# RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

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## RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

### Question 1: Write minors and co-factors of each element of first column of the following matrices and hence evaluate determinant.

Solution:

i) Let Mij and Cij  represents minor and co-factor of element. They are placed in ith  row and jth column.

Here, a11 = 5

Minor of a11 = M11 = -1

Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.

Minor of a12 = M12 = 0

Minor of a21 = M21 = 20

Minor of a22 = M22 = 0

As M12 and M22 are zero so we don’t consider them. Hence we have got only two minors for this determinant.

M11 = -1 & M21  = 20

Now, co-factors for the determinants are

C11 = (-1)1+1 x M11                           {∵Cij =(-1)1+1 x Mij}

= (+1)x(-1)

= -1

C21 = (-1)2+1 x M21

= (-1)3 x 20

= -20

Evaluating the determinant,

|A| = a11 x C11 + a21 x C21

=5 x (-1) + 0 x (-20)

= -5

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith  row and jth column.

Minor of a11 = M11 = 3

Note: In 2×2 matrix, minor is obtained for a particular element, by deleting that row and column were element is present.

Minor of a21 = M21 = 4

Now, co-factors for the determinants are

C11 = (-1)1+1 x M11                           {∵Cij =(-1)i+j x Mij}

= (+1) x 3

= 3

C21 = (-1)2+1 x M21

= (-1)3 x 4

= -4

Evaluating the determinant,

|A| = a11 x C11 + a21 x C21

=-1 x 3 + 2 x (-4)

=-11

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith row and jth column.

Cij = (-1)i+j x Mij

Given,

We have,

M11 = -1×2 – 5×2

M11 = -12

M21 = -3×2 – 5×2

M21 = -16

M31 = -3×2 – (-1) x 2

M31 = -4

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1x-12

= -12

C21 = (-1)2+1 x M21

= (-1)3 x -16

= 16

C31 = (-1)3+1 x M31

= (1)4 x (-4)

= -4

To evaluate the determinant expand along first column,

|A| = a11 x C11 + a21 x C21+ a31 x C31

=1x(-12) + 4×16 + 3x(-4)

= -12 + 64 – 12

= 40

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith  row and jth column.

Also, Cij = (-1)i+j x Mij

Given,

We have,

M11 = b x ab – c x ca

M11 = ab2 – ac2

M21 = a x ab – c x bc

M21 = a2b – c2b

M31 = a x ca – b x bc

M31 = a2c – b2c

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1 x (ab2 – ac2)

= ab2 – ac2

C21 = (-1)2+1 x M21

= (-1)3 x (a2b – c2b)

= c2b – a2b

C31 = (-1)3+1 x M31

= (1)4 x (a2c – b2c)

= a2c – b2c

To evaluate the determinant expand along first column,

|A| = a11 x C11 + a21 x C21+ a31 x C31

=1 x (ab2 – ac2) + 1 x (c2b – a2b) + 1 x (a2c – b2c)

= ab2 – ac2 + c2b – a2b + a2c – b2c

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith row and jth column.

Cij = (-1)i+j x Mij

Given,

We have,

M11 = 5×1 – 7×0

M11 = 5

M21 = 2×1 – 7×6

M21 = -40

M31 = 2×0 – 5×6

M31 = -30

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1×5

= 5

C21 = (-1)2+1 x M21

= (-1)3 x -40

= 40

C31 = (-1)3+1 x M31

= (1)4 x (-30)

= -30

To evaluate the determinant expand along first column,

|A| = a11 x C11 + a21 x C21+ a31 x C31

=0x5 + 1×40 + 3x(-20)

= 0 + 40 – 90

= 50

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith  row and jth column.

