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# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.1

• Last Updated : 02 Feb, 2021

### 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 donâ€™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                           {âˆµCij =(-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                           {âˆµCij =(-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 â€“ 5×2

M11 = -12

M21 = -3×2 â€“ 5×2

M21 = -16

M31 = -3×2 â€“ (-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 â€“ 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 â€“ c x ca

M11 = ab2 â€“ ac2

M21 = a x ab â€“ c x bc

M21 = a2b â€“ c2b

M31 = a x ca â€“ b x bc

M31 = a2c â€“ b2c

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1 x (ab2 â€“ ac2)

= ab2 â€“ ac2

C21 = (-1)2+1 x M21

= (-1)3 x (a2b â€“ c2b)

= c2b – a2b

C31 = (-1)3+1 x M31

= (1)4 x (a2c â€“ b2c)

= a2c â€“ b2c

To evaluate the determinant expand along first column,

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

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

= ab2 â€“ ac2 + c2b – a2b + a2c â€“ 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 â€“ 7×0

M11 = 5

M21 = 2×1 â€“ 7×6

M21 = -40

M31 = 2×0 â€“ 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 â€“ 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 â€“ f x f

M11 = bc â€“ f2

M21 = h x c â€“ f x g

M21 = hc â€“ fg

M31 = h x f â€“ b x g

M31 = hf â€“ bg

Co-factors of the determinant are as follows,

C11 = (-1)1+1 x M11

= 1x (bc â€“ f2)

= bc â€“ 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 â€“ f2) + h x (fg â€“ hc) + g x (hf – bg)

= abc â€“ af2 + hgf â€“ h2c + ghf â€“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) â€“ 1(1 x 0 â€“ (-1) x 1) + (-2)(1 x 5 â€“ (-1) x (-1))

M11 = -9

M21 = -1(-1 x 0 – 5 x 1) â€“ 0(1 x 0 â€“ (-1) x 1) + (1 x 5 â€“ (-1) x (-1))

M21 = 9

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

M31 = -9

M41 = -1(1 x 1 â€“ (-1) x (-2)) â€“ 0(0 x 1 â€“ 1 x (-2)) + 1(0 x (-1) â€“ 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 â€“ 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,

âˆ£Aâˆ£ = cos15Â°Ã—cos75Â°+sin15Â°Ã—sin75Â°

As per formula

cos(Aâˆ’B)=cosAcosB+sinAsinB

Substitute this in |A| so we get,

âˆ£Aâˆ£ = cos(75âˆ’15)Â°

âˆ£Aâˆ£ = cos60Â°

âˆ£Aâˆ£ = 0.5

Solution:

âˆ£Aâˆ£ = (a+ib)(aâˆ’ib)âˆ’(c+id)(âˆ’c+id)

Expanding the brackets we get,

âˆ£Aâˆ£=(a+ib)(aâˆ’ib)+(c+id)(câˆ’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, âˆ£ABâˆ£=âˆ£Aâˆ£âˆ£Bâˆ£

Cross multiplying the terms  in |A|

âˆ£Aâˆ£ = 2(17Ã—12âˆ’5Ã—20)âˆ’3(13Ã—12âˆ’5Ã—15)+7(13Ã—20âˆ’15Ã—17)

= 2(204âˆ’100)âˆ’3(156âˆ’75)+7(260âˆ’255)

= 2Ã—104âˆ’3Ã—81+7Ã—5

= 208âˆ’243+45

= 0

Now âˆ£Aâˆ£2=âˆ£Aâˆ£Ã—âˆ£Aâˆ£

âˆ£Aâˆ£2=0

### Question 4: Show that,

Solution:

Method 1:

Given,

Let the given determinant as A,

Using sin(a+B) = sinAÃ—cosB+cosAÃ—sinB

âˆ£Aâˆ£ = sin10Â°Ã—cos80Â°+cos10Â°Ã—sin80Â°

âˆ£Aâˆ£ = sin(10+80)Â°

âˆ£Aâˆ£ = sin90Â°

âˆ£Aâˆ£ = 1

Method 2:

