## Hello Students! In this post, you will find the complete** **NCERT Solutions for Maths Class 12 Exercise 4.4.

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**NCERT Solutions for Maths Class 12 Exercise 4.1**

**NCERT Solutions for Maths Class 12 Exercise 4.2**

**NCERT Solutions for Maths Class 12 Exercise 4.3**

**NCERT Solutions for Maths Class 12 Exercise 4.5**

**NCERT Solutions for Maths Class 12 Exercise 4.4**

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

**Maths Class 12 Ex 4.4 Question 1.**

**Solution:**

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

A_{11} = (–1)^{1 + 1} (4) = 4; A_{12} = (–1)^{1 + 2 }(3) = –3

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

Adj A = Transpose of

**Maths Class 12 Ex 4.4 Question 2.**

**Maths Class 12 Ex 4.4 Question 2.**

**Solution:**

**Maths Class 12 Ex 4.4 Question 3.**

**Maths Class 12 Ex 4.4 Question 4.**

**Maths Class 12 Ex 4.4 Question 5.**

**Maths Class 12 Ex 4.4 Question 6.**

**Maths Class 12 Ex 4.4 Question 7.**

**Maths Class 12 Ex 4.4 Question 8.**

**Maths Class 12 Ex 4.4 Question 9.**

**Solution:**

|A| = 2(–1 – 0) – 1(4 – 0) + 3(8 – 7) = –2 – 4 + 3 = –3

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

C_{11} = –1, C_{12} = –4, C_{13} = 1, C_{21} = 5, C_{22} = 22, C_{23} = –11, C_{31} = 3, C_{32} = 12, C_{33} = –6

**Solution:**

|A| = 2(–1 – 0) – 1(4 – 0) + 3(8 – 7) = –2 – 4 + 3 = –3

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

_{11}= –1, C

_{12}= –4, C

_{13}= 1, C

_{21}= 5, C

_{22}= 22, C

_{23}= –11, C

_{31}= 3, C

_{32}= 12, C

_{33}= –6

**Maths Class 12 Ex 4.4 Question 10.**

**Solution:**

|A| = 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = 2 + 9 – 12 = –1 ≠ 0

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

C_{11} = 2, C_{12} = –9, C_{13} = –6, C_{21} = 0, C_{22} = –2, C_{23} = –1, C_{31} = –1, C_{32} = 3, C_{33} = 2

**Solution:**

|A| = 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = 2 + 9 – 12 = –1 ≠ 0

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

_{11}= 2, C

_{12}= –9, C

_{13}= –6, C

_{21}= 0, C

_{22}= –2, C

_{23}= –1, C

_{31}= –1, C

_{32}= 3, C

_{33}= 2

**Maths Class 12 Ex 4.4 Question 11.**

**Solution:**

First find |A| = –cos² Î± – sin² Î± = –(cos² Î± + sin² Î±) = –1 ≠ 0

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

C_{11} = –1, C_{12} = 0, C_{13} = 0, C_{21} = 0, C_{22} = –cos *a*, C_{23} = –sin *a*, C_{31} = 0, C_{32} = –sin *a*, C_{33} = cos *a*

**Solution:**

First find |A| = –cos² Î± – sin² Î± = –(cos² Î± + sin² Î±) = –1 ≠ 0

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

_{11}= –1, C

_{12}= 0, C

_{13}= 0, C

_{21}= 0, C

_{22}= –cos

*a*, C

_{23}= –sin

*a*, C

_{31}= 0, C

_{32}= –sin

*a*, C

_{33}= cos

*a*

**Maths Class 12 Ex 4.4 Question 12.**

**Maths Class 12 Ex 4.4 Question 13.**

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

^{-1}.

**Solution:**

**Maths Class 12 Ex 4.4 Question 14.**

**Hence, find A**^{-1}.

**Solution:**

**Hence, find A**

^{-1}.

**Solution:**

**Maths Class 12 Ex 4.4 Question 15.**

**Maths Class 12 Ex 4.4 Question 16.**

**Maths Class 12 Ex 4.4 Question 17.**

**Maths Class 12 Ex 4.4 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

**Solution:**

|A| ≠ 0

⇒ A^{-1} exists

⇒ AA^{-1} = I

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

⇒ |A||A^{-1}| = I

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

Hence, option (B) is correct.

Related Links:

**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

**Solution:**

|A| ≠ 0

⇒ A^{-1} exists

⇒ AA^{-1} = I

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

⇒ |A||A^{-1}| = I

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

Hence, option (B) is correct.

Related Links:

**NCERT Solutions for Maths Class 12 Exercise 4.1**

**NCERT Solutions for Maths Class 12 Exercise 4.2**