Looking for NCERT Solutions for Class 12 Maths Exercise 3.1? If yes then you are in the right place. Here we have provided you with the Question and Answers of Exercise 3.1. These solutions are written by expert teachers and are so accurate to rely on.

Chapter | 3. Matrics |

Exercise | 3.1 |

Class | Twelve |

Subject | Maths |

Category | NCERT Solutions for Class 12 |

NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.

## NCERT Solutions for Class 12 Maths Exercise 3.1

### Matrics Solutions

**Q 1) In the matrix ** **, write:**

**(i) The order of the matrix.****(ii) The number of elements.****(iii) Write the elements **

**Answer) **

**(i) **There are 3 horizontal lines (rows) and 4 vertical lines (columns) in the given matrix A.

Therefore, Order of the matrix is 3 x 4.

**(ii)** The number of elements in the matrix A is 3 x 4 = 12.

**(iii)** a_{13} Element in first row and third column = 19

a_{21} Element in second row and first column = 35

a_{33} Element in third row and third column = -5

a_{24 }Element in second row and fourth column = 12

a_{23 }Element in second row and third column = 5/2

**Q 2) If a matrix has 24 elements, what are possible orders it can order? What, if it has 13 elements?**

**Answer)** We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4)

Hence, the possible orders of a matrix having 24 elements are:

1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

**Q 3) If a matrix has 18 elements, what are the possible orders it can have? What if has 5 elements?**

**Answer)** We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)

Hence, the possible orders of a matrix having 18 elements are:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

**Q 4) Construct a 2 x 2 matrix A = [a**_{ij }] whose elements are given by:

_{ij }] whose elements are given by:

**Answer)**

**Q 5) Construct a 3 x 4 matrix, whose elements are given by:**

**Answer)**

**Q 6) Find the values of x,y and z from the following equations:**

**Answer)**

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x = 1, y = 4 = z = 3

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x + y = 6, xy = 8 and 5 + z = 5

Now, 5 + z = 5 ⇒ z = 0

We know that:

(x – y)^{2} = (x + y)^{2} – 4xy

⇒ (x – y)^{2} = 36 – 32 = 4

⇒x – y = ± 2

Now, when x – y = 2 and x + y = 6 , we get x = 4, y = 2

When x – y = -2 and x + y = 6, we get x = 2, y = 4

∴ x = 4, y = 2, and z = 0, or x = 2, y = 4 and z = 0

**Q 7) Find the values of a,b, c and d from the equation**

**Answer)**

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

a − b = −1 … (1)

2a − b = 0 … (2)

2a + c = 5 … (3)

3c + d = 13 … (4)

From (2), we have:

b = 2a

Then, from (1), we have:

a − 2a = −1

⇒ a = 1

⇒ b = 2

Now, from (3), we have:

2 ×1 + c = 5

⇒ c = 3

From (4) we have:

3 ×3 + d = 13

⇒ 9 + d = 13 ⇒ d = 4

∴a = 1, b = 2, c = 3, and d = 4

**Q** **8) A = [a**_{ij}]_{m x n }is a square matrix if:

_{ij}]

_{m x n }is a square matrix if:

**(A) m < n (B) m > n (C) m = n (D) None of these**

**Answer)**

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. A = [a_{ij}]_{m x n}

option (C) is correct.

**Q 9) Which of the given values of x and y make the following pairs of matrices equal:**

**Answer)**

We find that on comparing the corresponding elements of the two matrices, we get two different values of x , which is not possible.

Hence, it is not possible to find the values of x and y for which the given matrices are equal.

Therefore, option (B) is correct.

**Q 10) The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:**

**(A) 27****(B) 18****(C) 81****(D) 512**

**Answer)**

The correct answer is D.

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 2^{9} = 512

That’s it. These were the solutions of NCERT Class 12 Maths Exercise 3.1. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.