = \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\)
\( M = \left[
\begin{matrix}
2&6 \cr
4&8 \cr
\end{matrix}
\right]
\) + N = \( \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\)
N = \( \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\) – \( \left[
\begin{matrix}
2&6 \cr
4&8 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
0 – 2&0 – 6 \cr
0 – 4&0 – 8 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
-2&-6 \cr
-4&-8 \cr
\end{matrix}
\right]
\)

Therefore, the matrix is \( N = \left[
\begin{matrix}
-2&-6 \cr
-4&-8 \cr
\end{matrix}
\right]
\)

Question 11. If \( A = \left[
\begin{matrix}
2&0&4 \cr
0&4&6 \cr
2&0&0 \cr
\end{matrix}
\right]
\), \( B = \left[
\begin{matrix}
0&-2&0 \cr
-4&0&6 \cr
0&2&4 \cr
\end{matrix}
\right]
\), and \( C = \left[
\begin{matrix}
4&6&2 \cr
0&0&-6 \cr
2&2&-2 \cr
\end{matrix}
\right]
\), find A + B + C.

Solution:

Given matrices are \( A = \left[
\begin{matrix}
2&0&4 \cr
0&4&6 \cr
2&0&0 \cr
\end{matrix}
\right]
\), \( B = \left[
\begin{matrix}
0&-2&0 \cr
-4&0&6 \cr
0&2&4 \cr
\end{matrix}
\right]
\), and \( C = \left[
\begin{matrix}
4&6&2 \cr
0&0&-6 \cr
2&2&-2 \cr
\end{matrix}
\right]
\)
Firstly add A + B = \( \left[
\begin{matrix}
2&0&4 \cr
0&4&6 \cr
2&0&0 \cr
\end{matrix}
\right]
\) + \( \left[
\begin{matrix}
0&-2&0 \cr
-4&0&6 \cr
0&2&4 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
2 + 0&0 – 2&4 + 0 \cr
0 – 4&4 + 0&6 + 6 \cr
2 + 0&0 + 2&0 + 4 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
2&-2&4 \cr
-4&4&12 \cr
2&2&4 \cr
\end{matrix}
\right]
\)
Now, add A + B + C = \( \left[
\begin{matrix}
2&-2&4 \cr
-4&4&12 \cr
2&2&4 \cr
\end{matrix}
\right]
\) + \(  \left[
\begin{matrix}
4&6&2 \cr
0&0&-6 \cr
2&2&-2 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
2 + 4&-2 + 6&4 + 2 \cr
-4 + 0&4 + 0&12 – 6 \cr
2 + 2&2 + 2&4 – 2 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
6&4&6 \cr
-4&4&6 \cr
4&4&2 \cr
\end{matrix}
\right]
\)

Therefore, the required matrix is A + B + C = \( \left[
\begin{matrix}
6&4&6 \cr
-4&4&6 \cr
4&4&2 \cr
\end{matrix}
\right]
\)

Question 12. If \( A =\left[
\begin{matrix}
13&25 \cr
7&8 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
14&16 \cr
2&3 \cr
\end{matrix}
\right]
\), then prove A and B matrices obey commutative property of addition.

Solution:

Given matrices are \( A =\left[
\begin{matrix}
13&25 \cr
7&8 \cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
14&16 \cr
2&3 \cr
\end{matrix}
\right]
\).
Both matrices are of the same order i.e, 2 × 2. Now, add the elements of matrix A with the respective elements of matrix B.
\( A + B =\left[
\begin{matrix}
13 + 14&25 + 16 \cr
7+ 2&8 + 3 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
27&41 \cr
9&11 \cr
\end{matrix}
\right]
\)
Now, add the elements of matrix B with the respective elements of matrix A.
\( B + A =\left[
\begin{matrix}
14 + 13&16 + 25 \cr
2 + 7&3 + 8 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
27&41 \cr
9&11 \cr
\end{matrix}
\right]
\)

Therefore, A + B = B + A. A and B matrices obey the commutative property of addition.

Question 13. Find the Additive Identity of Matrix \( A =\left[
\begin{matrix}
4&5 \cr
12&3 \cr
\end{matrix}
\right]
\)

Solution:

Given matrix is \( A =\left[
\begin{matrix}
4&5 \cr
12&3 \cr
\end{matrix}
\right]
\)
Add the null matrix with the given matrix A.
\( A =\left[
\begin{matrix}
4&5 \cr
12&3 \cr
\end{matrix}
\right]
\)+ \( \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
4 + 0&5 + 0 \cr
12 + 0&3 + 0 \cr
\end{matrix}
\right]
\) = \( A =\left[
\begin{matrix}
4&5 \cr
12&3 \cr
\end{matrix}
\right]
\) = A

Therefore, A + O = A.

Question 14. If the given matrix \( A =\left[
\begin{matrix}
12&14 \cr
15&25 \cr
\end{matrix}
\right]
\) obeys an Additive Inverse of Matrix, find the inverse of the matrix A.

Solution:

Given matrix is \( A =\left[
\begin{matrix}
12&14 \cr
15&25 \cr
\end{matrix}
\right]
\) obeys an Additive Inverse of Matrix.
Multiply the given matrix A with -1 to get the -A.
(-1) \( \left[
\begin{matrix}
12&14 \cr
15&25 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
-12&-14 \cr
-15&-25 \cr
\end{matrix}
\right]
\)  = -A.
\( A =\left[
\begin{matrix}
12&14 \cr
15&25 \cr
\end{matrix}
\right]
\) + \( A =\left[
\begin{matrix}
-12&-14 \cr
-15&-25 \cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
0&0 \cr
0&0 \cr
\end{matrix}
\right]
\)
A + (-A) = 0.

Therefore, the answer is \( A =\left[
\begin{matrix}
12&14 \cr
15&25 \cr
\end{matrix}
\right]
\)

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