## NCERT Solutions for Class 12 Maths Exercise 4.2 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

Using the properties of determinants and without expanding in Exercise 1 to 7, prove that:

1.

**Ans.**Given:

Operating

0 = 0 [ C_{1} and C_{2} are identical]

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

### 2. = 0

**Ans.**

Operating

=

=> = 0 = R.H.S.

[ All entries of one column here first are zero]

### 3.

**Ans.**

operating

=∣∣∣∣235789637281∣∣∣∣=|276338725981|

=

= 9 x 0 = 0 [ two columns are identical] Proved.

### 4.

**Ans.**

=

Operating

=

= = 0 [ two columns are identical] Proved.

### 5.

**Ans.**L.H.S. =

operating

=

=

=

[operating ]

=

[operating ]

=

[operating ]

=

[Interchanging and ]

=

[Interchanging and ]

=

= = R.H.S.

### 6. = 0

**Ans.**Let

[Taking common from each row]

=>

Interchanging rows and columns in the determinants on R.H.S.,

Proved.

### 7.

**Ans.**L.H.S. =

Taking common from respectively,

=

[operating ]

=

=

=

= = = R.H.S.

### 8. (i)

(ii)

**Ans.**(i) L.H.S. =

and ,

=

∣∣∣∣∣100ab−ac−aa2(b−a)(b+a)(c−a)(c+a)∣∣∣∣∣|1aa20b−a(b−a)(b+a)0c−a(c−a)(c+a)|

Taking (b-a) and (c-a) common from R_{2} and R_{3 }respectively.

=(b−a)(c−a)∣∣∣∣∣100a11a2(b+a)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa201(b+a)01(c+a)|

R2→R2−R3R2→R2−R3

=(b−a)(c−a)∣∣∣∣∣100a01a2(b−c)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa200(b−c)01(c+a)|

[Expanding along 1^{st} column]

=(b−a)(c−a)∣∣∣01(b+a)(c+a)∣∣∣=(b−a)(c−a)|0(b+a)1(c+a)|

=(b−a)(c−a)(0−(b−c))=(b−a)(c−a)(0−(b−c))

=(b−a)(c−a)(c−b)=(b−a)(c−a)(c−b)

= = R.H.S. Proved.

(ii) L.H.S. =

operating and

=

=

=

=

=

=

=

=

=

= = R.H.S.

### 9.

**Ans.**

L.H.S. =

[Multiplying by respectively]

=1xyz∣∣∣∣∣x2y2z2x3y3z3xyzxyzxyz∣∣∣∣∣=1xyz|x2x3xyzy2y3xyzz2z3xyz|

=

Taking xyz commong from C_{3}

=

=

[operating ]

=

=

=

=

=

=

=

=

=

=

=

= = R.H.S.

### 10. (i)

(ii)

**Ans.**(i) L.H.S. =

=

Taking 5x+4 common from R_{1}

=

[operating and ]

=

=

= = R.H.S.

(ii) L.H.S. =

=

Taking 3y+k common from C_{1}

=

[operating and ]

=

=

= =

= R.H.S. Proved.

### 11. (i)

(ii)

**Ans.**(i) L.H.S. =

=

Taking a+b+c common from R_{1}

=

=

=

=

= = R.H.S. Proved.

(ii) L.H.S. =

=

Taking 2(x+y+z) common from C_{1}

=

=

=

=

= = R.H.S. Proved.

### 12.

**Ans.**L.H.S. =

= T

Taking 1+x+x^{2} common from R_{1}

=

=

=

=

=

=

=

=

=

= = R.H.S. Proved.

### 13.

**Ans.**L.H.S. =

=

=

=

=

=

= = R.H.S.

### 14.

**Ans.**L.H.S. =

Multiplying by respectively and then dividing the determinant by

=

Taking a,b, and c common from R_{1}, R_{2 }and R_{3} respectively.

=

=

=

=

=

= = R.H.S. Proved.

### Choose the correct answer in Exercises 15 and 16.

15. Let A be a square matrix of order 3 x 3, then is equal to:

(A)

(B)

(C)

(D)

**Ans.**Let A = be a square matrix of order 3 x 3. ……….(i)

= [From eq. (i)]

Therefore, option (C) is correct.

### 16. Which is 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.

**Ans.**Since, Determinant is a number associated to a square matrix.

Therefore, option (C) is correct.