Class 12 Maths Chapter 4 NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 4 Determinants are provided by SimplyAcad here for students to practise for their upcoming examinations, both : CBSE Class 12 board exams and entrance exams of science and engineering like JEE and NEET.

Detailed explained solutions of NCERT textbooks Chapter 4 Determinants are prepared by our subject experts to support students ace their exams. These solutions not only aid in creating a base but also help in clearing the basic concepts to enhance problem-solving abilities. Go through the solutions provided below to practise and make sure to revisit and revise regularly to get hold of the concepts.

Here are the main topics and sub-topics present in the NCERT Solutions Class 12 Maths Chapter 4 Determinants given below:

Section Name & Topic Name in Chapter 4 Determinants

4 Determinants

4.1 Introduction

4.2 Determinant

4.3 Properties of Determinants

4.4 Area of a Triangle

4.5 Adjoint and Inverse of a Matrix

4.6 Applications of Determinants and Matrices

Class 12 Determinants Exercise 4.1

Question 1 to 5

Evaluate the following determinants in Exercise 1 and 2.

Question 1

Solution:

$\text{= 2(-1) – 4(-5) = -2 + 20 = 18}$

---

Question 2

(i)

(ii)

Solution:

(i)

$\text{= (cosθ)(cosθ) – (-sinθ) (sinθ) = cos² θ + sin² θ = 1}$

(ii)

$\text{= (x² − x + 1)(x + 1) − (x − 1)(x + 1)}$ $\text{= x³ − x² + x + x² − x + 1 − (x² − 1)}$ $\text{= x³ + 1 − x² + 1}$ $\text{= x³ − x² + 2}$

---

Question 3

If

$A = \begin{pmatrix} 2 & -1 \\ 4 & -5 \end{pmatrix}$

,then show that

$|2A| = 4|A|$

Solution:

Given:

$A = \begin{pmatrix} 2 & -1 \\ 4 & -5 \end{pmatrix}$

Then,

$2A = 2 \times \begin{pmatrix} 2 & -1 \\ 4 & -5 \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ 8 & -10 \end{pmatrix}$

Hence, proved.

---

Question 4

If

$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$

, then show that

$3|A| = 27|A|$

Solution:

Given:

$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$

​​

Then,

$3A = 3 \times \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$

​​

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.

Hence, proved.

---

Question 5

Evaluate the determinants:

(i)

(ii)

(iii)

(iv)

Solution:

Evaluate the determinants:

(i) Given:

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

$\text{=}$

(ii) Given:

By expanding along the first row, we have:

$\text{=}$

(iii) Given:

Expanding along the first row,

(iv) Given:

Expanding along the first row,

---

Question 6 to 8

Question 6

If

$A = \begin{pmatrix} 2 & 5 \\ 3 & 1 \end{pmatrix}$

A=(23​51​), find

$|A|$

Solution:

Given:

$A = \begin{pmatrix} 2 & 5 \\ 3 & 1 \end{pmatrix}$

Expanding along the first row,

$|A| = (2)(1) – (5)(3) = 2 – 15 = -13$

---

Question 7

Find the value of

$x$

(i)

(ii)

Solution:

(i) Given:

$2 \times 1 – 5 \times 4 = 2x \times x – 6 \times 4$ $2 – 20 = 2x² – 24$ $2x² = 6$ $x² = 3$ $x = \pm \sqrt{3}$

x=±3​

(ii)

$2 \times 5 – 4 \times 3 = x \times 5 – 2x \times 3$ $10 – 12 = 5x – 6x$ $-2 = -x$ $x = 2$

---

Question 8

If

$$ $<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;">then\; x\; is\; equal\; to:</span>$

(A) 6
(B) ± 6
(C) – 6
(D) 0

Solution:

Given:

$x \times x – 18 \times 2 = 6 \times 6 – 18 \times 2$ $x² – 36 = 36 – 36$ $x² – 36 = 0$ $x = \pm 6$

Therefore, option (B) is correct.

---

Class 12 Determinants Exercise 4.2

Using the properties of determinants and without expanding in Exercise 1 to 7, prove that:

Question 1 to 4

Question 1

$\left| \begin{array}{ccc} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{array} \right| = 0$

Solution:

L.H.S.:

$\left| \begin{array}{ccc} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{array} \right|$

​

Now, perform the operation

$C_3 \rightarrow C_3 – C_1$

C3​→C3​−C1​:

$= \left| \begin{array}{ccc} x & a & x + a – x \\ y & b & y + b – y \\ z & c & z + c – z \end{array} \right|$

​​

$= \left| \begin{array}{ccc} x & a & a \\ y & b & b \\ z & c & c \end{array} \right|$

​​

Now, perform the operation

$C_3 \rightarrow C_3 – C_2$ $= \left| \begin{array}{ccc} x & a & 0 \\ y & b & 0 \\ z & c & 0 \end{array} \right| = 0$

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Question 2

$\left| \begin{array}{ccc} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{array} \right| = 0$

