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# Class 12-science RD SHARMA Solutions Maths Chapter 6 - Determinants

## Determinants Exercise MCQ

### Solution 34

Applying C1 C1 + C2 + C3

Applying R2 R2 - R1 and R3 R3 - R1

But tan x = -2 is not possible as for  we get

Therefore,

Thus, only one real root exist.

### Solution 35

Now,

Hence,  is equal to 0.

### Solution 36

Area of a triangle

### Solution 37

For any matrix A of order n × n, |kA| = kn|A| where k is a scalar

|3A| = 33 |A| = 27 × 8 = 216

### Solution 1

Correct option: (d)

### Solution 2

Correct option: (d)

### Solution 3

Correct option: (d)

If A is a square matrix of order n then det(A) = a11C11+a21C21+a31C31

### Solution 4

Correct option: (b)

Minor of an element can never be equal to cofactor of the same element.

Cij=(-1)i+jMij

### Solution 5

Correct option: (e)

### Solution 6

Correct option: (a)

### Solution 7

Correct option: (a)

### Solution 9

Correct option: (a)

### Solution 10

Correct option: (a)

### Solution 11

Correct option: (d)

### Solution 12

Correct option: (b)

### Solution 13

Correct option: (a)

### Solution 14

Correct option: (c)

### Solution 15

Correct option: (c)

### Solution 16

Correct option: (b)

### Solution 17

Correct option: (b)

### Solution 18

Correct option:(a)

### Solution 19

Correct option: (a)

### Solution 20

Correct option: (b)

### Solution 21

Correct option: (d)

### Solution 22

Correct option: (c)

### Solution 23

Correct option: (a)

### Solution 24

Correct option: (c )

### Solution 25

Correct option: (d)

### Solution 26

Correct option: (d)

### Solution 27

Correct option: (a)

### Solution 28

Correct option: (a)

### Solution 29

Correct option: (b)

### Solution 30

Correct option: (a)

### Solution 31

Correct option: (c)

### Solution 32

Correct option: (d)

### Solution 33

Correct option: (c)

## Determinants Exercise Ex. 6.1

### Solution 10(i)

⇒ 2 × 1 - 5 × 4 = 2x × x - 6 × 4

⇒ 2 - 20 = 2x2 - 24

⇒ 2x2 = 6

⇒ x2 = 3

⇒ x =

## Determinants Exercise Ex. 6.2

### Solution 41

Consider the determinant

Applying R1 R1 - R2 and R2 R2 - R3

Taking (a - 1) common from R1 and R2

Now expanding this determinant along C3

### Solution 52(iv)

Given:

Applying R1 R1 - R3

Applying R2 R2 - R3

Taking (x - b) and (a - b) common from R1 and R2 respectively

Expanding along C1, we get

### Solution 52(x)

Given:

Applying C1 C1 + C2 + C3

Applying R1 R1 - R3 and R2 R2 - R3

### Solution 53

Given:

Applying C1 C1 - C2

Applying C2 C2 - C3

Expanding this determinant along R1, we get

### Solution 54

Given:

Applying R2 R2 - R1

Applying R3 R3 - R1

Expanding this determinant along C3, we get

Dividing throughout by xyz, we get

Hence, the value of  is 2.

### Solution 55

Consider the determinant

Applying R2 R2 - R3

Applying R1 R1 - R3

Taking 3 common from R1 and R2

Hence,

## Determinants Exercise Ex. 6VSAQ

### Solution 53

We know that, maximum value of sin is 1 as

### Solution 56

Given: Det (A-1) = (Det A)k

Given: AB = 2I

|AB| = |2I|

|B| |A| = 23 |I|

2|B| = 8

|B| = 4

### Solution 58

A square matrix A is said to be singular, if its determinant is 0, i.e. |A| = 0.