Cij = (-1)i+j x Mij

Given,

We have,

M11 = b x c – f x f

M11 = bc – f2

M21 = h x c – f x g

M21 = hc – fg

M31 = h x f – b x g

M31 = hf – bg

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1x (bc – f2)

= bc – f2

C21 = (-1)2+1 x M21

= (-1)3 x (hc – fg)

= fg – hc

C31 = (-1)3+1 x M31

= (1)4 x (hf – bg)

= hf – bg

To evaluate the determinant expand along first column,

|A| = a11 x C11 + a21 x C21+ a31 x C31

=a x (bc – f2) + h x (fg – hc) + g x (hf – bg)

= abc – af2 + hgf – h2c + ghf –bg2

Solution:

Let Mij and Cij  represents minor and co-factor of element. They are placed in ith row and jth column

Also, Cij = (-1)i+j x Mij

Given,

From the matrix we have,

M11 = 0(-1 x 0 – 5 x 1) – 1(1 x 0 – (-1) x 1) + (-2)(1 x 5 – (-1) x (-1))

M11 = -9

M21 = -1(-1 x 0 – 5 x 1) – 0(1 x 0 – (-1) x 1) + (1 x 5 – (-1) x (-1))

M21 = 9

M31 = -1(1 x 0 – 5 x (-2)) – 0(0 x 0 – (-1) x (-2)) + 1(0 x 5 – (-1) x 1)

M31 = -9

M41 = -1(1 x 1 – (-1) x (-2)) – 0(0 x 1 – 1 x (-2)) + 1(0 x (-1) – 1 x 1)

M41 = 0

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1x (-9)

= -9

C21 = (-1)2+1 x M21

= (-1)3 x 9

= -9

C31 = (-1)3+1 x M31

= (-1)4 x -9

= -9

C41 = (-1)4+1 x M41

= (-1)5 x 0

= 0

To evaluate the determinant expand along first column,

|A| = a11 x C11 + a21 x C21+ a31 x C31+ a41 x C41

=2 x (-9) + (-3) x (-9) + 1 x (-9) + 2 x 0

= -18 + 27 – 9

= 0

### Question 2: Evaluate following determinants

Solution:

Given, Cross multiplying the values inside the determinant,

|A| = (5x + 1) – (-7)x

|A| = 5x2 = 8x

Solution:

Given,  { Solution:

Given, A∣ = cos15°×cos75°+sin15°×sin75°

As per formula

cos(AB)=cosAcosB+sinAsinB

Substitute this in |A| so we get,

A∣ = cos(75−15)°

A∣ = cos60°

A∣ = 0.5

Solution:

∣A∣ = (a+ib)(a−ib)−(c+id)(−c+id)

Expanding the brackets we get,

∣A∣=(a+ib)(a−ib)+(c+id)(c−id)

|A| = a2-i2b2+c2-i2d2

We know i2 = -1

|A| = a2-1b2+c2-(-1)d2

|A| = a2+b2+c2+d2

### Question 3: Evaluate the following:

Solution:

In the given formula, ∣AB∣=∣A∣∣B

Cross multiplying the terms  in |A|

A∣ = 2(17×12−5×20)−3(13×12−5×15)+7(13×20−15×17)

= 2(204−100)−3(156−75)+7(260−255)

= 2×104−3×81+7×5

= 208−243+45

= 0

Now ∣A2=∣A∣×∣A

A2=0

### Question 4: Show that,

Solution:

Method 1:

Given,

Let the given determinant as A,

Using sin(a+B) = sinA×cosB+cosA×sinB

A∣ = sin10°×cos80°+cos10°×sin80°

A∣ = sin(10+80)°

A∣ = sin90°

A∣ = 1

Method 2:

A∣ = sin10°×cos80°+cos10°×sin80°

[∴cosθ sin(90−θ)]

A∣ = sin10°cos(90°−10°)+cos10°sin(90°−10°)

A∣ = sin10°sin10°+cos10°cos10°

A∣ = sin210°+cos210°

[∴sin2θ+cos2θ = 1]

A∣ = 1

### Question 5: Evaluate the following determinant by two methods.

Solution:

Method 1

Expanding along the first row

A∣ = 2(1×1−4×−2)−3(7×1−(−2)×−3)−5(7×4−1×(−3))

A∣ = 2(1+8)−3(7−6)−5(28+3)

A∣ = 2×9−3×1−5×31

A∣ = 18−3−155

A∣ = −140

Method 2

Here it is Sarus Method, we adjoin the first two columns.