âˆ£Aâˆ£ = sin10Â°Ã—cos80Â°+cos10Â°Ã—sin80Â°

[âˆ´cosÎ¸ = sin(90âˆ’Î¸)]

âˆ£Aâˆ£ = sin10Â°cos(90Â°âˆ’10Â°)+cos10Â°sin(90Â°âˆ’10Â°)

âˆ£Aâˆ£ = sin10Â°sin10Â°+cos10Â°cos10Â°

âˆ£Aâˆ£ = sin210Â°+cos210Â°

[âˆ´sin2Î¸+cos2Î¸ = 1]

âˆ£Aâˆ£ = 1

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

Solution:

Method 1

Expanding along the first row

âˆ£Aâˆ£ = 2(1Ã—1âˆ’4Ã—âˆ’2)âˆ’3(7Ã—1âˆ’(âˆ’2)Ã—âˆ’3)âˆ’5(7Ã—4âˆ’1Ã—(âˆ’3))

âˆ£Aâˆ£ = 2(1+8)âˆ’3(7âˆ’6)âˆ’5(28+3)

âˆ£Aâˆ£ = 2Ã—9âˆ’3Ã—1âˆ’5Ã—31

âˆ£Aâˆ£ = 18âˆ’3âˆ’155

âˆ£Aâˆ£ = âˆ’140

Method 2

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

Expanding along second column,

âˆ£Aâˆ£ = 2(1Ã—1âˆ’4Ã—(âˆ’2))âˆ’7(3Ã—1âˆ’4Ã—(âˆ’5))âˆ’3(3Ã—(âˆ’2)âˆ’1Ã—(âˆ’5))
âˆ£Aâˆ£ = 2(1+8)âˆ’7(3+20)âˆ’3(âˆ’6+5)
âˆ£Aâˆ£ = 2Ã—9âˆ’7Ã—23âˆ’3Ã—(âˆ’1)
âˆ£Aâˆ£ = 18âˆ’161+3
âˆ£Aâˆ£ = âˆ’140

### Question 6: Evaluate the following:

Solution:

âˆ£Aâˆ£ = 0(0âˆ’sinÎ²(âˆ’sinÎ²))âˆ’sinÎ±(âˆ’sinÎ±Ã—0âˆ’sinÎ²cosÎ±)âˆ’cosÎ±((âˆ’sinÎ±)(âˆ’sinÎ²)âˆ’0Ã—cosÎ±)
âˆ£Aâˆ£ = 0+sinÎ±sinÎ²cosÎ±âˆ’cosÎ±sinÎ±sinÎ²
âˆ£Aâˆ£ = 0

### Question 7:

Solution:

Expand C3, we have
âˆ£Aâˆ£ = sinÎ±(âˆ’sinÎ±sin2Î² âˆ’ cos2Î²sinÎ±) + cosÎ±(cosÎ±cos2Î² + cosÎ±sin2Î²)
âˆ£Aâˆ£ = sin2Î±(sin2Î² + cos2Î²) + cos2Î±(cos2Î² + sin2Î²)
âˆ£Aâˆ£ = sin2Î±(1) + cos2Î±(1)
âˆ£Aâˆ£ = 1

### Question 8: If   verify that âˆ£ABâˆ£ = âˆ£Aâˆ£âˆ£Bâˆ£

Solution:

Let’s take LHS,

âˆ£ABâˆ£ = âˆ’18âˆ’190
âˆ£ABâˆ£ = âˆ’208

Now taking RHS and calculating,

âˆ£Aâˆ£ = 2âˆ’10
âˆ£Aâˆ£ = âˆ’8
âˆ£Bâˆ£ = 20âˆ’(âˆ’6)
âˆ£Bâˆ£ = 26
âˆ£Aâˆ£âˆ£Bâˆ£ = âˆ’8Ã—26
âˆ£Aâˆ£âˆ£Bâˆ£ = âˆ’208
âˆ´LHS = RHS
Hence, it is proved.