Solution:

L.H.S.:

$\left| \begin{array}{ccc} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{array} \right|$

​​

Perform the operation

$C_1 \rightarrow C_1 + C_2 + C_3$ $= \left| \begin{array}{ccc} (a – b) + (b – c) + (c – a) & b – c & c – a \\ (b – c) + (c – a) + (a – b) & c – a & a – b \\ (c – a) + (a – b) + (b – c) & a – b & b – c \end{array} \right|$

=​(a−b)+(b−c)+(c−a)(b−c)+(c−a)+(a−b)(c−a)+(a−b)+(b−c)​b−cc−aa−b​c−aa−bb−c​​

$= \left| \begin{array}{ccc} 0 & b – c & c – a \\ 0 & c – a & a – b \\ 0 & a – b & b – c \end{array} \right|$

​​

Since the first column consists entirely of zeros, the determinant equals zero.

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Question 3

$\left| \begin{array}{ccc} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array} \right| = 0$

Solution:

L.H.S.:

$\left| \begin{array}{ccc} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array} \right|$

​

Perform the operation

$C_3 \rightarrow C_3 – C_1$ $= \left| \begin{array}{ccc} 2 & 7 & 63 \\ 3 & 8 & 72 \\ 5 & 9 & 81 \end{array} \right|$

​​

Now, perform the operation

$C_2 \rightarrow C_2 – C_1$

and

$C_3 \rightarrow C_3 – C_1$ $= 9 \times \left| \begin{array}{ccc} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \end{array} \right| = 9 \times 0 = 0$

Since the second and third columns are identical, the determinant equals zero.

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Question 4

$\left| \begin{array}{ccc} 1 & b & c \\ a(b+c) & 1 & c \\ a(b+c) & b & 1 \end{array} \right| = 0$

Solution:

L.H.S.:

$\left| \begin{array}{ccc} 1 & b & c \\ a(b+c) & 1 & c \\ a(b+c) & b & 1 \end{array} \right|$

​

Perform the operation

$C_3 \rightarrow C_3 – C_2$

C3​→C3​−C2​:

$= \left| \begin{array}{ccc} 1 & b & c-b \\ a(b+c) & 1 & c-1 \\ a(b+c) & b & 1-b \end{array} \right|$

​

Now, perform the operation

$C_2 \rightarrow C_2 – C_1$ $= \left| \begin{array}{ccc} 1 & b-1 & c-b \\ a(b+c) & 1 & c-1 \\ a(b+c) & b-1 & 1-b \end{array} \right|$

​​

Expanding along the first row, we get:

$= (1) \times \left| \begin{array}{cc} 1 & c-1 \\ b-1 & 1-b \end{array} \right| – (b-1) \times \left| \begin{array}{cc} a(b+c) & c-1 \\ a(b+c) & 1-b \end{array} \right| + (c-b) \times \left| \begin{array}{cc} a(b+c) & 1 \\ a(b+c) & b \end{array} \right| = 0$

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Question 5 to 7

Question 5

$\left| \begin{array}{ccc} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array} \right| = 2 \times \left| \begin{array}{ccc} a & p & x \\ b & q & y \\ c & r & z \end{array} \right|$

​​Solution:

L.H.S.:

$\left| \begin{array}{ccc} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array} \right|$

​

Perform the operation

$R_1 \rightarrow R_1 + R_2 + R_3$ $= \left| \begin{array}{ccc} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array} \right|$

​​

Factor out the common 2 from the first row:

$= 2 \times \left| \begin{array}{ccc} a+b+c & p+q+r & x+y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array} \right|$

​​Now, perform the operation

$R_1 \rightarrow R_1 – R_2$

and

$R_3 \rightarrow R_3 – R_1$ $= 2 \times \left| \begin{array}{ccc} a & p & x \\ b & q & y \\ c & r & z \end{array} \right|$

​​

Therefore, L.H.S. = R.H.S.

Hence Proved.

Question 6

$\left| \begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array} \right| = 0$

Solution:

Let

$\Delta = \left| \begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array} \right|$

Taking (-1) common from every row:

$\Delta = (-1)^3 \left| \begin{array}{ccc} 0 & -a & b \\ a & 0 & c \\ -b & -c & 0 \end{array} \right|$

​​

Interchange rows and columns:

$\Delta = – \left| \begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array} \right|$

​

Now,

$\Delta = -\Delta$

, which implies

$\Delta + \Delta = 0$ $2\Delta = 0$

2Δ=0, thus

$\Delta = 0$

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Question 7

$\left| \begin{array}{ccc} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{array} \right| = 4a^2b^2c^2$

Solution 7:

L.H.S.:

$\left| \begin{array}{ccc} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{array} \right|$

​​

Taking common

$a$

a from Row 1,

$b$

b from Row 2, and

$c$

c from Row 3:

$= abc \left| \begin{array}{ccc} -a & b & c \\ b & -b & c \\ c & c & -c \end{array} \right|$

​​

Now perform the operations

$R_1 \rightarrow R_1 + R_2 + R_3$ $= abc \times 2c \left| \begin{array}{ccc} -a & b & c \\ b & -b & c \\ c & c & -c \end{array} \right|$

​​

Expanding along the first column:

$= 2abc \left| \begin{array}{cc} b & c \\ c & -c \end{array} \right|$

​​

$= 4a^2b^2c^2$

Therefore, L.H.S. = R.H.S.

