## 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 + c^{2}) – (ab – a^{2} – b^{2} + ab)}

= (a + b + c)(ab – ca – bc + c^{2} – ab + a^{2} + b^{2} – ab)

= (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

= a^{3} + b^{3} + c^{3} – 3abc – R.H.S. Proved

Question 10.

Find

Solution:

Question 11.

If ω is a cube root, then find

Solution:

= 1(1 – ω^{2}) – 1(1 – ω^{3}) + ω^{2}(ω – ω^{2})

= 1 – ω^{2} – 1 + ω^{3} + ω^{3} – ω^{4}

= 1 – ω^{2} – 1 + 1 + 1 – ω^{3}.ω (∵ ω^{3} = 1)

= 3 – ω^{2} – 1 – ω (∵ ω^{3} = 1)

= 3 – (1 + ω + ω^{2}) = 3 – 0 = 3 (∵ 1 + ω + ω^{2} = 0)

Question 12.

Prove that

Solution:

Question 13.

If in determinant img, A_{1}, B_{1}, C_{1}, ……….. etc are co-factors of elements a_{1}, b_{1}, C_{1}, … respectively, then prove that :

Solution:

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