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| Board | SCERT, Kerala |
| Text Book | NCERT Based |
| Class | Plus Two |
| Subject | Math's Textbook Solution |
| Chapter | Chapter 3 |
| Exercise | Ex 3.2 |
| Chapter Name | Matrices |
| Category | Plus Two Kerala |
Kerala Syllabus Plus Two Math's Textbook Solution Chapter 3 Matrices Exercises 3.2
Chapter 3 : Matrices Solution
Chapter 3: Matrices Exercise 3.2
If 
and
then compute
.
Simplify 
If
, find values of x and y.
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
Let 
Find each of the following
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(i)
(ii)
(iii)
(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:
(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as:
Compute the following:
(i) 
(ii) 
(iii) 
(iv) 
In the addition of matrices , we add similar row and column matrix with another matrix . For example in first part a will be added with a , b with b , -b with b and a with a .
Compute the indicated products
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
If
, and
, then compute 
and
. Also, verify that 
Find X and Y, if
(i) 
and
(ii) 
and
(i)
Adding equations (1) and (2), we get:
(ii)
Multiplying equation (3) with (2), we get:
Multiplying equation (4) with (3), we get:
From (5) and (6), we have:
Now,
Comparing the corresponding elements of these two matrices, we have:
∴x = 3 and y = 3
Solve the equation for x, y, z and t if
Comparing the corresponding elements of these two matrices, we get:
Given
, find the values of x, y, z and w.
Comparing the corresponding elements of these two matrices, we get:
If
, show that
.
(i) 
(ii) 
(i)
(ii)
Find 
if
We have A2 = A × A
If
, prove that 
If 
and
, find k so that 
If
and I is the identity matrix of order 2, show that 
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 1,800 (b) Rs 2,000
(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x).
It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.
Therefore, in order to obtain an annual total interest of Rs 1800, we have:
Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.
(b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x).
Therefore, in order to obtain an annual total interest of Rs 2000, we have:
Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.
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.
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.
Assume X, Y, Z, W and P are matrices of order
, and 
respectively. The restriction on n, k and p so that 
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
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 will be defined.
Assume X, Y, Z, W and P are matrices of order
, and 
respectively. If n = p, then the order of the matrix 
is
(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n
The correct answer is B.
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.
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Chapter 3: Matrices Exercise 3.2 Textbook Solution
Chapter 3: Matrices Exercise 3.2 Textbook Solution- Preview
Plus Two Math's Chapter Wise Textbook Solution PDF Download
- Chapter 1: Relations and Functions
- Chapter 2: Inverse Trigonometric Functions
- Chapter 3: Matrices
- Chapter 4: Determinants
- Chapter 5: Continuity and Differentiability
- Chapter 6: Application of Derivatives
- Chapter 7: Application of Integrals
- Chapter 8: Integrals
- Chapter 9: Differential Equations
- Chapter 10: Vector Algebra
- Chapter 11: Three Dimensional Geometry
- Chapter 12: Linear Programming
- Chapter 13: Probability
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