 # NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part VI)

Find Class 12 Maths NCERT Solutions for Chapter 3 Matrices. In this article, you will find solutions of exercise 3.2 from question number 17 to question number 22. These questions are important for CBSE Class 12 Maths board exam & engineering entrance exams. NCERT Solutions for CBSE Class 12th Mathematics, Chapter 3: Matrices are available here. In this article, you will find solutions of exercise 3.2 (from question number 17 to question number 22). These questions are related to operations on Matrices. These questions are also important for Class 12 Maths board exam 2018 and other engineering entrance exams like JEE Mains, JEE Advanced etc.

NCERT Solutions for CBSE Class 12th Maths, Chapter 3: Matrices (Exercise 3.2) from question number 11 to 16 are given below:

Question 17: Solution 17: Download NCERT Solutions for Class 12 Maths (Chapter 3 - Matrices) in PDF format

NCERT Exemplar Questions: CBSE Class 12 Mathematics – Chapter 3

Questtion 18: Solution 18: Question 19:

A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1800

(b) Rs 2000

Solution 19: Question 20: The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Solution 20: Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Question 21: The restriction on n, k and p so that PY + WY will be defined are:

(A) k = 3, p = n

(B) k is arbitrary, p = 2

(C) p is arbitrary, k = 3

(D) k = 2, p = 3

Solution 21:

Order of matrix P = p × k

Order of matrix Y= 3 × k

Consequently, PY will be of the order = p × k

Order of matrix W = n × 3 Order of matrix Y = 3 × k

Matrix WY is well-defined and is of the order n × k

Matrices PY and WY can be added only when their orders are the same.

However,

Order of PY = p × k Order of WY = n × k

Therefore, p = n

Thus, k = 3 and p = n are the restrictions on n, k, and p so that PY + WY will be defined.

Question 22: If n = p, then the order of the matrix 7X – 5Z is:

(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n

Solution 22:

Order of matrix X = 2 × n

Therefore, Order of matrix 7X is also of the same order.

Order of matrix Z = 2 × p, i.e., 2 × n [Since n = p]

Therefore, Order of matrix 5Z is also of the same order.

Now, Order of matrices 7X and 5Z = 2 × n

Thus, Matrix 7X − 5Z is well-defined and is of the order 2 × n.