Class 12 RD Sharma Solutions Chapter 6 Exercise 6.1 (original) (raw)

Last Updated : 23 Jul, 2025

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 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 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 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 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 Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth 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 Mij and Cij 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 Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth 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 Mij and Cij represents minor and co-factor of element. They are placed in ith row and jth 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 Mij and Cij 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} 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_∣ = __cos_15°×__cos_75°+__sin_15°×__sin_75°

As per formula

_cos(_A_−__B)=_cosAcosB+_sinAsinB

Substitute this in |A| so we get,

∣__A_∣ = _cos(75−15)°

∣__A_∣ = __cos_60°

∣__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| = 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:

\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_∣ = __sin_10°×__cos_80°+__cos_10°×__sin_80°

∣__A_∣ = _sin(10+80)°

∣__A_∣ = __sin_90°

∣__A_∣ = 1

**Method 2:

∣__A_∣ = __sin_10°×__cos_80°+__cos_10°×__sin_80°

[∴_cosθ = sin(90−__θ)]

∣__A_∣ = _sin_10°__cos(90°−10°)+_cos_10°__sin(90°−10°)

∣__A_∣ = __sin_10°__sin_10°+__cos_10°__cos_10°

∣__A_∣ = __sin_210°+__cos_210°

[∴_sin_2__θ+_cos_2__θ = 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: