﻿ Class 12 NCERT Math Solution
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TOPICS
Exercise - 3.2

Question-1 :- Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA

Solution :-
```(i) ```
```(ii) ```
```(iii) ```
```(iv) ```
```(v) ```

Question-2 :-  Compute the following: Solution :-
```(i) ```
```(ii) ```
```(iii) ```
```(iv) ```

Question-3 :-  Compute the indicated products: Solution :-
```(i) ```
```(ii) ```
```(iii) ```
```(iv) ```
```(v) ```
```(vi) ```

Question-4 :- then compute (A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Solution :-
``` ```

Question-5 :- Solution :-
``` ```

Question-6 :- Solution :-
``` ```

Question-7 :-  Find X and Y, if Solution :-
```(i) ```
```(ii) ```

Question-8 :- Solution :-
``` ```

Question-9 :- Solution :-
``` ```

Question-10 :-  Solve the equation for x, y, z and t, if Solution :-
``` ```

Question-11 :- Solution :-
``` ```

Question-12 :- find the values of x, y, z and w.

Solution :-
``` ```

Question-13 :- Solution :-
``` ```

Question-14 :-  Show that : Solution :-
```(i) ```
```(ii) ```

Question-15 :- Solution :-
``` ```

Question-16 :- Solution :-
``` ```

Question-17 :- Solution :-
``` ```

Question-18 :- Solution :-
``` ```

Question-19 :-  A trust fund has ₹ 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 ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) ₹ 1800   (b) ₹ 2000

Solution :-
```(i) Let ₹ x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (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 ₹ 1800, we have: Thus, 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.
```
```(ii) Let ₹ x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (30000 − x).
Therefore, in order to obtain an annual total interest of ₹ 2000, we have: Thus, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest
₹ 5000 in the first bond and the remaining ₹ 25000 in the second bond.
```

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 ₹ 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 ₹ 80, ₹ 60, and ₹ 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 ₹ 20160 from the sale of all these books.
```

Question-21 :-  Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. 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 :-
```  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.
```

Question-22 :-  Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. 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 :-
```  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.