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The content of this article has been merged with Chapter 3 Matrices- Exercise 3.2 as per the revised syllabus of NCERT.
Chapter 3 of the Class 12 NCERT Mathematics Part I textbook, titled "Matrices," introduces the fundamental concepts and operations related to matrices. This chapter covers various matrix operations, including addition, subtraction, multiplication, and finding determinants and inverses. Exercise 3.2 | Set 2 focuses on applying these concepts to solve problems involving matrix operations and properties.
This section provides detailed solutions for Exercise 3.2 | Set 2 from Chapter 3 of the Class 12 NCERT Mathematics Part I textbook. The exercise includes problems that require applying matrix operations and properties to solve various tasks. Solutions are presented step-by-step to help students understand and apply matrix concepts effectively.
Solution:
Given:
Equating corresponding entries, we have
2x - y = 10 -(1)
3x + y = 5 -(2)
Adding eq.(1) and (2), we have 5x = 15 ⇒ x = 3
Putting x = 3 in eq.(2)
9 + y = 5 ⇒ y = -4
Therefore, x = 3 and y = -4
Solution:
Given:
Equating corresponding entries, we have
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
and 3y = 6 + x + y
⇒ 2y = 6 + 2
⇒ 2y = 8
⇒ y = 4
and 3z = -1 + z + w ⇒ 2z - w = - 1 -(1)
and 3w = 2w + 3 ⇒ w = 3
Putting w = 3 in eq(i), 2z - 3 = -1
⇒ 2z = 2 ⇒ z = 1
Therefore, x = 2, y = 4, z = 1, w = 3
Solution:
= F(x + y)
= F(x) F(y) = F(x + y)
Solution:
(i) L.H.S =
R.H.S =
Therefore, from (1) and (2), we get
i.e. L.H.S. ≠ R.H.S
(ii) L.H.S =
Multiply both the matrices
R.H.S.=
Therefore,
L.H.S. ≠ R.H.S.
i.e.
Solution:
Solution:
= 0 (Zero matrix)
= R.H.S.
Hence Proved
Solution:
Given:
Equating corresponding entries, we have
3k - 2 = 1
3k = 3
k = 1
and 4k = 4
k = 1
and -4 = -2k - 2
2k = 2
k = 1
Therefore, k = 1
Solution:
L.H.S. = R.H.S.
Hence, Proved.
Solution:
Let invested in the first bond = Rs x
Then, the sum of money invested in the second bond = ₹(30000 – x)
It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.
Thus, in order to obtain an annual total interest of ₹1800, we get:
⇒ 5x/100 + 7(30000 − x)/100 = 1800
⇒ 5x + 210000 -7x = 180000
⇒ 210000 -2x = 180000
⇒ 2x = 210000 – 180000
⇒ 2x = 30000
⇒ x = 15000
Therefore, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.
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 =
Hence, the interest obtained after one year can be expressed in matrix representation as:
⇒ 5x/100 + 7(30000 − x)/100 = 2000
⇒ 5x + 210000 − 7x = 200000
⇒ 210000 − 2x = 200000
⇒ 2x = 210000 – 200000
⇒ 2x = 10000
⇒ x = 5000
Therefore, in order to obtain an annual total interest of ₹2000, the trust fund should invest ₹5000 in the first bond and the remaining ₹(30000 − 5000) = ₹25000 in the second bond.
Solution:
Let the number of books as 1 × 3 matrix = B =
Let the selling prices of each book is a 3 × 1 matrix S =
Therefore, Total amount received by selling all books = BS =
Therefore, Total amount received by selling all the books = Rs 20,160
(A) k = 3, p = n (B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3 (D) k = 2, p = 3
Solution:
Since, Matrices P and Y are of the orders p × k and 3 × k respectively.
Therefore, matrix PY will be defined if k = 3.
Then, PY will be of the order p × k = p × 3.
Matrices W and Y are of the orders n × 3 and 3 × k = 3 × 3 respectively.
As, 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 × 3.
Matrices PY and WY can be added only when their orders are the same.
Therefore, PY is of the order p × 3 and WY is of the order n × 3.
Thus, we must have p = n.
Therefore, k = 3 and p = n are the restrictions on n, k and p so that PY + WY will be defined.
Therefore, answer is (A)
(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 order 2 × p = 2 × n -(∵ p = n)
Then, 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.
Therefore, answer is (B)
Chapter 3 of the Class 12 NCERT Mathematics Part I textbook, "Matrices," focuses on various matrix operations such as addition, subtraction, multiplication, and finding determinants and inverses. Exercise 3.2 | Set 2 provides problems that apply these concepts to matrix operations. Solutions are detailed to help students understand and perform matrix operations accurately.
