Contents

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths 4 Determinants

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Determinants |

Exercise |
Ex 4.1, Ex 4.2, Ex 4.3, Ex 4.4,Ex 4.5,Ex 4.6 |

Number of Questions Solved |
68 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 4 Determinants

### Chapter 4 Determinants Exercise 4.1

**Question 1.**

**Evaluate the following determinant:**

**Solution:**

= 2x(-1)-(-5)x(4)

=-2+20

=18

**Question 2.**

**(i) **

**(ii) **

**Solution:**

(i)

= cosθ cosθ – (sinθ)(-sinθ)

= cos²θ + sin²θ

= 1

**Question 3.**

**If then show that |2A|=|4A|**

**Solution:**

=>

L.H.S = |2A|

=

= – 24

**Question 4.**

** , then show that |3A| = 27|A|**

**Solution:**

3A =

=

**Question 5.**

**Evaluate the following determinant:**

**(i) **

**(ii) **

**(iii) **

**(iv) **

**Solution:**

(i)

**Question 6.**

**If , find |A|**

**Solution:**

|A| =

= 1(-9+12)-1(-18+15)-2(8-5)

= 0

**Question 7.**

**Find the values of x, if**

**(i) **

**(ii)**

**Solution:**

(i)

=> 2 – 20 = 2x² – 24

=> x² = 3

=> x = ±√3

(ii)

or

2 × 5 – 4 × 3 = 5 × x – 2x × 3

=>x = 2

**Question 8.**

**If , then x is equal to**

**(a) 6**

**(b) +6**

**(c) -6**

**(d) 0**

**Solution:**

(b)

=> x² – 36 = 36 – 36

=> x² = 36

=> x = ± 6

### Chapter 4 Determinants Exercise 4.2

**Using the property of determinants and without expanding in Q 1 to 5, prove that**

**Question 1.**

**Solution:**

L.H.S =

(C1 = C3 and C2 = C3)

= 0 + 0

= 0

= R.H.S

**Question 2.**

**Solution:**

L.H.S =

**Question 3.**

**Solution:**

**Question 4.**

**Solution:**

L.H.S =

**Question 5.**

**Solution:**

L.H.S = ∆ =

**By using properties of determinants in Q 6 to 14, show that**

**Question 6.**

**Solution:**

L.H.S = ∆ = …(i)

**Question 7.**

**Solution:**

L.H.S =

**Question 8.**

**(a) **

**(b) **

**Solution:**

(a) L.H.S =

**Question 9.**

**Solution:**

Let ∆ =

Applying R1–>R1 – R2, R2–>R2 – R3

**Question 10.**

**(a) **

**(b) **

**Solution:**

(a) L.H.S =

**Question 11.**

**(a) **

**(b) **

**Solution:**

(a) L.H.S =

=

**Question 12.**

**Solution:**

L.H.S =

**Question 13.**

**Solution:**

L.H.S =

**Question 14.**

**Solution:**

Let ∆ =

This may be expressed as the sum of 8 determinants

**Question 15.**

**If 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|**

**Solution:**

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**

**Solution:**

Option (c) is correct

### Chapter 4 Determinants Exercise 4.3

**Question 1.**

**Find the 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)**

**Solution:**

(i) Area of triangle =

= [1(0-3)+1(18-0)]

= 7.5 sq units

**Question 2.**

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

**Solution:**

The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)

**Question 3.**

**Find the value 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).**

**Solution:**

(i) Area of ∆ = 4 (Given)

= [-2k+8]

= -k+4

Case (a): -k + 4 = 4 ==> k = 0

Case(b): -k + 4 = -4 ==> k = 8

Hence, k = 0,8

(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)

**Question 4.**

**(i) Find the equation of line joining (1, 2) and (3,6) using determinants.**

**(ii) Find the equation of line joining (3,1), (9,3) using determinants.**

**Solution:**

(i) Given: Points (1,2), (3,6)

