# Worksheet on Understanding Matrix | Matrices Worksheet with Answers PDF

Worksheet on Understanding Matrix consists of all problems related to addition, subtraction, multiplication of matrices, etc. Solve all the Matrix Addition Problems with Solutions on your own to know how to find matrices. Various tricks and tips are also included in this Understanding Matrices Worksheet PDF for easy solving of matrix questions.

Quickly go through the entire article and gain a high knowledge of matrix problems. Complete matrices information is given in the Matrices Worksheet for Grade 10 PDF articles. This worksheet help students solve problems based on the properties of matrices.

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### Understanding Matrices Worksheet with Answers PDF

Check out the below problems to learn the matrices concepts deeply. The different matrix problems along with their answers are given below.

Problem 1:  The matrix $$A = \left[ \begin{matrix} 5 & 7\cr 6 & 9\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 3 & 2\cr 4 & 6\cr \end{matrix} \right]$$. What is the added value of Matrices A and B?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 5 & 7\cr 6 & 9\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 3 & 2\cr 4 & 6\cr \end{matrix} \right]$$
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is $$A +B= \left[ \begin{matrix} 5+3 & 7+2\cr 6+4 & 9+6\cr \end{matrix} \right]$$
Therefore, the addition matrices value of A +B is $$\left[ \begin{matrix} 8 & 9\cr 10 & 15\cr \end{matrix} \right]$$

Problems 2: Multiply the given matrix by 4. The matrix is,
$$A = \left[ \begin{matrix} 2 & 4\cr 6 & 3\cr \end{matrix} \right]$$ What is the final value of Matrix?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 2 & 4\cr 6 & 3\cr \end{matrix} \right]$$
Now, multiply given matrix by 4.
After, multiplication the matrix is $$A = \left[ \begin{matrix} 2 *4 & 4*4\cr 6*4 & 3*4\cr \end{matrix} \right]$$
Therefore, the final matrix value of A is $$A = \left[ \begin{matrix} 8 & 16\cr 24 & 12\cr \end{matrix} \right]$$

Problem 3: Subtract X and Y where
$$X = \left[ \begin{matrix} 8 & 6\cr 5 & -4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -2\cr 2 & 1\cr \end{matrix} \right]$$

Solution:

Given matrices are $$X = \left[ \begin{matrix} 8 & 6\cr 5 & -4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -2\cr 2 & 1\cr \end{matrix} \right]$$
Both the matrices have the same order of 2 × 2. So, we will subtract the elements of the first matrix with the respective elements of the second matrix.
So, the matrices X-Y is $$= \left[ \begin{matrix} 8 & 6\cr 5 & -4\cr \end{matrix} \right]$$ – $$\left[ \begin{matrix} -4 & -2\cr 2 & 1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8-(-4) & 6-(-2) \cr 5-2 & -5-1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 12 & 8\cr 3 & -6\cr \end{matrix} \right]$$
Thus, the subtraction of X and Y is $$\left[ \begin{matrix} 12 & 8\cr 3 & -6\cr \end{matrix} \right]$$

Problem 4: The matrix is,
$$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$  and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$. What is the final product value of two matrices?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$
Now, we need to find out the final product value of two matrices.
After the multiplication of two matrices is  $$\left[ \begin{matrix} (1*6) + (0*4) & (1*8) + (0*3) \cr (2*6) + (4*4) & (2*8) + (4*3) \cr \end{matrix}\right]$$
$$\left[ \begin{matrix} 6+0 & 8+0\cr 12+16 & 16+12\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$
Therefore, the final product value of the two matrices is $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$

Problem 5: For the matrix $$A = \left[ \begin{matrix} 6 & 13\cr -5 & -7\cr 2 & 4/cr \end{matrix} \right]$$, answer the following.
(i) What is the order of the matrix A?
(ii) Find the (2, 1), (1, 2) and (3, 2) elements.
(iii) Is it a rectangular matrix or a square matrix?

Solution:

In the given question, the matrix is given.
Now, we will find the following.
(i) The order of matrix A is 3 × 2.

(ii) In the given matrix, the (2, 1)th element is -5.
Next, the (1, 2)th element is 13.
The (3, 2)th element is 4.

(iii) The matrix A will have 3 rows and 2 columns.
The number of rows ≠ the number of columns.
Therefore matrix A is a rectangular matrix.

Problem 6: Write the elements of the sum matrix C = A+B is explicitly by addition of matrices A and B of dimension 1 × 2 whose elements are given as a11 = 4, a12 = 6 and b11 = -2, and b12 = -8.

Solution:

Given the a and b values.
Now, we need to find the value of C using the addition method.
The addition of matrices is defined as when the matrices have the same order and we will add the respective elements of the two matrices A and B.
Now, adding the corresponding elements, we have
C11= a11 + b11  = 4 + (-2) = 2.
Next, the value is C12= a12 + b12 = 6 + (-8) = -2.
Therefore, the value of C11 is 2, and the value of C12 is -2.

Problem 7: Determine the element of the second row and third column of the matrix X + Y using the addition of matrices definition. If x23 = -16 is the element of X and the y23 = 20 is the element of Y.

Solution:

As given in the question, X23 = -16 is the element of X and Y23= 20 is the element of Y.
Now, we need to add X23and Y23 to find the element of the second row and third column of the matrix X + Y.
So, the finding value is, X23+ Y23
= -16 + 20 = 4.
Therefore, the element in the second row and third column of X + Y is 4.

