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

**Question 1. Evaluate the determinants in Exercises 1 and 2. Answer: **

Use formula

= 2(−1) − 4(−5) = − 2 + 20 = 18

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

** Answer **Use formula

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

Use formula

= (x

= x

= x

= x

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

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 |Answer:**

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 |**

**Answer:**

**Question 7. Find values of x, if**

**Solution:** (i)

2(1) - 5(4) = 2x(x) – 6(4)

- 2 - 20 = 2x
^{2}– 24

- 18 = 2x^{2} – 24

- 18 + 24 = 2x^{2}

6 = 2x^{2}

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)

X^{2} – 36 = 36 – 36

X^{2} = 36

x = ±6

Hence option B is correct

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**]**

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=> (b-a)(c-a)(c

=> (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)

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**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 | **

**Answer:**

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

**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)**

**Answer:**

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

**Answer:**

**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:**

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.**

**Answer:**

**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 **

**Answer:**

**Question **

**Answer:**

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**Answer:**

Answer part (ii)

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**Answer:**

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**Answer:**

**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

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**Question Verify that (AB) ^{–1} = B^{–1} A^{–1}.**

**Answer:**

**Hence (AB) ^{–1} = B^{–1} A^{–1}**

**Question show that A ^{2} – 5A + 7I = O. Hence find A^{–1}.**

**Answer:**

**Question 14. For the matrix find the numbers a and b such that A ^{2} + aA + bI = O.**

**Answer:**

**Question 15. For the matrix Show that A ^{3}– 6A^{2} + 5A + 11 I = O. Hence, find A^{–1}.**

**Answer:**

**Question **

**Answer:**

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|

**Answer:**We have the formula

Use property of determinant A.adj A = AI

Take mode both sides we get

|A.adj A| = |AI|

As A is matrix of 3x3 hence

|AI| = A^{3}I

Hence

|A| |adj A| = |A|^{3}

ide by |A| both side we get

|adj A| = |A^{2}|

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

**Answer:**We have the formula

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

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

**Answer:**

Equations has unique solutions. Hence equations are consistence

Question 2. 2x – y = 5 and x + y = 4

**Answer:**

Equations has unique solutions. Hence equations are consistence

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

**Answer:**

Hence equations are consistence

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

**Answer:**

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

**Answer:**

** **

** **

** ****Multiply both matrices we get **

** **

** ****Hence it is inconsistence **

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

**Answer:**

|A|≠0

Hence equations are consistence

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

**Answer:**

Hence answer is x =2 and y = -3

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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.

**Answer:**

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

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