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

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

**Solution:**

i)Let M_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column.Here, a

_{11}= 5Minor of a

_{11}= M_{11}= -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 a

_{12}= M_{12}= 0Minor of a

_{21}= M_{21}= 20Minor of a

_{22}= M_{22}= 0As M

_{12}and M_{22}are zero so we donâ€™t consider them. Hence we have got only two minors for this determinant.M

_{11}= -1 & M_{21}= 20Now, co-factors for the determinants are

C

_{11}= (-1)^{1+1}x M_{11}{âˆµCij =(-1)1+1 x Mij}= (+1)x(-1)

= -1

C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x 20= -20

Evaluating the determinant,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}=5 x (-1) + 0 x (-20)

= -5

**Solution:**

Let M

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column.Minor of a

_{11}= M_{11}= 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 a

_{21}= M_{21}= 4Now, co-factors for the determinants are

C

_{11}= (-1)^{1+1}x M_{11}{âˆµCij =(-1)i+j x Mij}= (+1) x 3

= 3

C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x 4= -4

Evaluating the determinant,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}=-1 x 3 + 2 x (-4)

=-11

**Solution:**

Let M

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column.

C_{ij}= (-1)^{i+j}x M_{ij}Given,

We have,

M

_{11}= -1×2 â€“ 5×2M

_{11}= -12M

_{21}= -3×2 â€“ 5×2M

_{21}= -16M

_{31}= -3×2 â€“ (-1) x 2M

_{31}= -4Co-factors of the determinant are as follows,

C

_{11}= (-1)^{1+1}x M_{11}= 1x-12

= -12

C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x -16= 16

C

_{31}= (-1)^{3+1}x M_{31}= (1)

^{4}x (-4)= -4

To evaluate the determinant expand along first column,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}+ a_{31}x C_{31}=1x(-12) + 4×16 + 3x(-4)

= -12 + 64 â€“ 12

= 40

**Solution:**

Let M

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in ith row and jth column.

Also, C_{ij}= (-1)^{i+j}x M_{ij}Given,

We have,

M

_{11}= b x ab â€“ c x caM

_{11}= ab^{2}â€“ ac^{2}M

_{21}= a x ab â€“ c x bcM

_{21}= a^{2}b â€“ c^{2}bM

_{31}= a x ca â€“ b x bcM

_{31}= a^{2}c â€“ b^{2}cCo-factors of the determinant are as follows,

C

_{11}= (-1)^{1+1}x M_{11}= 1 x (ab

^{2}â€“ ac^{2})= ab

^{2}â€“ ac^{2}C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x (a^{2}b â€“ c^{2}b)= c

^{2}b – a^{2}bC

_{31}= (-1)^{3+1}x M_{31}= (1)

^{4}x (a^{2}c â€“ b^{2}c)= a

^{2}c â€“ b^{2}cTo evaluate the determinant expand along first column,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}+ a_{31}x C_{31}=1 x (ab

^{2}â€“ ac^{2}) + 1 x (c^{2}b – a^{2}b) + 1 x (a^{2}c â€“ b^{2}c)= ab

^{2}â€“ ac^{2}+ c^{2}b – a^{2}b + a^{2}c â€“ b^{2}c

**Solution:**

Let M

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column.

C_{ij}= (-1)^{i+j}x M_{ij}Given,

We have,

M

_{11}= 5×1 â€“ 7×0M

_{11}= 5M

_{21}= 2×1 â€“ 7×6M

_{21}= -40M

_{31}= 2×0 â€“ 5×6M

_{31}= -30Co-factors of the determinant are as follows,

C

_{11}= (-1)^{1+1}x M_{11}= 1×5

= 5

C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x -40= 40

C

_{31}= (-1)^{3+1}x M_{31}= (1)

^{4}x (-30)= -30

To evaluate the determinant expand along first column,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}+ a_{31}x C_{31}=0x5 + 1×40 + 3x(-20)

= 0 + 40 â€“ 90

= 50

**Solution:**

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column.

C_{ij}= (-1)^{i+j}x M_{ij}Given,

We have,

M

_{11}= b x c â€“ f x fM

_{11}= bc â€“ f^{2}M

_{21}= h x c â€“ f x gM

_{21}= hc â€“ fgM

_{31}= h x f â€“ b x gM

_{31}= hf â€“ bgCo-factors of the determinant are as follows,

C

_{11}= (-1)^{1+1}x M_{11}= 1x (bc â€“ f

^{2})= bc â€“ f

^{2}C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x (hc – fg)= fg – hc

C

_{31}= (-1)^{3+1}x M_{31}= (1)

^{4}x (hf – bg)= hf – bg

To evaluate the determinant expand along first column,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}+ a_{31}x C_{31}=a x (bc â€“ f

