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The 4th Chapter of NCERT Exemplar Solutions for Class 12 Mathematics is Determinants. This chapter covers topics such as introduction to determinants, then a determinant matrix of order one, two and three, properties of the determinant, minors and cofactor, adjoint and the inverse of a matrix, applications of matrices and solution of the system of linear equations using the inverse of a matrix. Further, students can make use of the solutions PDF ofNCERT Exemplar Solutions for Class 12 Maths Chapter 4 Determinants from the link given below to learn and understand all the concepts of determinants in an easy manner.
Access answers to NCERT Exemplar Solutions For Class 12 Maths Chapter 4 Determinants
Exercise 4.3 Page No: 77
Short Answer (S.A.)
Using the properties of determinants in Exercises 1 to 6, evaluate:
= (x2 – 2x + 2) . (x + 1) – (x – 1) . 0
= x3 – 2x2 + 2x + x2 – 2x + 2
= x3 – x2 + 2
[Expanding along first column]
= (x + y + z) . 1[3y(3z + x) + (3z)(x – y)]
= (x + y + z)(3yz + 3yx + 3xz)
= 3(x + y + z)(xy + yz + zx)
Lastly, expanding along R1, we have
= (a + b + c) [1 x 0 + (a + b + c)2]
= (a + b + c)3
Using the properties of determinants in Exercises 7 to 9, prove that:
10. If A + B + C = 0, then prove that
11. If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x1, y1),
(x2, y2), (x3, y3), then
12. Find the value of q satisfying
13. If , then find values of x.
14. If a1, a2, a3, …, ar are in G.P., then prove that the determinant
is independent of r.
Hence, the determinant is independent of r.
15. Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
Given points are (a + 5, a – 4), (a – 2, a + 3) and (a, a).
Now, we have to prove that these points do not lie on a straight line.
So, if we prove that these points form a triangle then it can’t line on a straight line.
Hence, the given points form a triangle and can’t lie on a straight line.
16. Show that the DABC is an isosceles triangle if the determinant
17. Find A–1 if and show that A-1 = (A2 – 3I)/ 2.
Long Answer (L.A.)
18. If A = , find A-1.
Using A–1, solve the system of linear equations
x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7.
19. Using matrix method, solve the system of equations
3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.
Given system of equations are:
3x + 2y – 2z = 3
x + 2y + 3z = 6 and
2x – y + z = 2
AX = B
20. Given find BA and use this to solve the system of
equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17.
21. If a + b + c ¹ 0 and then prove that a = b = c.
22. Prove that is divisible by a + b + c and find the quotient.
23. If x + y + z = 0, prove that
Objective Type Questions (M.C.Q.)
Choose the correct answer from given four options in each of the Exercises from 24 to 37.
24. If then, value of x is
(A) 3 (B) ± 3 (C) ± 6 (D) 6
Option (C) ± 6
From the given,
On equating the determinants, we have
2x2 – 40 = 18 + 14
2x2 = 72
x2 = 36
Thus, x = ± 6
25. The value of determinant
(A) a3 + b3 + c3 (B) 3 bc (C) a3 + b3 + c3 – 3abc (D) none of these
Option (C) a3 + b3 + c3 – 3abc
26. The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(A) 9 (B) 3 (C) – 9 (D) 6
Option (B) 3
We know that, the area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
27. The determinant equals
(A) abc (b–c) (c – a) (a – b) (B) (b–c) (c – a) (a – b)
(C) (a + b + c) (b – c) (c – a) (a – b) (D) None of these
28. The number of distinct real roots of = 0 in the interval -π/4 ≤ x ≤ π/4 is
(A) 0 (B) 2 (C) 1 (D) 3
Option (C) 1
29. If A, B and C are angles of a triangle, then the determinant is equal to
(A) 0 (B) -1 (C) 1 (D) None of these
Option (C) 0