NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 Matrices

Class 12 Maths Chapter 3 Exercise 3.2 Matrices:- NCERT Solutions for Class 12 Maths Chapter 3: Matrices offer comprehensive answers to all questions in Exercise 3.2. Our faculty have meticulously crafted these solutions to address any doubts students might encounter while working through the problems in the NCERT Maths textbook.

 This chapter covers the definition and types of matrices, matrix operations such as addition, subtraction, and multiplication, and concepts like the transpose of a matrix, symmetric and skew-symmetric matrices. Exercise 3.2 specifically focuses on the various properties and operations related to matrices, challenging students to apply these concepts through a series of problems. Check out the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 Matrices below.

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NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 Matrices

Go through the Class 12 Maths Chapter 3 Exercise 3.2 Matrices NCERT Solutions below:- 

Question1. Let A = . Find each of the following:

(i) A + B

(ii) A – B

(iii) 3A – C

(iv) AB

(v) BA

Solution :
(i) 

(ii)

(iii)

(iv)

(v) 

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Question2. Compute the following:

(i)

(ii)

(iii)

(iv)

Solution :

(ii)

(iii)

(iv)

Question3. Compute the indicated products:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution :
(i)

(ii)

(iii) =

(iv)=

(v)=

(vi)=

Question4. If   then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Solution :

L.H.S. = R.H.S.  Proved.

Question5. Ifthen compute 3A – 5B.

Solution :

Question6. Simplify:

Solution :

Question7. Find X and Y, if:

(i) 

(ii)

Solution :
(i) 

(ii)

 

Question8. Find X if Y =

Solution :

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Question9. Find x and y if

Solution :

Question10. Solve the equation for x,y,z and t if

Solution :

Question11. Iffind the values of x and y

Solution :

Comparing the corresponding elements of these two matrices, we get:

2x − y = 10 and 3x + y = 5

Adding these two equations, we have:

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 − 3x

⇒ y = 5 − 9 = −4

∴x = 3 and y = −4

Question12. Given: find the values of x ,y,z and w

Solution :

 

Question13. If show that F(x) F(y) = F(x + y)

Solution :

Question14. Show that:

(i)

(ii)

Solution :
(i)

(ii)

Question15. Find A2 – 5A + 6I if A = .

Solution:

Question16. If A = prove that A3 – 6A2 + 7A + 2I = 0.

Solution :
L.H.S. = A3 – 6A2 + 7A + 2I

= R.H.S.  Proved.

Question17. If  find k so that A2 = kA – 2I

Solution :

Question18. If  and I is the identity matrix of order 2, show that I + A = (I  – A)

Solution :

Question19. A trust fund has ` 30,000 that must be invested in two different types of bond. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 in two types of bonds, if the trust fund must obtain an annual interest of (a) ` 1800, (b) ` 2000.

Solution :
(a) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x ​30,000 − x​]

Annual interest obtained is Rs. 1800. We know, Interest = PTR/100

Here, the time is one year and thus T = 1.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒  5x + 210000 − 7x = 180000

⇒ – 2x = – 30000

∴  x = 15000

Amount invested in the first bond = x = Rs. 15000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 15000) = Rs. 15000

∴ The trust has to invest Rs. 15000 each in both the bonds in order to obtain an annual interest of Rs. 1800.

(b) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x ​30,000 − x​]

Annual interest obtained is Rs. 2000.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒  5x + 210000 − 7x = 200000

⇒  −2x = − 10000

∴  x = 5000

Amount invested in the first bond = x = Rs. 5000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 5000) = Rs. 25000

∴ The trust has to invest Rs.5000 in the first bond and Rs.25000 in the second bond in order to obtain an annual interest of Rs.2000.

Question20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ` 80, ` 60 and ` 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Solution :
The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.

The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

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Question21. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k repectively. The restriction on n,k and p so that PY + WY will be define are:

A. k = 3, p = n

B. k is arbitrary, p = 2

C. p is arbitrary, k = 3

D. k = 2, p = 3

Solution :
Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k.

Matrices W and Y are of the orders n × 3 and 3 × k respectively.

Since the number of columns in W is equal to the number of rows in Y, 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, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n.

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

Therefore, option (A) is correct.

Question22. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k repectively. If n = p then order of matrix 7X – 5Z is:

(A) p × 2

(B) 2 × n   

(C) n × 3

(D)p × n

Solution :

Matrix X is of the order 2 × n.

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

Matrix Z is of the order 2 × p, i.e., 2 × n [Since n = p]

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

Now, both the matrices 7X and 5Z are of the order 2 × n.

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

The correct answer is B.

Class 12 Maths Chapter 3 Exercise 3.2 Matrices Summary

Types of Matrices

• Square Matrix: A matrix with the same number of rows and columns (e.g., a 2x2 matrix or a 3x3 matrix).

• Row Matrix: A matrix with just one row (e.g., 1x3 matrix).

• Column Matrix: A matrix with just one column (e.g., 3x1 matrix).

• Diagonal Matrix: A square matrix where all elements except those on the main diagonal are zero (e.g., entries like a11, a22, a33 on a 3x3 diagonal matrix).

• Scalar Matrix: A diagonal matrix where all the diagonal elements are the same constant value (e.g., all diagonal entries are 5 in a 3x3 scalar matrix).

• Identity Matrix: A square matrix with 1s along the main diagonal and 0s elsewhere (e.g., a 2x2 identity matrix would be [[1, 0], [0, 1]]).

Properties of Matrices

• Addition: Applicable only for matrices with the same dimensions (equal rows and columns). Corresponding items between matrices are added.

• Subtraction: Similar to addition, applicable only for matrices with the same dimensions. Corresponding items are subtracted.

• Scalar Multiplication: Each term of the matrix is multiplied by a constant scalar value.

• Matrix Multiplication: Applicable for specific dimensions. The resulting product matrix will have dimensions based on the number of columns in the first matrix and the number of rows in the second matrix.

• Transpose: Reverses the rows and columns of a matrix.

Special Matrices

• Symmetric Matrix: A square matrix where the elements above the diagonal are mirrored by the elements below the diagonal (i.e., aij =aji for all i and j).

• Skew-Symmetric Matrix: A square matrix where the elements above the diagonal are negative counterparts of the elements below the diagonal (i.e., aij = -aji for all i ≠ j) and the diagonal elements are zero (i.e., aii = 0 for all i).

 Class 12 Maths Chapter 3 Exercise 3.2 Matrices FAQs

Q1. What is a matrix?

Ans. A matrix is a rectangular array of numbers arranged in rows and columns.

Q2. What are the different types of matrices?

Ans. Types include row matrix, column matrix, square matrix, diagonal matrix, scalar matrix, identity matrix, and zero matrix.

Q3. How do you add two matrices?

Ans. To add two matrices, add the corresponding elements of each matrix. Both matrices must have the same dimensions.

Q4. What is matrix multiplication?

Ans. Matrix multiplication involves multiplying rows of the first matrix by columns of the second matrix. The number of columns in the first matrix must equal the number of rows in the second matrix.

Q5. How is the transpose of a matrix defined?

Ans. The transpose of a matrix is obtained by interchanging its rows and columns.