^{2}) + h x (fg â€“ hc) + g x (hf – bg)= abc â€“ af

^{2}+ hgf â€“ h^{2}c + ghf â€“bg^{2}

**Solution:**

Let M

_{ij}and C_{ij}represents minor and co-factor of element. They are placed in i^{th}row and j^{th}column

Also, C_{ij}= (-1)^{i+j}x M_{ij}Given,

From the matrix we have,

M

_{11}= 0(-1 x 0 – 5 x 1) â€“ 1(1 x 0 â€“ (-1) x 1) + (-2)(1 x 5 â€“ (-1) x (-1))M

_{11}= -9M

_{21}= -1(-1 x 0 – 5 x 1) â€“ 0(1 x 0 â€“ (-1) x 1) + (1 x 5 â€“ (-1) x (-1))M

_{21}= 9M

_{31}= -1(1 x 0 – 5 x (-2)) â€“ 0(0 x 0 â€“ (-1) x (-2)) + 1(0 x 5 â€“ (-1) x 1)M

_{31}= -9M

_{41}= -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)M

_{41}= 0Co-factors of the determinant are as follows,

C

_{11}= (-1)^{1+1}x M_{11}= 1x (-9)

= -9

C

_{21}= (-1)^{2+1}x M_{21}= (-1)

^{3}x 9= -9

C

_{31}= (-1)^{3+1}x M_{31}= (-1)

^{4}x -9= -9

C

_{41}= (-1)^{4+1}x M_{41}= (-1)

^{5}x 0= 0

To evaluate the determinant expand along first column,

|A| = a

_{11}x C_{11}+ a_{21}x C_{21}+ a_{31}x C_{31}+ a_{41}x C_{41}=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| = 5x

^{2}= 8x

**Solution:**

Given,

{

**Solution:**

Given,

âˆ£

Aâˆ£ =cos15Â°Ã—cos75Â°+sin15Â°Ã—sin75Â°As per formula

cos(Aâˆ’B)=cosAcosB+sinAsinBSubstitute 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| = a^{2}-i^{2}b^{2}+c^{2}-i^{2}d^{2}

We know i^{2}= -1

|A| = a^{2}-1b^{2}+c^{2}-(-1)d^{2}

|A| = a^{2}+b^{2}+c^{2}+d^{2}

### Question 3: Evaluate the following:

**Solution:**

Inthegivenformula, 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

Nowâˆ£Aâˆ£^{2}=âˆ£Aâˆ£Ã—âˆ£Aâˆ£âˆ£

Aâˆ£^{2}=0

### Question 4: Show that,

**Solution:**

Method 1:Given,

LetthegivendeterminantasA,

Usingsin(a+B) =sinAÃ—cosB+cosAÃ—sinBâˆ£

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

Expandingalongthefirstrowâˆ£

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—31âˆ£

Aâˆ£ = 18âˆ’3âˆ’155âˆ£

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±sin^{2}Î² âˆ’ cos^{2}Î²sinÎ±) + cosÎ±(cosÎ±cos^{2}Î² + cosÎ±sin^{2}Î²)

âˆ£Aâˆ£ = sin2Î±(sin^{2}Î² + cos^{2}Î²) + cos^{2}Î±(cos^{2}Î² + sin^{2}Î²)

âˆ£Aâˆ£ = sin^{2}Î±(1) + cos^{2}Î±(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 = 2x

^{2}âˆ’24

âˆ’18 = 2x^{2}âˆ’24

2x^{2}= 6

Taking the square root,

x^{2}= 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’x^{2}= 3âˆ’8

âˆ’x^{2}= âˆ’8

x^{2}= 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

x^{2}+ 3x + 2 âˆ’ x^{2}+4x âˆ’ 3 = 13

7xâˆ’1 = 13

7x = 14

x = 2

**Solution:**

2x(x)âˆ’5(8) = 6(3)âˆ’5(8)

2x^{2}âˆ’40 = 18âˆ’40

2x^{2}= 18

x^{2}= 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

x^{2}(8âˆ’1)âˆ’x(0âˆ’3)+1(0âˆ’6)

8x^{2}âˆ’x^{2}+3xâˆ’6 = 28

7x^{2}+3xâˆ’6 = 28

7x^{2}+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)

^{2}âˆ’1] âˆ’ 1[xâˆ’1âˆ’1] + 1[1âˆ’x+1]

âˆ£Aâˆ£ = (xâˆ’1)(x^{2}+1âˆ’2xâˆ’1) âˆ’ 1(xâˆ’2) + 1(2âˆ’x)Expanding the brackets to factorize

|A| = (xâˆ’1)(x^{2}âˆ’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)(x^{2}âˆ’xâˆ’2) = 0

There are two cases,

Case1:

(xâˆ’2) = 0

x = 2

Case2:

x^{2}âˆ’xâˆ’2 = 0

x^{2}âˆ’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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