Hence Proved.

---

Class 12 Determinants Exercise 4.3

Question 1 to 3

Question 1

Find the area of the triangle with vertices at the points given in each of the following:

(i)

$(1, 0), (6, 0), (4, 3)$

(ii)

$(2, 7), (1, 1), (10, 8)$

(iii)

$(−2, −3), (3, 2), (−1, −8)$

Solution 1:

(i) The area of the triangle with vertices

$(1, 0), (6, 0), (4, 3)$

is given by the relation,

$\text{Area} = \frac{1}{2} \left| 1(0-3) + 6(3-0) + 4(0-0) \right| = \frac{1}{2} \times 18 = 9 \, \text{sq. units}$

(ii) The area of the triangle with vertices

$(2, 7), (1, 1), (10, 8)$

is given by the relation,

$\text{Area} = \frac{1}{2} \left| 2(1-8) + 1(8-7) + 10(7-1) \right| = \frac{1}{2} \times 24 = 12 \, \text{sq. units}$

(iii) The area of the triangle with vertices

$(−2, −3), (3, 2), (−1, −8)$

(−2,−3),(3,2),(−1,−8)

$\text{Area} = \frac{1}{2} \left| -2(2+8) + 3(-8+3) + -1(-3-2) \right| = \frac{1}{2} \times 62 = 31 \, \text{sq. units}$

---

Question 2

Show that the points

$A(a,b + c), B(b, c + a), C(c, a+b)$

are collinear.

Solution 2:

To show that the points are collinear, the area of the triangle formed by them must be zero.

Therefore, points

$A$

A,

$B$

and

$C$

are collinear.

---

Question 3

Find values of

$k$

k if the area of the triangle is 4 sq. units and vertices are:

(i)

$(k, 0), (4, 0), (0, 2)$

(k,0),(4,0),(0,2)
(ii)

$(−2, 0), (0, 4), (0, k)$

Solution 3:

(i) We know that the area of a triangle whose vertices are

$(x_1, y_1), (x_2, y_2), (x_3, y_3)$

(x1​,y1​),(x2​,y2​),(x3​,y3​) is the absolute value of the determinant

$\Delta$

, where

$\Delta = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$

When

$-k + 4 = -4$ $k = 8$

When

$-k + 4 = 4$

−k+4=4,

$k = 0$

Hence,

$k = 0, 8$

(ii) The area of the triangle with vertices

$(−2, 0), (0, 4), (0, k)$

is given by the relation,

$\Delta = \frac{1}{2} \left| -2(4-k) + 0(k-0) + 0(0-4) \right|$ $\therefore k – 4 = \pm 4$

∴k−4=±4

When

$k − 4 = −4$ $k = 0$

When

$k − 4 = 4$ $k = 8$

k=8.

Hence,

$k = 0, 8$

---

Question 4 to 5

Question 4

(i) Find the equation of the line joining

$(1, 2)$

(1,2) and

$(3, 6)$

(3,6) using determinants.

(ii) Find the equation of the line joining

$(3, 1)$

(3,1) and

$(9, 3)$

(9,3) using determinants.

Solution 4:

(i) Let

$P(x, y)$

P(x,y) be any point on the line joining the points

$(1, 2)$

(1,2) and

$(3, 6)$

Then, the area of the triangle that could be formed by these points is zero.

Hence, the equation of the line joining the given points is

$y = 2x$

y=2x.

(ii) Let

$P(x, y)$

P(x,y) be any point on the line joining points

$A(3, 1)$

A(3,1) and

$B(9, 3)$

. Then, the points

$A$

,

$B$

, and

$P$

are collinear. Therefore, the area of the triangle

$ABP$

ABP will be zero.

Hence, the equation of the line joining the given points is

$x − 3y = 0$

Question 5

If the area of the triangle is 35 square units with vertices

$(2, −6), (5, 4),$

and(k,4), then k is:

(A) 12
(B) −2
(C) −12, −2
(D) 12, −2

Solution 5:

The area of the triangle with vertices

$(2, −6), (5, 4),$

and

 

(k,4) is given by the relation,

It is given that the area of the triangle is

$\pm 35$

Therefore, we have:

$25 – 5k = \pm 35$ $5(5 – k) = \pm 35$ $5 – k = \pm 7$

When

$5 − k = −7$ $k = 5 + 7 = 12$

When

$5 − k = 7$ $k = 5 − 7 = −2$

Hence,

$k = 12, −2$

.

Therefore, option (D) is correct.