RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants

Here we provide RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 6.1 Solutions Chapter 6 Determinants book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter6
Exercise6.1
CategoryRD Sharma Solutions

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.

i) A= \begin{bmatrix} 5 &20\\ 0&-1\\ \end{bmatrix}

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

ii) A= \begin{bmatrix}    -1 & 4  \\    2 & 3  \\ \end{bmatrix}

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

iii) A= \begin{bmatrix}    1 & -3 &2  \\    4 & -1 & 2\\    3 & 5 & 2 \\ \end{bmatrix}

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,

      A= \begin{bmatrix}    1 & -3 &2  \\    4 & -1 & 2\\    3 & 5 & 2 \\ \end{bmatrix}

We have,

M_{11}= \begin{bmatrix}    -1 & 2  \\    5 & 2 \\ \end{bmatrix}

M11 = -1×2 – 5×2

M11 = -12

M_{21}= \begin{bmatrix}    -3 & 2  \\    5 & 2 \\ \end{bmatrix}

M21 = -3×2 – 5×2

M21 = -16

M_{31}= \begin{bmatrix}    -3 & 2  \\    -1 & 2 \\ \end{bmatrix}

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  

iv) A= \begin{bmatrix}    1 & a &bc  \\    1 & b & ca\\    1 & c & ab\\ \end{bmatrix}

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,

 A= \begin{bmatrix}    1 & a &bc  \\    1 & b & ca\\    1 & c & ab\\ \end{bmatrix}

We have,

M_{11}= \begin{bmatrix}    b & ca  \\    c & ab\\ \end{bmatrix}

M11 = b x ab – c x ca

M11 = ab2 – ac2

M_{21}= \begin{bmatrix}    a & bc  \\    c & ab\\ \end{bmatrix}

M21 = a x ab – c x bc

M21 = a2b – c2b

M_{31}= \begin{bmatrix}    a & bc  \\    b & ca\\ \end{bmatrix}

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

v) A= \begin{bmatrix}    0 & 2 &6  \\    1 & 5 & 0\\    3 & 7 & 1 \\ \end{bmatrix}

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,

 A= \begin{bmatrix}    0 & 2 &6  \\    1 & 5 & 0\\    3 & 7 & 1 \\ \end{bmatrix}

We have,

M_{11}=\begin{bmatrix}    5 & 0 \\    7 & 1\\ \end{bmatrix}

M11 = 5×1 – 7×0

M11 = 5

M_{21}=\begin{bmatrix}    2 & 6 \\    7 & 1\\ \end{bmatrix}

M21 = 2×1 – 7×6

M21 = -40

M_{31}=\begin{bmatrix}    2 & 6 \\    5 & 0\\ \end{bmatrix}

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

vi) A= \begin{bmatrix}    a & h & g  \\    h & b & f\\    g & f & c \\ \end{bmatrix}

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,

A= \begin{bmatrix}    a & h & g  \\    h & b & f\\    g & f & c \\ \end{bmatrix}

We have,

M_{11}= \begin{bmatrix}    b & f   \\     f&c\\ \end{bmatrix}

M11 = b x c – f x f

M11 = bc – f2

M_{21}= \begin{bmatrix}    h & g  \\    f & c\\ \end{bmatrix}

M21 = h x c – f x g

M21 = hc – fg

M_{31}= \begin{bmatrix}    h & g  \\    b & f\\ \end{bmatrix}

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

vii) A= \begin{bmatrix}    2 & -1 & 0 & 1 \\    -3 & 0 & 1 & -2 \\    1 & 1 & -1 & 1 \\    2 & -1 & 5 & 0  \\ \end{bmatrix}

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,  

 A= \begin{bmatrix}    2 & -1 & 0 & 1 \\    -3 & 0 & 1 & -2 \\    1 & 1 & -1 & 1 \\    2 & -1 & 5 & 0  \\ \end{bmatrix}

From the matrix we have,

 M_{11}= \begin{bmatrix}     0 & 1 & -2 \\     1 & -1 & 1 \\     -1 & 5 & 0  \\ \end{bmatrix}

