ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise 8.1

 Exercise 8.1

Question 1

(i) [2151]

(ii)[2 3 – 7]

(iii) [301]

(iv) [240017]

(v) [278120]

(vi) [000000]

Sol :

(i) It is square matrix of order 2

(ii) It is row matrix of order 1 × 3

(iii) It is column matrix of order 3 × 1

(iv) It is matrix of order 3 × 2

(v) It is matrix of order 2 × 3

(vi) It is zero matrix of order 2 × 3


Question 2

(i) If a matrix has 4 elements, what are the possible order it can have ?

(ii) If a matrix has 8 elements, what are the possible order it can have ?

Sol :

(i) It can have 1 × 4, 4 × 1 or 2 × 2 order

(ii) It can have 1 × 8, 8 × 1,2 × 4 or 4 × 2 order


Question 3

Construct a 2 x 2 matrix whose elements aij are given by

(i) aij=2ij

(ii) aij=i.j

Sol :

(i) It can be [1032]

(ii) It can be [1224]


Question 4

Find the values of x and y if : [2x+y3x2y]=[54]

Sol :

Comparing corresponding elements,

2x + y = 5 …(i)

3x – 2y = 4 …(ii)

Multiply (i) by 2 and (ii) by ‘1’ we get

4x + 2y = 10, 3x – 2y = 4

Adding we get, 7x = 14 ⇒ x = 2

Substituting the value of x in (i)

2 x 2 + y = 5 ⇒ 4 + y = 5

y = 5 – 4 = 1

Hence x = 2, y = 1


Question 5

Find the value of x if [3x+yy2yx3]=[1253]

Sol :

[3x+yy2yx3]=[1253]

Comparing the corresponding terms, we get.

-y = 2

⇒ y = -2

3x+y=13x+1y

3x=1(2)=1+2=3

x=33=1

Hence, x=1, y=-2


Question 6

If [x+34y4x+y]=[5439], find values of x and y

Sol :

[x+34y4x+y]=[5439]

Comparing the corresponding terms, we get.

x + 3 = 5

⇒ x = 5 – 3 = 2

⇒ y – 4 = 3

⇒ y = 3 + 4 = 7

x = 2, y = 7


Question 7

Find the values of x, y and z if

[x+2635z]=[5y2+y320]

Sol :

Comparing the corresponding elements of equal determinents,

x + 2 = -5

⇒ x = -5 – 2 = -7

x=7,5z=20

z=205=4

y2+y=6

y2+y6=0

y2+3y2y6=0

y(y+3)2(y+3)=0

(y+3)(y2)=0

Either y+3=0

then y=-3 or y-2=0, then y=2

Hence, x=-7, y=-3,2,z=-4


Question 8

Find the values of x, y, a and b if

[x2ya+2b3ab]=[3151]

Sol :

Comparing corresponding elements

x – 2 = 3, y = 1

x = 3 + 2 = 5

a + 2b = 5 ……(i)

3a – b = 1 ……..(ii)

Multiplying (i) by 1 and (ii) by 2 
a+2 b=5, 6a-2b=2
Adding, we get 7a=7 
⇒ a=1
Substituting the value of a in..(i)
1+2 b=5 
⇒ 2b-5-1=4 
⇒ b=2
Hence x=5, y=1, a=1, b=2

Question 9

Find the values of a, b, c and d if

[a+b35+cab]=[6d18]

Sol :
[a+b35+cab]=[6d18]

Comparing the corresponding terms, we get.
3 = d ⇒ d = 3
⇒ 5 + c = – 1
⇒ c = -1 – 5
⇒ c = -6
a + b = 6 and ab = 8

(ab)2=(a+b)24ab

=(6)24×8=3632=4=(±2)2
ab=±2
(i) If a-b=2
a+b=6
Adding , we get 2a=8
⇒a=4

a+b=6
⇒4+b=6
⇒b=6-4=2
∴a=4, b=2

(ii) If a-b=-2
a+b=6
Adding , we get 2a=4
a=42=2
a+b=62+b=6b=622=4
a=2,b=4
Hence a=4,b=2, or a=2,b=4
c=6 and d=3

Question 10

Find the values of x, y, a and b, if
[3x+4y2x2ya+b2ab1]=[224551]
Sol :
Comparing the corresponding terms, we get.
3x + 4y = 2 ……(i)
x – 2y = 4 …….(ii)
Multiplying (i) by 1 and (ii) by 2
3x+4y=2
2 x-4 y=8

Adding we get, 5x=10
⇒x=2
3×2+4y=2,6+4y=2,4y=26=4
y=-1

x=2,y=1
a+b=5..(iii)
2a-b=-5...(iv)

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2