# NCERT Solutions class 12 Maths Exercise 3.4 Ch 3 Matrices

## NCERT Solutions for Class 12 Maths Exercise 3.4 NCERT Solutions for Class 12 Maths Chapter 3 – Matrices – FREE PDF Download

NCERT Class 12 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 12 Maths Chapter 3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. Class 12 Maths Chapter 3 – Matrices solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Matrices Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 3 – Matrices

Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.

1.

Ans. Let A =

Since A = IA

=

Ans. Let A =

### 3.

Ans. Let A =

Since A = IA

= =>[7231]=>[7−3−21]

### 4.

Ans. Let A =

Since A = IA

=>[1213]=[2110]A=>[1−123]=[−2−110]A

[R2R22R1][R2→R2−2R1]

=

### 5.

Ans. Let A =

Since A = IA

=>[1101]=[4311]A=>[1011]=[4−1−31]A

=>[1001]=[4712]A=>[1001]=[4−1−72]A

= [4712][4−1−72]

### 6.

Ans. Let A =

Since A = IA

=>[1235]=[0110]A=>[1325]=[0110]A

=

### Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 7 to 14.

7.

Ans. Let A =

Since A = IA

[R2R25R1][R2→R2−5R1]

=

### 8.

Ans. Let A =

Since A = IA

=

### 9.

Ans. Let A =

Since A = IA

=

### 10.

Ans. Let A =

Since A = IA

[R1(1)R1][R1→(−1)R1]

=>[1012]=[1413]A=>[1−10−2]=[−1−1−43]A

=

### 11.

Ans. Let A =

Since A = IA

=>[1226]=[0110]A=>[1−22−6]=[0110]A

=

### 12.

Ans. Let A =

Since A = IA

Here, all entries in second row of left side are zero.

does not exist.

### 13.

Ans. Let A =

Since A = IA

=

### 14.

Ans. Let A =

Since A = IA

Here, all entries in second row of left side are zero.

does not exist.

15.

Ans. Let A =

We know that A = IA,

[R2R2+R3][R2→R2+R3]

### 16.

Ans. Let A =

Since, A = IA

=>⎡⎣⎢⎢100010001⎤⎦⎥⎥=⎡⎣⎢⎢⎢⎢1253525425125351125925⎤⎦⎥⎥⎥⎥A=>[100010001]=[1−25−35−254251125−35125925]A

=

### 17.

Ans. Let A =

Since, A = IA

=>⎡⎣⎢⎢100121253⎤⎦⎥⎥=⎡⎣⎢⎢250120001⎤⎦⎥⎥A=>[1120−2−5013]=[−2105−20001]A

=>⎡⎣⎢⎢100112235⎤⎦⎥⎥=⎡⎣⎢⎢205102010⎤⎦⎥⎥A=>[1120130−2−5]=[−2100015−20]A

=

### 18.  Matrices A and B will be inverse of each other only if:

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans. By definition of inverse of square matrix,

Option (D) is correct.