Class 12-commerce 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 8
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 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 2 (i)
Solution 2 (ii)
Solution 2 (iii)
Solution 2 (iv)
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 10(v)
Solution 10(vi)
Solution 11
Solution 12(i)
Solution 12(ii)
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 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 1(viii)
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 2(v)
Solution 2(vi)
Solution 2(vii)
Solution 2(viii)
Solution 2(ix)
Solution 2(x)
Solution 2(xi)
Solution 2(xii)
Solution 2(xiii)
Solution 2(xiv)
Solution 2(xv)
Solution 2(xvi)
Solution 2(xvii)
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52(i)
Solution 52(ii)
Solution 52(iii)
Solution 52(v)
Solution 52(vi)
Solution 52(vii)
Solution 52(viii)
Solution 52(ix)
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. 6.3
Solution 1 (i)
Solution 1 (ii)
Solution 1 (iii)
Solution 1 (iv)
Solution 2 (i)
Solution 2(ii)
Solution 2 (iii)
Solution 2 (iv)
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12 (i)
Solution 12 (ii)
Solution 13 (i)
Solution 13 (ii)
Determinants Exercise Ex. 6.4
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Determinants Exercise Ex. 6.5
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Determinants Exercise Ex. 6VSAQ
Solution 53
We know that,
maximum value of sin is 1 as 
Solution 54
Solution 55
Solution 56
Given: Det (A-1) = (Det A)k
Solution 57
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.
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