Problem 8:  Find the transpose of a column matrix $$A = \left[ \begin{matrix} 6 \cr 18 \cr 9\cr \end{matrix} \right]$$

Solution:

The given matrix is $$A = \left[ \begin{matrix} 6\cr 18 \cr 9\cr \end{matrix} \right]$$
To find the transpose of the given column matrix, the elements of the column matrix must be written as row elements.
A=[6 18 9]
Therefore, the transpose of the given column matrix is A=[6 18 9].

Problem 9: Example of a null matrix having two rows and three columns.

Solution:

In the given question, a null matrix has two rows and three columns.
Actually, the null matrix will have all the elements are zero.
So, all the elements of two rows and three columns in a null matrix are zero.
$$A = \left[ \begin{matrix} 0 & 0 & 0\cr 0 & 0 & 0\cr \end{matrix} \right]$$
Therefore, the answer is$$A = \left[ \begin{matrix} 0 & 0 & 0\cr 0 & 0 & 0\cr \end{matrix} \right]$$

Problem 10: Write the elements of the matrix C = A – B explicitly. If A = [6 12 28] and B = [2 18 34] using matrix subtraction formula?

Solution:

The given matrices are A = [6 12 28] and B = [2 18 34].
Now, we will find the value of the C.
The dimensions of the matrices A and B are the same, that is, 1 × 3. We can perform subtraction of matrices as two matrices have the same order.
First, subtract the elements of the first matrix with the respective elements of the second matrix. i.e.,
A – B = [6-2 12-18 28-34] = [4 -6 -6]
Hence, the elements of C = A – B are C11 = 2, C12 = -8, C13 = -6.

Problem 11: Find 3C where
$$B= \left[ \begin{matrix} 7 & 8\cr 9 & 12\cr \end{matrix} \right]$$

Solution:

The given matrix is $$C = \left[ \begin{matrix} 7 & 8\cr 9 & 12\cr \end{matrix} \right]$$
To find the value of 3C, we need to multiply the value of 3 by all the elements of matrix C.
i.e., 3C= $$\left[ \begin{matrix} 3*7 & 3*8\cr 3*9 & 3*12\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 21 & 24\cr 27 & 36\cr \end{matrix} \right]$$
Therefore, the required matrix is 3C $$=\left[ \begin{matrix} 21 & 24\cr 27 & 36\cr \end{matrix} \right]$$

Problem 12:  If $$A =\left[ \begin{matrix} a-4 & 0\cr 0 & b-5\cr \end{matrix} \right]$$ is a unit matrix, then find the values of a and b.

Solution:

Given matrix is $$A =\left[ \begin{matrix} a-4 & 0\cr 0 & b-5\cr \end{matrix} \right]$$ which is equal to $$A =\left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$
Now, find the values of a and b.
First, compare the elements of the matrices.
a – 4 = 1
a = 1 + 4
So, the value of a is 5.
Next, the value of b is,
b – 5 = 1
b = 5 + 1
Hence, the value of b is 6.
Therefore, the values of a and b are 5 and 6.

Problem 13: Find the values of ‘m’ and ‘n’ in the given matrix A such that A is a strictly upper triangular matrix.
$$A =\left[ \begin{matrix} 3m & 4\cr n & 5\cr \end{matrix} \right]$$

Solution:
Given the matrix is $$A =\left[ \begin{matrix} 3m & 4\cr n & 5\cr \end{matrix} \right]$$
Now, we will find the values of the m and n.
For the strictly upper triangular matrix, the elements below the diagonal are zero and the elements of the main diagonal are zero.
So, we must have 3m = 0
i.e., m = 0 and n = 0.
Therefore, the values of m and n are 0.

Problem 14:
The matrix is,
$$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$  and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$. What is the final product value of two matrices?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 1 & 0\cr 2 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 6 & 8\cr 4 & 3\cr \end{matrix} \right]$$
Now, we will find the final product value of two matrices.
After, multiplication of two matrix is  $$\left[ \begin{matrix} (1*6) + (0*4) & (1*8) + (0*3) \cr (2*6) + (4*4) & (2*8) + (4*3) \cr \end{matrix}\right]$$
$$\left[ \begin{matrix} 6+0 & 8+0\cr 12+16 & 16+12\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$
Therefore, the final product value of the two matrix is $$\left[ \begin{matrix} 6 & 8\cr 28 & 28\cr \end{matrix} \right]$$

Problem 15:  Check if the matrices $$A = \left[ \begin{matrix} 2 & 4 & 6\cr 8 & 10 & 12\cr 14 & 16 & 18 \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 1 & 3 & 5\cr 7 & 9 & 11\cr 13 & 15 & 17\cr \end{matrix} \right]$$ are equal or not using equality of matrices definition.

Solution:

Given that, the matrices are $$A = \left[ \begin{matrix} 2 & 4 & 6\cr 8 & 10 & 12\cr 14 & 16 & 18 \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 1 & 3 & 5\cr 7 & 9 & 11\cr 13 & 15 & 17\cr \end{matrix} \right]$$
The order of the two matrices is equal i.e, 3 × 3. The same number of rows and columns. So, the first condition for matrix equality is satisfied.
Now, check the elements present in the given matrices.
The (1, 1)th element is 2 and 1.
The (1, 2)th element is 4 and 3.
The elements present in the two matrices are not equal.
Therefore, the given matrices are not equal.

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