Question-1 :-
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
(i)![]()
(ii)![]()
(iii)![]()
(iv)![]()
(v)![]()
Question-2 :-
Compute the following:
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(ii)![]()
(iii)![]()
(iv)![]()
Question-3 :-
Compute the indicated products:
(i)![]()
(ii)![]()
(iii)![]()
(iv)![]()
(v)![]()
(vi)![]()
Question-4 :-
then compute (A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Question-5 :-
Question-6 :-
Question-7 :-
Find X and Y, if
(i)![]()
(ii)![]()
Question-8 :-
Question-9 :-
Question-10 :-
Solve the equation for x, y, z and t, if
Question-11 :-
Question-12 :-
find the values of x, y, z and w.
Question-13 :-
Question-14 :-
Show that :
(i)![]()
(ii)![]()
Question-15 :-
Question-16 :-
Question-17 :-
Question-18 :-
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
(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
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. The correct answer is A.
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
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.