Expanding along second column,

∣A∣ = 2(1×1−4×(−2))−7(3×1−4×(−5))−3(3×(−2)−1×(−5))
∣A∣ = 2(1+8)−7(3+20)−3(−6+5)
∣A∣ = 2×9−7×23−3×(−1)
∣A∣ = 18−161+3
∣A∣ = −140

### Question 6: Evaluate the following:

Solution:

∣A∣ = 0(0−sinβ(−sinβ))−sinα(−sinα×0−sinβcosα)−cosα((−sinα)(−sinβ)−0×cosα)
∣A∣ = 0+sinαsinβcosα−cosαsinαsinβ
∣A∣ = 0

### Question 7:

Solution:

Expand C3, we have
∣A∣ = sinα(−sinαsin2β − cos2βsinα) + cosα(cosαcos2β + cosαsin2β)
∣A∣ = sin2α(sin2β + cos2β) + cos2α(cos2β + sin2β)
∣A∣ = sin2α(1) + cos2α(1)
∣A∣ = 1

### Question 8: If  verify that ∣AB∣ = ∣A∣∣B∣

Solution:

Let’s take LHS, ∣AB∣ = −18−190
∣AB∣ = −208

Now taking RHS and calculating,

∣A∣ = 2−10
∣A∣ = −8
∣B∣ = 20−(−6)
∣B∣ = 26
∣A∣∣B∣ = −8×26
∣A∣∣B∣ = −208
∴LHS = RHS
Hence, it is proved.

### Question 9: If , then show that ∣3A∣ = 27∣A∣.

Solution:

Evaluate along the first column, Now every element with 3, = 3(36−0) − 0 + 0
= 108
Now, according to the question,
∣3A∣ = 27∣A∣
Substituting the values we get,
108 = 27(4)
108 = 108
Hence, proved.

### Question 10: Find the values of x, if:

Solution:

2−20 = 2x2−24
−18 = 2x2−24
2x2 = 6
Taking the square root,
x2 = 3
x = ±√3

Solution:

2 × 5 − 3 × 4 = 5 × x − 3 × 2x
10 − 12 = 5x − 6x
−2 = −x
x = 2

Solution:

3(1)−x(x) = 3(1)−2(4)
3−x2 = 3−8
−x2 = −8
x2 = 8
x = ±2√2

Solution:

3x(4)−7(2) = 10
12x−14 = 10
12x = 24
x = 24/12
​x = 2

Solution:

Cross multiplying elements from LHS,
(x+1)(x+2)−(x−3)(x−1) = 12+1
x2 + 3x + 2 − x2+4x − 3 = 13
7x−1 = 13
7x = 14
x = 2

Solution:

2x(x)−5(8) = 6(3)−5(8)
2x2−40 = 18−40
2x2 = 18
x2 = 9
x = ±3

### Question 11: Find integral value of x, if

Solution:

Here we have to take the determinant of the 3×3 matrix
x2(8−1)−x(0−3)+1(0−6)
8x2−x2+3x−6 = 28
7x2+3x−6 = 28
7x2+3x−34 = 0
Factorization of the above equation we get,
(7x+17)(x−2) = 0
x = 2
Integral value of x is 2. Thus, x = −17/7 is not an integer.

### Question 12: For what value of x the matrix A is singular?

Solution:

Matrix A is singular if,
∣A∣ = 0 Cross−multiply the elements in the determinant,
8 + 8x − 21 + 7x = 0
15x − 13 = 0
15x = 13
x = 13/15

Solution:

Matrix A is singular if ∣A∣=0
Expanding along first row,

∣A∣ = (x−1)[(x−1)2−1] − 1[x−1−1] + 1[1−x+1]
∣A∣ = (x−1)(x2+1−2x−1) − 1(x−2) + 1(2−x)

Expanding the brackets to factorize
|A| = (x−1)(x2−2x) − x + 2 + 2 − x
|A| = (x-1) × x × (x-2) + (4-2x)
|A| = (x−1)× x ×(x−2) + 2(2−x)
|A| = (x−1)× x ×(x−2) − 2(x−2)
[∴ Take (x−2) as common]
|A| = (x−2)[x(x−1)−2]
Since A is a singular matrix, so ∣A∣ = 0
(x−2)(x2−x−2) = 0
There are two cases,
Case1:
(x−2) = 0
x = 2
Case2:
x2−x−2 = 0
x2−2x + x−2 = 0
x(x−2) + 1(x−2) = 0
(x−2)(x+1) = 0
x = 2,−1
∴ x = 2 or −1

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