### Question 9: If  , then show that âˆ£3Aâˆ£ = 27âˆ£Aâˆ£.

Solution:

Evaluate along the first column,

Now every element with 3,

= 3(36âˆ’0) âˆ’ 0 + 0
= 108
Now, according to the question,
âˆ£3Aâˆ£ = 27âˆ£Aâˆ£
Substituting the values we get,
108 = 27(4)
108 = 108
Hence, proved.

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

Solution:

2âˆ’20 = 2x2âˆ’24
âˆ’18 = 2x2âˆ’24
2x2 = 6
Taking the square root,
x2 = 3
x = Â±âˆš3

Solution:

2 Ã— 5 âˆ’ 3 Ã— 4 = 5 Ã— x âˆ’ 3 Ã— 2x
10 âˆ’ 12 = 5x âˆ’ 6x
âˆ’2 = âˆ’x
x = 2

Solution:

3(1)âˆ’x(x) = 3(1)âˆ’2(4)
3âˆ’x2 = 3âˆ’8
âˆ’x2 = âˆ’8
x2 = 8
x = Â±2âˆš2
â€‹

Solution:

3x(4)âˆ’7(2) = 10
12xâˆ’14 = 10
12x = 24
x = 24/12
â€‹x = 2

Solution:

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

Solution:

2x(x)âˆ’5(8) = 6(3)âˆ’5(8)
2x2âˆ’40 = 18âˆ’40
2x2 = 18
x2 = 9
x = Â±3

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

Solution:

Here we have to take the determinant of the 3Ã—3 matrix
x2(8âˆ’1)âˆ’x(0âˆ’3)+1(0âˆ’6)
8x2âˆ’x2+3xâˆ’6 = 28
7x2+3xâˆ’6 = 28
7x2+3xâˆ’34 = 0
Factorization of the above equation we get,
(7x+17)(xâˆ’2) = 0
x = 2
Integral value of x is 2. Thus, x = âˆ’17/7 is not an integer.

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

Solution:

Matrix A is singular if,
âˆ£Aâˆ£ = 0

Crossâˆ’multiply the elements in the determinant,
8 + 8x âˆ’ 21 + 7x = 0
15x âˆ’ 13 = 0
15x = 13
x = 13/15

Solution:

Matrix A is singular if âˆ£Aâˆ£=0
Expanding along first row,

âˆ£Aâˆ£ = (xâˆ’1)[(xâˆ’1)2âˆ’1] âˆ’ 1[xâˆ’1âˆ’1] + 1[1âˆ’x+1]
âˆ£Aâˆ£ = (xâˆ’1)(x2+1âˆ’2xâˆ’1) âˆ’ 1(xâˆ’2) + 1(2âˆ’x)

Expanding the brackets to factorize
|A| = (xâˆ’1)(x2âˆ’2x) âˆ’ x + 2 + 2 âˆ’ x
|A| = (x-1) Ã— x Ã— (x-2) + (4-2x)
|A| = (xâˆ’1)Ã— x Ã—(xâˆ’2) + 2(2âˆ’x)
|A| = (xâˆ’1)Ã— x Ã—(xâˆ’2) âˆ’ 2(xâˆ’2)
[âˆ´ Take (xâˆ’2) as common]
|A| = (xâˆ’2)[x(xâˆ’1)âˆ’2]
Since A is a singular matrix, so âˆ£Aâˆ£ = 0
(xâˆ’2)(x2âˆ’xâˆ’2) = 0
There are two cases,
Case1:
(xâˆ’2) = 0
x = 2
Case2:
x2âˆ’xâˆ’2 = 0
x2âˆ’2x + xâˆ’2 = 0
x(xâˆ’2) + 1(xâˆ’2) = 0
(xâˆ’2)(x+1) = 0
x = 2,âˆ’1
âˆ´ x = 2 or âˆ’1

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