RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2

Rajasthan Board RBSE Class 12 Maths Chapter 4 Determinants Ex 4.2

Question 1.
, then find l : m.
Solution:

⇒ l x 3 – 2 x m = 0
⇒ 3l – 2m= 0
⇒ 3l = 2m
⇒ \(\frac { l }{ m }\) = \(\frac { 2 }{ 3 }\)
So, l : m = 2 : 3

Question 2.
Find the minors of elements of second row of determinant
.
Solution:

Question 3.
Find the value of determinant

Solution:

Question 4.
Write the effect on value of determinant, if first and third columns of any determinant are interchanged.
Solution:
Sign of determinant will be changed.

Question 5.
Prove that

Solution:

Question 6.
Find the value of determinant
.
Solution:

Question 7.
Solve the equation :

Solution:

Now, according to question, changed form of given eq.
i.e., eq. (i) is equal to zero.
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⇒ 1(- 145 + 143) – 1(- 58 + 13x + 104) + 2( – 22 – 5x +40) = 0
– 2 + 13x – 46 – 10x + 36 = 0
⇒ 3x – 12 = 0
⇒ \(\frac { 12 }{ 3 }\) = 4.

Question 8.
Prove without expansion that,

Solution:
Let

Question 9.
Prove that,

Solution:

= (a + b + c)[0 – 0 + 1{(a – c)(b – c) – (a – b) (b – a)}]
= (a + b + c){(ab – ca – bc + c²) – (ab – a² – b² + ab)}
= (a + b + c)(ab – ca – bc + c² – ab + a² + b² – ab)
= (a + b + c)(a² + b² + c² – ab – bc – ca)
= a³ + b³ + c³ – 3abc – R.H.S. Proved

Question 10.
Find

Solution:

Question 11.
If ω is a cube root, then find

Solution:

= 1(1 – ω²) – 1(1 – ω³) + ω²(ω – ω²)
= 1 – ω² – 1 + ω³ + ω³ – ω⁴
= 1 – ω² – 1 + 1 + 1 – ω³.ω (∵ ω³ = 1)
= 3 – ω² – 1 – ω (∵ ω³ = 1)
= 3 – (1 + ω + ω²) = 3 – 0 = 3 (∵ 1 + ω + ω² = 0)