Exercise 3.2

cos x = - 2 , x lies in third quadrant.

Solve the following pair of linear equations by the substitution method.

(i) $x + y = 14$ $x - y = 4$

(ii) $s - t = 3$ $6s + 2t = 3$

(iii) $3x - y = 3$ $9x - 3y = 9$

(iv) $0.2x + 0.3y = 1.3$ $0.4x + 0.5y = 2.3$

(v) $2x + 3y = 0$ $3x - 8y = 0$

(vi) $3x + 5y = 22$ $3x - 2y = 6$

Let 2 4 1 3 2 5A , B , C3 2 2 5 3 4 −     = = =      −      Find each of the following: (i) A + B (ii) A - B (iii) 3A - C (iv) AB (v) BA

x z / y t −     + =          

cot (- 15π / 4 )

If / 2 1 10 3 1 5x y −     + =           , find the values of x and y.

+     = +     − +      , find the values of x, y, z and w.

If / cos sin 0 / F ( ) sin cos 0 0 0 1 x x

−   =      , show that F(x) F(y) = F(x + y).

Show that (i) 5 1 2 1 2 1 5 1 / 6 7 3 4 3 4 6 7 − −        ≠               (ii) 1 2 3 1 1 0 1 1 0 1 2 3 / 0 1 0 0 1 1 0 1 1 0 1 0 / 1 1 0 2 3 4 2 3 4 1 1 0 − −                − ≠ −                      

Find A2 - 5A + 6I, if 2 0 1

1 1 0 /    =     − 

If / 1 0 2

2 0 3 /    =      , prove that A3 - 6A2 + 7A + 2I = 0

If 3 2 1 0A and I=4 2 0 1 −   =    −    , find k so that A 2 = kA - 2I

If / 0 tan 2A tan 02 α −  =  α     and I is the identity matrix of order 2, show that I + A = (I - A) cos sin sin cos α − α   α α 

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

sin x = 5 , x lies in second quadrant.

Compute the following: (i) a b a b

    +   −    (ii) 2 2 2 2 / 2 2 2 2 / 2 2 / 2 2 a b b c ab bc

  + +  +   − −+ +     (iii) 1 4 6 12 7 6 / 8 5 16 8 0 5 / 2 8 5 3 2 4 − −        +           (iv) 2 2 2 2 / 2 2 2 2 / cos sin sin cos / sin cos cos sin

    +          

Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of $m$ for which $y = mx + 3$.

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. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

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

If n = p, then the order of the matrix 7X - 5Z is: (A) p × 2 (B) 2 × n (C) n × 3 (D) p × n 3.5. Transpose of a Matrix In this section, we shall learn about transpose of a matrix and special types of matrices such as symmetric and skew symmetric matrices. Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A′ or (AT). In other words, if A = [aij]m × n, then A′ = [aji]n × m. For example, if 2 3 3 2 3 5 3 3 0 A 3 1 , then A 15 10 1 5 × ×       = ′ =  −    −     3.5.1 Properties of transpose of the matrices We now state the following properties of transpose of matrices without proof. These may be verified by taking suitable examples. For any matrices A and B of suitable orders, we have (i) (A′)′ = A, (ii) (kA)′ = kA′ (where k is any constant) (iii) (A + B)′ = A′ + B′ (iv) (A B)′ = B′ A′ Example 20 If 2 1 23 3 2A and B 1 2 44 2 0   − = =      , verify that (i) (A′)′ = A, (ii) (A + B)′ = A′ + B′, (iii) (kB)′ = kB′, where k is any constant. Solution (i) We have A = ( ) 3 4 3 3 2 3 3 2A 3 2 A A 4 2 0 4 2 02 0        ′′ ′⇒ = ⇒ = =           Thus (A′)′ = A (ii) We have A = 3 3 2 , 4 2 0       B = 2 1 2 5 3 1 4A B1 2 4 5 4 4  −  −⇒ + =       Therefore (A + B)′ = 5 5 3 1 4 4 4     −     Now A′ = 3 4 2 1 3 2 , B 1 2 , 2 0 2 4       ′ = −          So A′ + B′ = 5 5 3 1 4 4 4     −     Thus (A + B)′ = A′ + B′ (iii) We have kB = k 2 1 2 2 2 1 2 4 2 4

− −   =       Then (kB)′ = 2 2 1 2 1 2 B 2 4 2 4 k k

k k         ′− = − =           Thus (kB)′ = kB′ Example 21 If [ ] A 4 , B 1 3 6 −   = = −     , verify that (AB) ′ = B′A′. Solution We have A = [ ] 4 , B 1 3 6 −    = −     then AB = [ ] 4 1 3 6 −    −     = 2 6 12 4 12 24 5 15 30 − −   −   −  Now A′ = [-2 4 5] , 1    ′ =    −  B′A′ = [ ] 1 2 4 5 3 2 4 5 6 12 15 (AB) 6 12 24 30 −        ′− = − =       − − −    Clearly (AB)′ = B′A′

Compute the indicated products. (i) a b a b

−       −    (ii) 1 2/3 /   /        [2 3 4] (iii) 1 2 1 2 3 2 3 2 3 1 −           (iv) 2 3 4 1 3 5 / 3 4 5 0 2 4 / 4 5 6 3 0 5 −                   (v) 2 1 1 0 13 2 1 2 11 1         −  −  (vi) 2 33 1 3 1 01 0 2 3 1 − −      −    

cot x = 4 3 , x lies in third quadrant.

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is $26$ and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by $18^\circ$. Find them.

(iii) The coach of a cricket team buys $7$ bats and $6$ balls for $\text{₹ } 3800$. Later, she buys $3$ bats and $5$ balls for $\text{₹ } 1750$. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10 \text{ km}$, the charge paid is $\text{₹ } 105$ and for a journey of $15 \text{ km}$, the charge paid is $\text{₹ } 155$. What are the fixed charges and the charge per $\text{km}$? How much does a person have to pay for travelling a distance of $25 \text{ km}$?

(v) A fraction becomes $\dfrac{2}{3}$, if $2$ is added to both the numerator and the denominator. If, $3$ is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

If / 1 2 3 3 1 2 4 1 2 A 5 0 2 , B 4 2 5 and C 0 3 2 1 1 1 2 0 3 1 2 3 − −           = = =           − −      , then compute (A+B) and (B - C). Also, verify that A + (B - C) = (A + B) - C.

sec x = 5 , x lies in fourth quadrant.

If / 2 5 2 31 13 3 5 5 1 2 4 1 2 4A and B3 3 3 5 5 5 7 2 7 6 223 3 5 5 5                = =                   , then compute 3A - 5B.

tan x = - 12 , x lies in second quadrant. Find the values of the trigonometric functions in Exercises 6 to 10.

sin 765° 7. cosec (- 1410°)

Simplify cos sin sin coscos + sinsin cos cos sin θ θ θ − θ   θ θ   − θ θ θ θ   

Find X and Y, if (i) 7 0 3 0X + Y and X - Y2 5 0 3    = =       (ii) 2 3 2 22X + 3Y and 3X 2Y4 0 1 5 −   = + =   −   

tan 19π / 3 9. sin (- 11π 3 )

Find X, if Y = 3 2 1 4      and 2X + Y = 1 0 3 2    − 

y / x      + =          