# NCERT Solutions Class 12 Maths Chapter 4 DETERMINANTS

Ncert Solutions for class 12 subject Maths Chapter 4 DETERMINANTSin pdf Best Free NCERT Solutions for class 1 to 12 in pdf NCERT Solutions, cbse board, Maths, ncert Solutions for Class 12 Maths, class 12 Maths ncert solutions, DETERMINANTS, Class 12, ncert solutions chapter 4 DETERMINANTS, class 12 Maths, class 12 Maths ncert solutions, Maths ncert solutions class 12, Ncert Solutions Class 12 Mathematics Chapter 4 DETERMINANTS

## Exercise 4.1 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question 1. Evaluate the determinants in Exercises 1 and 2.

Use formula

= 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2. Evaluate the determinants in Exercises 1 and 2.

Use formula

= (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1
(ii)

Use formula

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

Question 3. If , then show that | 2A | = 4 | A |

LHS

|2A| =

Use formula

=2(4) – 4(8)
= 8 – 32
= - 24

R.H.S
4|A| = 4 = 4(1 x 2 - 2 x 4 )
= 4(2 – 8)
= 4(-6)
= - 24
L.H.S = R.H.S
Hence proved

Question 4. If, then show that | 3 A | = 27 | A |

LHS
=>|3A| =
Use formula

=>|3A| = 3(36 - 0) – 0(0 -0) + 3(0-0)
= 108
R.H.S
= 27{1(4 - 0) – 0(0 -0) + 1(0-0)}
= 27 x 4
= 108

LHS = RHS
Hence proved

Question 5. Evaluate the determinants

Solution: (i)

(ii)

(III)

(IV)

Question 6. If , find | A |

Question 7. Find values of x, if

Solution: (i)
2(1) - 5(4) = 2x(x) – 6(4)

1. 2 - 20 = 2x2 – 24

- 18 = 2x2 – 24
- 18 + 24 = 2x2
6 = 2x2
3 = x2

(ii)
2 (5) – 3(4) = x(5) – 2x(3)
10 – 12 = 5x – 6x
-2 = - x
x = 2

Question 8. If , then x is equal to
(A) 6 (B) ± 6 (C) – 6 (D) 0

Solution:

x(x) – 2(18) = 6(6) – 18(2)
X2 – 36 = 36 – 36
X2 = 36
x = ±6
Hence option B is correct

## Exercise 4.2 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question

Question

Question

Question

Question

Question

]

Question

Question

Question

=> (b-a)(c-a)(c2 –b2 + ac –ab +a2 –a2)
=> (b-a)(c-a){(c –b)(c+b) + a(c –b)}
=>(b-a)(c-a)(c-b)(a+b+c)
=> (a-b)(b-c)(c-a)(a+b+c)

Hence proved

Question

Question

Question

Question

Question

Question

Question

Question 15.Let A be a square matrix of order 3 × 3, then | kA| is equal to
(A) k| A| (B) k2 | A| (C) k3 | A| (D) 3k | A |

So option C is correct

Question 16. Which of the following is correct?
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these

Answer:We can calculate the determinant of a square matrix only so that Determinant is a number associated to a square matrix. Option (C) is correct

## Exercise 4.3 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question 1. 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)

Question 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Question 3. Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Answer is k = 0 and 8

Question 4.
(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.

Question 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

## Exercise 4.4 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question

Question

Question

Question

Question

Answer:Value of determinant is always equal to sum of product of rows or columns with their corresponding cofactor
In option D there is first column operation

So D is correct option

## Exercise 4.5 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question Verify that (AB)–1 = B–1 A–1.

Hence (AB)–1 = B–1 A–1

Question show that A2 – 5A + 7I = O. Hence find A–1.

Question 14. For the matrix find the numbers a and b such that A2 + aA + bI = O.

Question 15. For the matrix Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1.

Question

Question 17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
(A) |A| (B) |A|2 (C) |A|3 (D) 3|A|

Use property of determinant A.adj A = AI
Take mode both sides we get
As A is matrix of 3x3 hence
|AI| = A3I
Hence
ide by |A| both side we get
Hence option B is correct

Question 18. If A is an invertible matrix of order 2, then det (A–1) is equal to
(A) det (A) (B)1/det (A) (C) 1 (D) 0

AA-1 = I
Take determinant both side we get
|A ||A-1| = 1
Divide by |A| both side we get
|A-1| = 1/|A |
Hence option B is correct

## Exercise 4.6 Chapter 4 Class 12 Maths DETERMINANTS Ncert Solutions

Question 1. x + 2y = 2 and 2x + 3y = 3

Equations has unique solutions. Hence equations are consistence

Question 2. 2x – y = 5 and x + y = 4

Equations has unique solutions. Hence equations are consistence

Question 3. x + 3y = 5 and 2x + 6y = 8

Hence equations are consistence

Question 4. x + y + z = 1 , 2x + 3y + 2z = 2 and ax + ay + 2az = 4

Equation third is ided by a we get
x + y + z = 1
2x + 3y + 2z = 2
x + y + 2z = 4/a

Hence |A| ≠0
So Equations has unique solutions. Hence equations are consistence

Question 3x–y – 2z = 2, 2y – z =-1 and –3x – 5y = 3

Multiply both matrices we get

Hence it is inconsistence

Question 6. 5x – y + 4z = 5,2x + 3y + 5z = 2 and 5x – 2y + 6z = –1

|A|≠0
Hence equations are consistence

Question 7. 5x + 2y = 4 and 7x + 3y = 5

Hence answer is x =2 and y = -3

Question 8. 2x – y = –2 and 3x + 4y = 3

Question 9. 4x – 3y = 3 and 3x – 5y = 7

Question 10. 5x + 2y = 3 and 3x + 2y = 5

Question 11. 2x + y + z = 1, x – 2y – z =3/2 and 3y – 5z = 9

Question 12. x – y + z = 4, 2x + y – 3z = 0 and x + y + z = 2

Question 13. 2x + 3y +3 z = 5, x – 2y + z = – 4 and 3x – y – 2z = 3

Question 14. x – y + 2z = 7,3x + 4y – 5z = – 5 and 2x – y + 3z = 12

Question 15. If , find A–1. Using A–1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3

Question 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Let cost of each Kg of onion = Rs x
Cost of each Kg of wheat = Rs y
And cost of each Kg of rice = Rs z
Given that cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60
So
4x + 3y + 2 z = 60 …(1)
Given that The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90
So
2x + 4y + 6z = 90
Divide equation by 2 we get
X + 2y + 3z = 45 … (2)
Given that The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.
So
6x + 2y + 3z = 70 … (3)

So our set of equation is
4x + 3y + 2 z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70

### NCERT Books Free Pdf Download for Class 5, 6, 7, 8, 9, 10 , 11, 12 Hindi and English Medium

 Mathematics Biology Psychology Chemistry English Economics Sociology Hindi Business Studies Geography Science Political Science Statistics Physics Accountancy