# 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 don’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}{∵Cij =(-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}{∵Cij =(-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 – 5×2M

_{11}= -12M

_{21}= -3×2 – 5×2M

_{21}= -16M

_{31}= -3×2 – (-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 – 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 – c x caM

_{11}= ab^{2}– ac^{2}M

_{21}= a x ab – c x bcM

_{21}= a^{2}b – c^{2}bM

_{31}= a x ca – b x bcM

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

C

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

^{2}– ac^{2})= ab

^{2}– ac^{2}C

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

^{3}x (a^{2}b – c^{2}b)= c

^{2}b – a^{2}bC

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

^{4}x (a^{2}c – b^{2}c)= a

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

^{2}– ac^{2}+ c^{2}b – a^{2}b + a^{2}c – 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 – 7×0M

_{11}= 5M

_{21}= 2×1 – 7×6M

_{21}= -40M

_{31}= 2×0 – 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 – 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 – f x fM

_{11}= bc – f^{2}M

_{21}= h x c – f x gM

_{21}= hc – fgM

_{31}= h x f – b x gM

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

C

_{11}= (-1)^{1+1}x M_{11}= 1x (bc – f

^{2})= bc – 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 – f

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

^{2}+ hgf – h^{2}c + ghf –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) – 1(1 x 0 – (-1) x 1) + (-2)(1 x 5 – (-1) x (-1))M

_{11}= -9M

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

_{21}= 9M

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

_{31}= -9M

_{41}= -1(1 x 1 – (-1) x (-2)) – 0(0 x 1 – 1 x (-2)) + 1(0 x (-1) – 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 – 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,

∣

A∣ =cos15°×cos75°+sin15°×sin75°As per formula

cos(A−B)=cosAcosB+sinAsinBSubstitute 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| = 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, ∣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∣A∣^{2}=∣A∣×∣A∣∣

A∣^{2}=0

### Question 4: Show that,

**Solution:**

Method 1:Given,

LetthegivendeterminantasA,

Usingsin(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

Expandingalongthefirstrow∣

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(−sinαsin^{2}β − cos^{2}βsinα) + cosα(cosαcos^{2}β + cosαsin^{2}β)

∣A∣ = sin2α(sin^{2}β + cos^{2}β) + cos^{2}α(cos^{2}β + sin^{2}β)

∣A∣ = sin^{2}α(1) + cos^{2}α(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 = 2x

^{2}−24

−18 = 2x^{2}−24

2x^{2}= 6

Taking the square root,

x^{2}= 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−x^{2}= 3−8

−x^{2}= −8

x^{2}= 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

x^{2}+ 3x + 2 − x^{2}+4x − 3 = 13

7x−1 = 13

7x = 14

x = 2

**Solution:**

2x(x)−5(8) = 6(3)−5(8)

2x^{2}−40 = 18−40

2x^{2}= 18

x^{2}= 9

x = ±3

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

**Solution:**

Here we have to take the determinant of the 3×3 matrix

x^{2}(8−1)−x(0−3)+1(0−6)

8x^{2}−x^{2}+3x−6 = 28

7x^{2}+3x−6 = 28

7x^{2}+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)(x^{2}+1−2x−1) − 1(x−2) + 1(2−x)Expanding the brackets to factorize

|A| = (x−1)(x^{2}−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)(x^{2}−x−2) = 0

There are two cases,Case1:

(x−2) = 0

x = 2Case2:

x^{2}−x−2 = 0

x^{2}−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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