NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths 4 Determinants Ex 4.5.

- Determinants Class 12 Ex 4.1
- Determinants Class 12 Ex 4.2
- Determinants Class 12 Ex 4.3
- Determinants Class 12 Ex 4.4
- Determinants Class 12 Ex 4.6

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Determinants |

Exercise |
Ex 4.5 |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 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.

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