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

M11 = -9

 M_{21}= \begin{bmatrix}     -1 & 0 & 1 \\     1 & -1 & 1 \\     -1 & 5 & 0  \\ \end{bmatrix}

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

M21 = 9

 M_{31}= \begin{bmatrix}     -1&0&1 \\     0 & 1 & -2 \\     -1 & 5 & 0  \\ \end{bmatrix}

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

M31 = -9

 M_{41}= \begin{bmatrix}     -1& 0 & -1 \\     0 & 1 & -2 \\     1 & -1 &1   \\ \end{bmatrix}

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

i)A= \begin{vmatrix} x& -7\\ z&5x+1 \end{vmatrix}

Solution:

Given, A= \begin{vmatrix} x& -7\\ z&5x+1 \end{vmatrix}

Cross multiplying the values inside the determinant,

|A| = (5x + 1) – (-7)x

|A| = 5x2 = 8x

ii) A= \begin{vmatrix} cos\theta& -sin\theta \\ sin\theta&cos\theta \end{vmatrix}

Solution:

Given, A= \begin{vmatrix} cos\theta& -sin\theta \\ sin\theta&cos\theta \end{vmatrix}

|A|=cos\theta \times sin\theta-(-sin\theta)\times sin\theta                           {\therefore cos^2\theta + sin^2\theta = 1

|A|=cos^2\theta+sin^2\theta \\ |A|=1
iii) A= \begin{vmatrix} cos15\degree & -sin15\degree\\ sin75\degree & cos75\degree \\ \end{vmatrix}

Solution:

Given, A= \begin{vmatrix} cos15\degree & -sin15\degree\\ sin75\degree & cos75\degree \\ \end{vmatrix}

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

iv) A= \begin{vmatrix} a+ib & c+id \\ -c+id & a-ib\\ \end{vmatrix}

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:

\begin{vmatrix} 2&3&7\\ 13&17&5\\ 15&20&12\\ \end{vmatrix}^2

Solution:

In the given formula, ∣AB∣=∣A∣∣B

\\ |A| = \begin{vmatrix} 2&3&7\\ 13&17&5\\ 15&20&12\\ \end{vmatrix}

Cross multiplying the terms  in |A| 

\\ |A| = 2 \begin{vmatrix} 17&5\\ 20&12\\ \end{vmatrix} -3 \begin{vmatrix} 13&5\\ 15&12\\ \end{vmatrix} +7 \begin{vmatrix} 13&17\\ 15&20\\ \end{vmatrix} \\

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,

\begin{vmatrix} sin10\degree & -cos10\degree \\ sin80\degree & cos 80\degree \\ \end{vmatrix}

Solution:

Method 1:

Given,

\\ \begin{vmatrix} sin10\degree & -cos10\degree \\ sin80\degree & cos 80\degree \\ \end{vmatrix}

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.

\begin{vmatrix} 2 &3&-5 \\ 7&1&-2 \\ -3&4&1\\ \end{vmatrix}

Solution:

Method 1

Expanding along the first row

\\ |A| = 2 \begin{vmatrix} 1&-2 \\ 4&1\\ \end{vmatrix} -3 \begin{vmatrix} 7&-2 \\ -3&1\\ \end{vmatrix} -5 \begin{vmatrix} 7&1 \\ -3&4\\ \end{vmatrix}

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 \begin{vmatrix} 1&-2 \\ 4&1 \\ \end{vmatrix} -7 \begin{vmatrix} 3&-5 \\ 4&1 \\ \end{vmatrix} -3 \begin{vmatrix} 3&-5 \\ 1&-2\\ \end{vmatrix}

∣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:

A = \begin{vmatrix} 0&sin\alpha & -cos\alpha \\ -sin\alpha &0 & sin\beta \\ cos\alpha & -sin\beta & 0 \\ \end{vmatrix}

Solution:

|A| =0 \begin{vmatrix} 0&sin\beta  \\ -sin\beta &0 \\ \end{vmatrix} -sin\alpha \begin{vmatrix} -sin\alpha &sin\beta \\ cos\alpha & 0 \\ \end{vmatrix}  -cos\alpha \begin{vmatrix} -sin\alpha &0 \\ cos\alpha&-sin\beta \\ \end{vmatrix}

