# Problems on Understanding Matrices | Matrix Problems and Solutions

Various tricks and tips are included in the Matrices Worksheet for easy solving of matrix questions. Check problems on matrices along with the solutions. Find the step-by-step procedure to know how to solve matrix problems. Refer to all the solutions present in the below sections to prepare for the exam.

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### Matrices Questions with Solutions

Problems 1: Multiply the given matrix by 3. The matrix is,
$$A = \left[ \begin{matrix} 1 & 2\cr 3 & 4\cr \end{matrix} \right]$$ What is the final value of Matrix?

Solution:
As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 1 & 2\cr 3 & 4\cr \end{matrix} \right]$$
Now, multiply given matrix by 3.
After, multiplication the matrix is $$A = \left[ \begin{matrix} 1 *3 & 2*3\cr 3*3 & 4*3\cr \end{matrix} \right]$$
Therefore, the final matrix value of A is $$A = \left[ \begin{matrix} 3 & 6\cr 9 & 12\cr \end{matrix} \right]$$

Problem 2: Matrix A is of 4×1 order. Matrix B is of 1×3 order. Both the products of AB and BA are defined or not?

Solution:
In the given question,
The Matrix A is of order 4×1.
The Matrix B is of order 1×3.
The product AB will be defined as the number of columns of A is the same as the number of rows of the B.
By using the multiplication of matrices rule, the product matrix hence obtained is of order 4×3.
Next, the product BA is not defined as the number of columns of B is not equal to the number of rows of A.
So, we can say that the matrix multiplication is not commutative, AB is not necessarily equal to BA and sometimes one of the products may also be not defined.
Hence, the product AB is defined but not BA.

Problem 3: 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 = 1, a12 = 4 and b11 = -1, 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 corresponding elements of the two matrices A and B.
Now, adding the corresponding elements, we have
C11= a11 + b11  = 1 + (-1) = 0
Next, the value is C12= a12 + b12 = 4 + (-8) = -4
Therefore, the value of C11 is 0, and the value of C12 is -4.

Problem 4: The matrix $$A = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$. What is the added value of Matrices A and B?

Solution:
As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$
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 $$A +B= \left[ \begin{matrix} 11+7 & 2+8\cr 3+6 & 4+5\cr \end{matrix} \right]$$
Therefore, the addition matrices value of A +B is $$\left[ \begin{matrix} 18 & 10\cr 9 & 9\cr \end{matrix} \right]$$

Problem 5: Determine the element of the second row and third column of the matrix X + Y using the addition of matrices definition. If x23 = -19 is the element of X and the y23 = 30 is the element of Y.

Solution:
As given in the question, X23 = -19 is the element of X and Y23= 30 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
= -19 + 30 = 11.
Therefore, the element in the second row and third column of A + B is 11.

Problem 6:  Find the transpose of a column matrix $$A = \left[ \begin{matrix} 3 \cr 1 \cr 7\cr \end{matrix} \right]$$

Solution:
The given matrix is $$A = \left[ \begin{matrix} 3 \cr 17 \cr \end{matrix} \right]$$
To find the transpose of the given column matrix, the elements of the column matrix must be written as row elements.
A=[3 17]
Therefore, the transpose of the given column matrix is A=[3 17]

Problem 7: What is the 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 8: Write the elements of the matrix C = A – B explicitly. If A = [4 10 18] and B = [2 18 24] using matrix subtraction formula?

Solution:
The given matrices are A = [4 10 18] and B = [2 18 24].
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 = [4-2 10-18 18-24]
= [2 -8 -6]
Hence, the elements of C = A – B are C11 = 2, C12 = -8, C13 = -6.

Problem 9: Subtract X and Y where
$$X = \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$

Solution:
Given matrices are $$X = \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$
Both the matrices have the same order of 2 × 2. Now, we can subtract the elements of the first matrix with the respective elements of the second matrix.
So, the matrices X-Y is $$= \left[ \begin{matrix} 3 & 8\cr 2 & -5\cr \end{matrix} \right]$$ – $$\left[ \begin{matrix} -4 & -3\cr 2 & 1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 3 +4 & 8+3 \cr 2-2 & -5-11\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 7 & 11\cr 0 & -6\cr \end{matrix} \right]$$
Thus, the subtraction of X and Y is $$\left[ \begin{matrix} 7 & 11\cr 0 & -6\cr \end{matrix} \right]$$

Problem 10: 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 11: Find 3B where
$$B= \left[ \begin{matrix} 4 & 5\cr 11 & 10\cr \end{matrix} \right]$$

Solution:
The given matrix is $$B = \left[ \begin{matrix} 4 & 5\cr 11 & 10\cr \end{matrix} \right]$$
To find the value of 3B, we need to multiply the value of 3 by all the elements of matrix B.
i.e., 3B= $$\left[ \begin{matrix} 3*4 & 3*5\cr 3*11 & 3*10\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 12 & 15\cr 33 & 30\cr \end{matrix} \right]$$
Therefore, the required matrix is 3B $$=\left[ \begin{matrix} 12 & 15\cr 33 & 30\cr \end{matrix} \right]$$

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