NCERT Solutions Class 12 Maths Chapter 4 - Determinants

 Exercise 4.1

Q1: Evaluate the determinants in Exercises 1 and 2.

Ans:
 = 2(−1) − 4(−5) = − 2 + 20 = 18

Q2: Evaluate the determinants in Exercises 1 and 2.

Ans:
(i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos² θ+ sin² θ = 1

(ii) = (x² − x + 1)(x + 1) − (x − 1)(x + 1)
= x³ − x² + x + x² − x + 1 − (x² − 1)
= x³ + 1 − x² + 1
= x³ − x² + 2

Q3:
Ans:
 .

Q4: 
Ans: The given matrix is  .


Q5: Evaluate the determinants

Ans: (i) Let
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

 

Q6: If
Ans:


Q7: Find values of x, if

Answer


Q8: If  , then x is equal to 
(A) 6 
(B) ±6 
(C) −6 
(D) 0
Ans: B

Hence, the correct answer is B.

Exercise 4.2

Q1: Find area of the triangle with vertices at the point 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)
Ans: (i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
(iii)The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,
Hence, the area of the triangle is |-15| = 15 sq units.

Q2: Show that points
A (a, b + c), B (b, c + a), C (c, a + b)  are collinear
Ans: Area of ΔABC is given by the relation,
= 0
Thus, the area of the triangle formed by points A, B, and C is zero. Hence, the points A, B, and C are collinear.  

Q3: Find values of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)
Ans: We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is the absolute value of the determinant (Δ), where
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
∴ -k + 4 = ±4
When −k + 4 = − 4, k = 8.
When −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,
∴ k - 4 = ±4
When k − 4 = − 4, k = 0.
When k − 4 = 4, k = 8.
Hence, k = 0, 8.

Q4: (i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Ans: (i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
Hence, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
Hence, the equation of the line joining the given points is x − 3y = 0.

Q5: If area of 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
Ans:(d)
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
When 5 − k = −7, k = 5 + 7 = 12.
When 5 − k = 7, k = 5 − 7 = −2.
Hence, k = 12, −2.
The correct answer is (d). 

Exercise 4.3

Q1: Write Minors and Cofactors of the elements of following determinants:

Ans: (i) The given determinant is Minor of element a_ij is M_ij.



Q2:

Ans:







Q3: Using Cofactors of elements of second row, evaluate Δ =.
Ans: The given determinant is
 .We have:



We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
∆ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7.

Q4: Using Cofactors of elements of third column, evaluate Δ=  

Ans: The given determinant is
.
We know that Δ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.