**Exercise 8.1**

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

**Solution:**

We have given that,

Matrix has 4 elements

It can have 1 × 4,4 × 1 or 2 × 2 order.

**(ii) If a matrix has 4 elements, what are the possible orders it can have?**

**Solution:**

We have given that,

Matrix has 4 elements,

It can have 1 × 8, 8 × 1, 2 × 4 or 4 × 2 order.

x = 2

Substituting the value of x in (i)

4 + y = 5

y = 5 – 4

y = 1

Hence x = 2 and y = 1

x = 5 – 3

x = 2

Again we have

y – 4 = 3

y = 3 + 4

y = 7

Hence x = 2 and y = 7

x + 2 = -5

x = -5 – 2

x = -7

Also, we have 5z = -20

z = -20/5

z = – 4

Again from given matrix we have

y^{2} + y – 6 = 0

The above equation can be written as

y^{2} + 3y – 2y – 6 = 0

y (y + 3) – 2 (y + 3) = 0

y + 3 = 0 or y – 2 = 0

y = -3 or y = 2

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

x – 2 = 3 and y = 1

x = 2 + 3

x = 5

Again, we have

a + 2b = 5 ... i

3a – b = 1 …ii

Multiply (i) by 1 and (ii) by 2

a + 2b = 5

6a – 2b = 2

Now by adding above equations we get

7a = 7

a =7/7

a = 1

Substituting the value of a in (i) we get

1 + 2b = 5

2b = 5 -1

2b = 4

b =4/2

b = 2

Comparing the corresponding terms, we get

3 = d

d = 3

Also we have

5 + c = -1

c = -1 – 5

c = -6

Also we have,

a + b = 6 and a b = 8

We know that,

(a – b)^{2} = (a + b)^{2} – 4ab

(6)^{2} – 32 = 36 – 32 = 4 = (± 2)^{2}

a – b = ± 2

If a – b = 2

a + b = 6

Adding the above two equations we get

2a = 4

a =4/2

a = 2

b = 6 – 4

b = 2

Again, we have a – b = -2

and a + b = 6

Adding above equations we get

2a = 4

a =4/2

a = 2

Also, b = 6 – 2 = 4

a = 2 and b = 4

Comparing the corresponding terms, we get

3x + 4y = 2 ….i

x – 2y = 4 …..ii

Multiply (i) by 1 and (ii) by 2

3x + 4y = 2

2x – 4y = 8

Adding the above equations we get

5x = 10

x =10/5

x = 2

Substituting the value of x in (i)

We get

6 + 4y = 2

4y = 2 – 6

4y = -4

y = -1

**Exercise 8.2**

**EXERCISE : 8.3**

⇒b(b-1)-2(b-1)=0

⇒(b-1)(b-2)=0

Either b-1=0,

Then, b=1 or b-2=0,

Then, b=2

Hence a=2,b=2 or 1

Now, compare both the elements

3a=4+a

a-a=4

2a=4

∴ a=2

3b=a+b+6

3b-b=2+6

2b=8

∴b=4

3d=3+2d

3d-2d=3

∴d=3

3c=c+d-1

3c-c=3-1

2c=2

∴c=1

Hence a=2,b=4,c=1 and d=3