# NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-2/

## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.2

Class 12 Maths Ex 4.2 Solutions NCERT Solutions Question 1.
$$\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right|$$ = 0
Solution:
L.H.S = $$\left| \begin{matrix} x & a & x \\ y & b & y \\ z & c & z \end{matrix} \right| +\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right|$$
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Ncert Solutions Class 12 Maths Chapter 4 Exercise 4.2 Question 2.
$$\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0$$
Solution:

Exercise 4.2 Class 12 NCERT Solutions Question 3.
$$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0$$
Solution:
$$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{matrix} \right|$$
$${ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }-{ 9C }_{ 2 }=0$$

Ex 4.2 Class 12 Maths NCERT Solutions  Question 4.
$$\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| =0$$
Solution:

Ex 4.2 Class 12 NCERT Solutions Question 5.
$$\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| =2\left| \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} \right|$$
Solution:

By using properties of determinants in Exercise 6 to 14, show that:

Question 6.
$$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$$
Solution:

Question 7.
$$\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$$
Solution:

Question 8.
(a) $$\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$$
(b) $$\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 3 } & { b }^{ 3 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(a+b+c)$$
Solution:
(a) $$\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$$

Taking out (b – a) from R2 and (c – a) from R3.
Expanding along C1
= (b – a)(c – a)$$\left|\begin{array}{ll} 1 & b+a \\ 1 & c+a \end{array}\right|$$
= (b – a) (c – a) (c + a – b – a) = (b – a) (c – a) (c – b)
= (-1 )(a – b)(c – a)(-1)(b – c)
= (a- b) (b – c) (c – a)

(b) $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|$$

Question 9.
$$\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| =(x-y)(y-z)(z-x)(xy+yz+zx)$$
Solution:
Let ∆ = $$\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right|$$
Applying R1 → R1 – R2, R2 → R2 – R3

Question 10.
(a) $$\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right| =(5x+4){ (4-x) }^{ 2 }$$
(b) $$\left| \begin{matrix} y+x & y & y \\ y & y+k & y \\ y & y & y+k \end{matrix} \right| ={ k }^{ 2 }(3y+k)$$
Solution:

Question 11.
(a) $$\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| ={ (a+b+c) }^{ 3 }$$
(b) $$\left| \begin{matrix} x+y+2z & \quad z & \quad z \\ x & \quad y+z+2x & \quad x \\ y & y & \quad z+x+2y \end{matrix} \right| ={ 2(x+y+z) }^{ 3 }$$
Solution:

Question 12.
$$\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| ={ { (1-x }^{ 3 }) }^{ 2 }$$
Solution:

Question 13.
$$\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| ={ (1+{ a }^{ 2 }+{ b }^{ 2 }) }^{ 3 }$$
Solution:

Question 14.
$$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$$
Solution:
Let
∆ = $$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right|$$
$$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab+0 & \quad ac+0 \\ ab+0\quad & \quad b^{ 2 }+1 & \quad bc+0 \\ ca+0\quad & \quad cb+0 & \quad { c }^{ 2 }+1 \end{matrix} \right|$$
This may be expressed as the sum of 8 determinants

Question 15.
If A be a square matrix of order 3×3, then | kA | is equal to
(a) k|A|
(b) k² |A|
(c) k³ |A|
(d) 3k|A|
Solution:
Option (c) is correct.

Question 16.
Which of 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
Solution:
Option (c) is correct.

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