# NCERT Solutions for Class 12 Maths Exercise 3.2

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## NCERT Solutions for Class 12 Maths Exercise 3.2

### Matrics Solutions

(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA

i)

#### Q12) Given:

find the values of x ,y,z and w

#### Q 16) Ifprove that A3 – 6A2 + 7A + 2I = 0.

L.H.S. = A3 – 6A2 + 7A + 2I

= R.H.S.      Proved.

#### Q 19) 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.

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.

#### Q 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.

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.

#### Q 21) 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

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.

#### Q 22) 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

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.