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

Question 7: 

\begin{vmatrix} cos\alpha cos\beta&cos\alpha sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin\alpha cos\beta& sin\alpha sin\beta & cos\alpha \\ \end{vmatrix}

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  A= \begin{bmatrix} 2&5 \\ 2&1 \\ \end{bmatrix} \ B= \begin{bmatrix} 4&-3 \\ 2&15\\ \end{bmatrix} verify that ∣AB∣ = ∣A∣∣B∣

Solution:

Let’s take LHS,

AB = \begin{bmatrix} 2&5 \\ 2&1 \\ \end{bmatrix} \times \begin{bmatrix} 4&-3 \\ 2&5 \\ \end{bmatrix}  \\ = \begin{bmatrix} 8+10 & -6+25 \\ 8+2 & -6+5 \\ \end{bmatrix} \\ =\begin{bmatrix} 18 & 19 \\ 10&-1 \\ \end{bmatrix}
∣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  A = \begin{bmatrix} 1&0&1 \\ 0&1&2 \\ 0&0&4\\ \end{bmatrix}, then show that ∣3A∣ = 27∣A∣.

Solution:

Evaluate along the first column,

|A|=1 \begin{vmatrix} 1&2 \\ 0&4\\ \end{vmatrix}-0 \begin{vmatrix} 0&1 \\ 0&4\\ \end{vmatrix}+0 \begin{vmatrix} 0&1 \\ 1&2\\ \end{vmatrix}
Now every element with 3,
 \\ |3A|=3 \begin{vmatrix} 3&6 \\ 0&12\\ \end{vmatrix}-0 \begin{vmatrix} 0&3 \\ 0&12\\ \end{vmatrix}+0 \begin{vmatrix} 0&3 \\ 3&6\\ \end{vmatrix}
= 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:

i)\begin{vmatrix} 2&4 \\ 5&1\\ \end{vmatrix}= \begin{vmatrix} 2x&4 \\ 6&x\\ \end{vmatrix} \\

Solution:

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

ii)\begin{vmatrix} 2&3 \\ 4&5\\ \end{vmatrix}= \begin{vmatrix} x&3 \\ 2x&5\\ \end{vmatrix} \\

Solution:

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

iii)\begin{vmatrix} 3&x \\ x&1\\ \end{vmatrix}= \begin{vmatrix} 3&2 \\ 4&1\\ \end{vmatrix} \\

Solution:

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

iv)\begin{vmatrix} 3x&7 \\ 2&4\\ \end{vmatrix}=10

Solution:

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

v)\begin{vmatrix} x+1&x-1 \\ x-3&x+2\\ \end{vmatrix}= \begin{vmatrix} 4&-1 \\ 1&3\\ \end{vmatrix} \\

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

vi)\begin{vmatrix} 2x&5 \\ 8&x\\ \end{vmatrix}= \begin{vmatrix} 6&5 \\ 8&3\\ \end{vmatrix} \\

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

\begin{vmatrix} x^2 & x& 1 \\ 0&2&1\\ 3&1&4\\ \end{vmatrix}=28

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?

i)A=\begin{vmatrix} 1+x&7 \\ 3-x&8\\ \end{vmatrix}=0

Solution:

Matrix A is singular if,
∣A∣ = 0
\\ |A| =\begin{vmatrix} 1+x&7 \\ 3-x&8\\ \end{vmatrix}=0
Cross−multiply the elements in the determinant,
8 + 8x − 21 + 7x = 0
15x − 13 = 0
15x = 13
x = 13/15

ii)A=\begin{vmatrix} x-1&1&1 \\ 1&x-1&1 \\ 1&1&x-1\\ \end{vmatrix}

Solution:

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

|A|=(x-1)\begin{vmatrix} x-1&1 \\ 1&x-1\\ \end{vmatrix} -1 \begin{vmatrix} 1&1 \\ 1&x-1\\ \end{vmatrix} +1 \begin{vmatrix} 1&1 \\ x-1&1\\ \end{vmatrix} \\

∣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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.