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Relations And Functions

Question
CBSEENMA12032435

Find ,x, y , a and b if

open square brackets table row cell 2 straight x minus 3 straight y end cell cell straight a minus straight b end cell 3 row 1 cell straight x plus 4 straight y end cell cell 3 straight a plus 4 straight b end cell end table close square brackets equals open square brackets table row 1 cell negative 2 end cell 3 row 1 6 29 end table close square brackets

Solution

We are given that 
open square brackets table row cell 2 straight x minus 3 straight y end cell cell straight a minus straight b end cell 3 row 1 cell straight x plus 4 straight y end cell cell 3 straight a plus 4 straight b end cell end table close square brackets equals open square brackets table row 1 cell negative 2 end cell 3 row 1 6 29 end table close square brackets

From the definition of equality of matrices, we have,
2    x – 3 v = 1    ...(I)
x + 4 y = 6 ...(2)
a – b= – 2 ...(3)
3a + 4 b = 29    ...(4)
(1) and (2) can be writta/i«s
2 x – 3 y – I = 0
x + 4 y – 6 = 0
therefore space space space space space space space space fraction numerator straight x over denominator 18 plus 4 end fraction equals fraction numerator straight y over denominator negative 1 plus 12 end fraction equals fraction numerator 1 over denominator 8 plus 3 end fraction
therefore space space space space space space space space space straight x over 22 equals straight y over 11 equals 1 over 11
therefore space space space space space space space space straight x equals 22 over 11 equals 2 comma space space space straight y space equals 11 over 11 equals 1
(3) and (4) can be written as
a–b+2-0
3 a + 4 b – 29 = 0
therefore space space space space space fraction numerator straight a over denominator 29 minus 8 end fraction equals fraction numerator straight b over denominator 6 plus 29 end fraction equals fraction numerator 1 over denominator 4 plus 3 end fraction space space rightwards double arrow space straight a over 21 equals straight b over 35 equals 1 over 7
therefore space space space space space straight a equals 21 over 7 equals 3 comma space space space space straight b equals 35 over 7 equals 5
∴ solution is x = 2, y= 1, a = 3, b = 5.

Some More Questions From Relations and Functions Chapter

Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R’ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Give an example of a relation which is

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by R = {(a,b) : a ≤ b3} is refleive, symmetric or transitive.