Equation of the line is

**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**

**Solution:**

(d) Area of ∆ =

= [50 – 10k] = 25 – 5k

∴ 25-5k = 35 or 25-5k = -35

-5k = 10 or 5k = 60

=> k = -2 or k = 12

### Chapter 4 Determinants Exercise 4.4

**Question 1.**

**Write the minors and cofactors of the elements of following determinants:**

**(i) **

**(ii) **

**Solution:**

(i) Let A =

M_{11} = 3, M_{12} = 0, M_{21} = – 4, M_{22 }= 2

For cofactors

**Question 2.**

**Write Minors and Cofactor of elements of following determinant**

**(i) **

**(ii) **

**Solution:**

(i) Minors M_{11} = = 1

**Question 3.**

**Using cofactors of elements of second row, evaluate**

**Solution:**

Given

**Question 4.**

**Using Cofactors of elements of third column, evaluate**

**Solution:**

Elements of third column are yz, zx, xy

**Question 5.**

**If and Aij is the cofactors of aij? then value of ∆ is given by**

**(a) a _{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}**

**(b) a**

_{1}_{1}A_{11}+a_{12}A_{21}+a_{13}A_{31}**(c) a**

_{2}_{1}A_{11}+a_{22}A_{12}+a_{23}A_{13}**(d) a**

_{1}_{1}A_{11}+a_{21}A_{21}+a_{31}A_{31}**Solution:**

Option (d) is correct.

### Chapter 3 Determinants Exercise 4.5

**Find the adjoint of each of the matrices in Questions 1 and 2.**

**Question 1.**

**Solution:**

Let C_{ij} be cofactor of a_{ij} in A. Then, the cofactors of elements of A are given by

C_{11} = (-1)^{1+1} (4) = 4; C_{12} = (-1)^{1+2 }(3) = -3

C_{21} = (-1)^{2+1} (2)= – 2; C_{22} = (-1)^{2+2} (1) = -1

Adj A =

=

**Question 2.**

**Solution:**

Similarly,

**Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.**

**Question 3.**

**Solution:**

|A| = 24

**Question 4.**

**Solution:**

A_{11} = 0, A_{12 }= – 11, A_{13 }= 0,

A_{21 }= – 3, A_{22 }= 1, A_{23 }= 1, A_{31} = – 2

A_{32 }= 8, A_{33 }= – 3

**Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:**

**Question 5.**

**Solution:**

So, A is a non-singular matrix and therefore it is invertible. Let c_{ij} be cofactor of a_{ij} in A. Then, the cofactors of elements of A are given by

**Question 6.**

**Solution:**

So, A is a non-singular matrix and therefore it is invertible. Let c_{ij} be cofactor of a_{ij} in A. Then, the cofactors of elements of A are given by

**Question 7.**

**Solution:**

|A| = 10

**Question 8.**

**Solution:**

So, A is a non-singular matrix and therefore it is invertible. Let c_{ij} be cofactor of a_{ij }in A. Then, the cofactors of elements of A are given by

**Question 9.**

**Solution:**

|A| =

= 2(-1-0)-1(4-0)+3(8-3)

So, A is non-singular matrix and therefore, it is invertible.

**Question 10.**

**Solution:**

|A| =

= 1(8-6)+1(0+9)+2(0-6)

= 2+9-12

= -1≠0

∴A is invertible and

**Question 11.**

**Solution:**

A =

adj A =

First find |A| = -cos²α-sin²α

=-1≠0

**Question 12.**

**Let , verify that (AB) ^{-1} = B^{-1}A^{-1}**

**Solution:**

Here |A| =

= 15-14

= 1≠0

**Question 13.**

**If show that A² – 5A + 7I = 0,hence find A ^{-1}**

**Solution:**

A =

A² =

**Question 14.**

**For the matrix A = find file numbers a and b such that A²+aA+bI²=0. Hence, find A ^{-1}.**

**Solution:**

A =

A²+aA+bI²=0

**Question 15.**

**For the matrix ****Show that A³-6A²+5A+11I _{3}=0.Hence find A^{-1}**

**Solution:**

A² =

**Question 16.**

**If show that A³-6A²+9A-4I=0 and hence, find A ^{-1}**

**Solution:**

We have

**Question 17.**

**Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:**

**(a) | A |**

**(b) | A |²**

**(c) | A |³**

**(d) 3 | A |**

**Solution:**

Let A =

Dividing by | A |, |Adj. A| = | A |²

Hence, Part (b) is the correct answer.

