Matrices%20and%20Determinants

Matrices%20and%20Determinants - Topic Wise Multiple Choice Questions (MCQ)

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Hindi Question

1 )

If the system of equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 3y – 4z = 0 has non-trivial solution, then (xy / z²) = ….

â€“ (5/6)

(6/5)

â€“ (6/5)

(5/6)

View Solution

Answer :

writing given equations in determinant form.

1	k	3
3	k	– 2	= 0
2	3	– 4

∴ 1(– 4k + 6) – k(– 12 + 4) + 3(9 – 2k) = 0

∴ – 4k + 6 + 12k – 4k + 27 – 6k = 0

∴ 2k = 33

∴ k = (33 / 2)

putting value of k in given equations

(x / 15) = [y / (– 2)] = (z / 6)

∴ (xy / z²) = [(– 30) / 36] = – (5/6).

2 )

n	_nP_n	_nC_n
n + 1	_n+1P_n+1	_n+1C_n+1	= …..
n + 2	_n+2P_n+2	_n+2C_n+2

(n2 + n + 1)n!

n(n + 1)!

(n + 1)n!

(n + 2)n!

View Solution

Answer :

	n	n!	1
D =	n + 1	(n + 1)!	1
	n + 2	(n + 2)!	1

as ⁿP_n = n!

ⁿC_n = 1

ⁿ⁺¹P_n+1 = (n + 1)!

ⁿ⁺¹C_n+1 = 1

ⁿ⁺²P_n+2 = (n + 2)!

ⁿ⁺²C_n+2 = 1 also (n + 1)! = (n + 1)n!

	n	1	1
∴ D = n!	n + 1	(n + 1)	1
	n + 2	(n + 1)(n + 2)	1

R₃ – R₂ and R₂ – R₁ gives,

	n	1	1
D = n!	1	n	0
	1	(n + 1)²	0

after expanding,

∴ D = n![n(0 – 0) – 1(0) + 1(n + 1)² – n]

= n![n² + 2n + 1 – n]

= n!(n² + n + 1)

3 )

Let a, b, c be such that b(a + c) ≠ 0. If

a	a + 1	a – 1		a + 1	b + 1	c – 1
– b	b + 1	b – 1	+	a – 1	b – 1	c + 1	= 0,
c	c – 1	c + 1		(– 1)ⁿ⁺² a	(– 1)ⁿ⁺¹ b	(– 1)ⁿc

then n is

zero

any even integer

any odd integer

any integer

View Solution

Answer :

Consider second determinant,

(– 1)ⁿ⁺² = (– 1)ⁿ (– 1)² = (– 1)ⁿ, (– 1)ⁿ⁺¹ = (– 1)ⁿ (– 1)

Hence D₂ can be written as

	a + 1	b + 1	c – 1
D₂ =	a – 1	b – 1	c + 1
	(– 1)ⁿa	– (– 1)ⁿb	(– 1)ⁿc

	a + 1	b + 1	c – 1
= (– 1)ⁿ	a – 1	b – 1	c + 1	(1)
	a	– b	c

	a	a + 1	a – 1
First determinant = D₁ =	– b	b + 1	b – 1
	c	c – 1	c + 1

Taking transpose we get

a	– b	c
a + 1	b + 1	c – 1
a – 1	b – 1	c + 1

R₂ ↔ R₁

	a + 1	b + 1	c – 1
D₁ = –	a	– b	c
	a – 1	b – 1	c + 1

R₂ ↔ R₃

	a + 1	b + 1	c – 1
∴ D₁ = (– 1) (– 1)	a – 1	b – 1	c + 1
	a	– b	c

from (1), D₂ = (– 1)ⁿ D₁

∴ given expression is D₁ + D₂ = 0

i.e. D₁ [1 + (– 1)ⁿ] = 0 for any odd integer

∵ b(a + c) ≠ 0

∴ D₁ ≠ 0

∴ n should be odd integer

4 )

sin(x + p)	sin(x + q)	sin(x + r)
sin(y + p)	sin(y + q)	sin(y + r)	= …
sin(z + p)	sin(z + q)	sin(z + r)

sin(x + y + z)

sin(p + q + r)

View Solution

Answer :

∵ sin(A + B) = sin A cos B + cos A sin B.

∴	sin x cos p + cos x sin p	sin x cos q + cos x sin q	sin x cos r + cos x sin r
	sin y cos p + cos y sin p	sin y cos q + cos y sin q	sin y cos r + cos y sin r
	sin z cos p + cos z sin p	sin z cos q + cos z sin q	sin z cos r + cos z sin r

Hence determinant can be written as

sin x	cos x		cos p	cos q	cos r
sin y	cos y	∙	sin p	sin q	sin r
sin x	cos z		0	0	0

= 0 × 0 = 0

5 )

3x – 8	3	3
3	3x – 8	3	= 0, then x = …..
3	3	3x – 8

(3/2), (3 / 11)

(3/2), (11 / 3)

(2/3), (11 / 3)

(2/3), (3 / 11)

View Solution

Answer :

	3x – 8	3	3
D =	3	3x – 8	3	= 0
	3	3	3x – 8

c₁ → c₁ + c₂ + c₃

	(3x – 8 + 3 + 3)	3	3
D =	(3 + 3x – 8 + 3)	3x – 8	3	= 0
	(3 + 3 + 3x – 8)	3	3x – 8

	3x – 2	3	3
∴ D =	3x – 2	3x – 8	3	= 0
	3x – 2	3	3x – 8

	1	3	3
∴ (3x – 2)	1	3x – 8	3	= 0
	1	3	3x – 8

∴ 3x – 2 = 0 and

∴ [(3x – 8)² – 9] – 3[3x – 8 – 3] + 3[3 – 3x + 8] = 0

∴ [(3x – 8)² – 9] – 3[3x – 11] + 3[11 – 3x] = 0

∴ [(3x – 8)² – 9] – 9x + 33 + 33 – 9x = 0

∴ 9x² – 48x + 55 – 18x + 66 = 0

∴ 9x² – 66x + 121 = 0

∴ x = [{66±√(66² – 4 × 9 × 121)} / 18] = (11 / 3)

also as 3x – 2 = 0 ⇒ x = (2/3)

hence

x = (2/3), (11 / 3)

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