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# NCERT Solutions class 12 Maths Exercise 4.5 Ch 4 Determinants

## NCERT Solutions for Class 12 Maths Exercise 4.5 Chapter 4 Determinants – FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants is a sure-shot way of obtaining the complete marks in the particular chapter for Board Exam 2019- 2020. CoolGyan provides you with Free PDF download of the same solved by Expert Teachers as per NCERT (CBSE) Book guidelines. They provide the students with precise and to the point answers which fetch very good marks in Board Exams. Download today the NCERT CBSE Solutions for Class 12 Maths Chapter 4 – Determinants to achieve your goal score. Class 12 Maths Chapter 4 – Determinants solved by Expert Teachers as per NCERT (CBSE) Book guidelines.

# NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

Find adjoint of each of the matrices in Exercise 1 and 2.

1.

Ans. Here A =

A11 = Cofactor of

A12 = Cofactor of

A21 = Cofactor of

A22 = Cofactor of

=∣∣∣4231∣∣∣=|4−3−21|′

=∣∣∣4321∣∣∣=|4−2−31|

### 2.

Ans. Here A =

=

=∣∣∣∣31261521115∣∣∣∣=|31−11−125−1625|

### Verify A (adj. A) = (adj. A) A=   in Exercise 3 and 4.

3.

Ans. Let A =

=   …..(i)

=     …..(ii)

And  =

Again      …..(iii)

From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A =

### 4.

Ans. Let A =

=

=∣∣∣∣0110311283∣∣∣∣=|032−11180−13|

……….(i)

……….(ii)

And

Also  =     ……….(iii)

From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A =

### Find the inverse of the matrix (if it exists) given in Exercise 5 to 11.

5.

Ans. Let A =

= 6-(-8) = 6+8 = 14014≠0

Matrix A is non-singular and hence  exist.

### 6.

Ans. Let A =

=

Matrix A is non-singular and hence  exist.

### 7.

Ans. Let A =

=

exists.

A11 = ,  A12 = ,

A13 = ,   A21 = ,

A22 = ,   A23 = ,

A31 = ,   A32 = ,

A33 =

=∣∣∣∣10001050242∣∣∣∣=|10−10205−4002|

### 8.

Ans. Let A =

=

exists.

A11 = , A12 = ,

A13 = , A21 = ,

A22 = , A23 = ,

A31 = ,   A32 = ,

A33 =

Ans. Let A =

=

exists.

A11 = , A12 = ,

A13 = , A21 = ,

A22 = , A23 = ,

A31 = ,  A32 = ,

A33 =

Ans. Let A =

exists.

A11 = ,  A12 = ,

A13 = , A21 = ,

A22 = , A23 = ,

A31 = , A32 = ,

A33 =

### 11.

Ans. Let A =

exists.

A11 = ,

A12 = , A13 = ,

A21 = ,   A22 = ,

A23 = ,  A31 = ,

A32 = , A33 =

### 12. Let A =  and B =  verify that

Ans. Given: Matrix A =

= 15 – 14 = 1  0

=

Matrix B =

= 54 – 56 =   0

Now AB =  =  =

=

Now L.H.S. =     ……….(i)

R.H.S. =

……….(ii)

From eq. (i) and (ii), we get

L.H.S. = R.H.S.

### 13. If A = , show that A2 – 5A + 7I = 0. Hence find

Ans. Given: A =

L.H.S. =

= R.H.S.

……(i)

To find: , multiplying eq. (i) by .

=

### 14. For the matrix A =  find numbers a and b such that

Ans. Given: A =

We have  ……….(i)

……………..(ii)

Here  satisfies  also, therefore

Putting  in eq. (i),

Here also  satisfies  , therefore

Therefore,  and

### 15. For the matrix A = , show that  Hence find

Ans. Given: A =

Now

L.H.S. =

=  = R.H.S.

Now,   to find , multiplying  by

=> 11A-1

### 16. If A = , verify that  and hence find

Ans. Given: A =

=

Now

L.H.S. =

=  = R.H.S.

Now,   to find , multiplying  by

=

### 17. Let A be a non-singular matrix of order 3 x 3. Then  is equal to:

(A)

(B)

(C)

(D)

Ans. If A is a non-singular matrix of order  then

Putting

Therefore, option (B) is correct.

### 18. If A is an invertible matrix of order 2, then det  is equal to:

(A) det A

(B)

(C) 1

(D) 0

Ans. Since

Therefore, option (B) is correct.