**Question 18.**

**If A is an invertible matrix of order 2, then det. (A ^{-1}) is equal to:**

**(a) det. (A)**

**(b)**

**(c) 1**

**(d) 0**

**Solution:**

|A|≠0

=> A

^{-1}exists => AA

^{-1}= I

|AA

^{-1}| = |I| = I

=> |A||A

^{-1}| = I

Hence option (b) is correct.

### Chapter 4 Determinants Exercise 4.6

**Examine the consistency of the system of equations in Questions 1 to 6:**

**Question 1.**

**x + 2y = 2**

**2x + 3y = 3**

**Solution:**

x + 2y = 2,

2x + 3y = 3

=>

=> AX = B

Now |A| =

= 3 – 4

= – 1 ≠ 0.

Hence, equations are consistent.

**Question 2.**

**2x – y = 5**

**x + y = 4**

**Solution:**

2x – y = 5,

x + y = 4

=>

=> AX = B

Now |A| =

= 2 + 1

= 3 ≠ 0.

Hence, equations are consistent.

**Question 3.**

**x + 3y = 5,**

**2x + 6y = 8**

**Solution:**

x + 3y = 5,

2x + 6y = 8

=>

=> AX = B

Now |A| =

= 6 – 6

= 0.

Hence, equations are consistent with no solution

**Question 4.**

**x + y + z = 1**

**2x + 3y + 2z = 2**

**ax + ay + 2az = 4**

**Solution:**

x + y + z = 1

2x + 3y + 2z = 2

x + y + z =

**Question 5.**

**3x – y – 2z = 2**

**2y – z = – 1**

**3x – 5y = 3**

**Solution:**

=> AX = B

**Question 6.**

**5x – y + 4z = 5**

**2x + 3y + 5z = 2**

**5x – 2y + 6z = -1**

**Solution:**

Given

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

= 5(18 + 10)+1(12 – 25)+4(-4-15)

= 140-13-76

= 51 ≠ 0

Hence equations are consistent with a unique

solution.

**Solve system of linear equations using matrix method in Questions 7 to 14:**

**Question 7.**

**5x + 2y = 4**

**7x + 3y = 5**

**Solution:**

The given system of equations can be written as

**Question 8.**

**2x – y = – 2**

**3x + 3y = 3**

**Solution:**

The given system of equations can be written

**Question 9.**

**4x – 3y = 3**

**3x – 5y = 7**

**Solution:**

The given system of equations can be written as

where

**Question 10.**

**5x + 2y = 3**

**3x + 2y = 5**

**Solution:**

The given system of equations can be written as

where

**Question 11.**

**2x + y + z = 1,**

**x – 2y – z = 3/2**

**3y – 5z = 9**

**Solution:**

The given system of equations are

2x + y + z = 1,

x – 2y – z = 3/2,

3y – 5z = 9

We know AX = B => X = A^{-1}B

**Question 12.**

**x – y + z = 4**

**2x + y – 3z = 0**

**x + y + z = 2.**

**Solution:**

The given system of equations can be written

**Question 13.**

**2x + 3y + 3z = 5**

**x – 2y + z = – 4**

**3x – y – 2z = 3**

**Solution:**

The given system of equations can be written as:

**Question 14.**

**x – y + 2z = 7**

**3x + 4y – 5z = – 5**

**2x – y + 3z = 12.**

**Solution:**

The given system of equations can be written

**Question 15.**

**If A = Find A ^{-1}. Using A^{-1}. **

**Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3**

**Solution:**

We have AX = B

where

**Question 16.**

**The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. 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 the cost of each item per kg by matrix method**

**Solution:**

Let cost of 1 kg onion = Rs x

and cost of 1 kg wheat = Rs y

and cost of 1 kg rice = Rs z

4x+3y+2z=60

2x+4y+6z=90

6x+2y+3z=70

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