Class 12 RD Sharma Solutions - Chapter 6 - Exercise 6.1

In this article, we will be going to solve the entire exercise 6.1 of our RD Sharma textbook.

In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix and the linear transformations it represents. Determinants have applications in various fields, including solving systems of linear equations, analyzing matrix invertibility, and finding eigenvalues.

What is Determinant?

The determinant of a square matrix A is denoted as \det(A)or |A|. For a 2x2 matrix, the determinant is calculated as:

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

\det(A) = ad - bc

Read More: Linear Algebra

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

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

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, a₁₁ = 5

Minor of a₁₁ = M₁₁ = -1

Note: In 2x2 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 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 a₁₁ = M₁₁ = 3

Note: In 2x2 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 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,

      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 = -1x2 – 5x2

M₁₁ = -12

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

M₂₁ = -3x2 – 5x2

M₂₁ = -16

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

M₃₁ = -3x2 – (-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) + 4x16 + 3x(-4)

       = -12 + 64 – 12

       = 40  

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

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,

      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 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,

      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 = 5x1 – 7x0

M₁₁ = 5

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

M₂₁ = 2x1 – 7x6

M₂₁ = -40

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

M₃₁ = 2x0 – 5x6

M₃₁ = -30

Co-factors of the determinant are as follows,

C₁₁ = (-1)¹⁺¹ x M₁₁                            

     = 1x5

     = 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 + 1x40 + 3x(-20)

       = 0 + 40 – 90

       = 50

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

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,

     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 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,  

         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(A−B)=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 ∣A∣²=∣A∣×∣A∣

∣A∣²=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

Practice Questions

1).Evaluate the determinant:

|3 -1|

|2 4|

2).If |a b| = 6, find the value of |2a 2b|

|c d| |2c 2d|

3).Solve for x:

|x+1 3| = 10

| 2 5|

4).Calculate the determinant:

| 1 2 -1|

|-2 0 3|

| 4 1 2|

5).If |a b| = 3 and |c d| = 4, find |ac bd|

|c d| |a b| |ca db|

6).Prove that |a+p b+p| = |a b|

|c+p d+p| |c d|

7).Evaluate:

|cos θ -sin θ|

|sin θ cos θ|

8).Find the value of k for which the following determinant is zero:

|k 2|

|3 k-1|

9).If |a b| = 5, find the value of |a² ab|

|c d| |ac bd|

10).Solve the system of equations using determinants:

2x + 3y = 7

4x - y = 5

Summary

Chapter 6 of RD Sharma Class 12 introduces determinants, which are numerical values associated with square matrices. Key points include:

• Definition of determinants
• Evaluation of 2x2 and 3x3 determinants
• Properties of determinants
• Applications in solving systems of linear equations