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.
img
⇒ 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)
Question 12.
Prove that
Solution:
Question 13.
If in determinant img, A1, B1, C1, ……….. etc are co-factors of elements a1, b1, C1, … respectively, then prove that :
Solution:
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