Matrices and Determinants

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

Topic Wise Matrices and Determinants MCQ for competitive exam Practice Multiple Choice Questions for Engg Bank JEE CET Entrance Online Test

Hindi Question

11 )

The homogeneous system of equations

2	α + β + γ + δ	αβ + γδ	x
α + β + γ + δ	α(α + β)(γ + δ)	αβ(γ + δ) + γδ(α + β)	y	= 0
αβ + γδ	αβ(γ + δ) + γδ(α + β)	2αβγδ	z

has non-trivial solutions only if…..

α + β + γ + δ = 0

for any α, β, γ, δ

αβ + γδ = 0

αβ(γ + δ) + γδ(α + β)

View Solution

Answer :

1	1	1	γ + δ	γδ
α + β	γ + δ	1	α + β	αβ
αβ	γδ	0	0	0

= [1(0) – 1(0) + 0] [1(0) – (γ + δ)(0) + γδ(0)]

= 0 × 0 = 0

Hence system has non-trivial solution for any value of α, β, γ, δ.

12 )

Let

	4	4k	k
A =	0	k	4k
	0	0	4

If det(A²) = 16 then |k| is…..

(1/4)

View Solution

Answer :

det(A²) = [det(A)]² = 16

	4	4k	k	²
∴	0	k	4k	= 16
	0	0	4

∴ [4(4k) – 4k(0) + k(0)]² = 16

∴ [16 k]² = 16

∴ k² = (1 / 16)

|k| = (1/4).

13 )

If 1, ω, ω² are cube roots of unity, then

a	a²	a³ – 1
a^ω	a^2ω	a^3ω – 1	= ……
a^(ω)2	a^2(ω)2	a^3(ω)2 – 1

a^2

a^3

View Solution

Answer :

given determinant can be written as.

a	a²	a³		a	a²	1
a^ω	a^2ω	a^3ω	–	a^ω	a^2ω	1
a^(ω)2	a^2(ω)2	a^3(ω)2		a^(ω)2	a^2(ω)2	1

	1	a	a²		1	a²	a
= aa^ω a^(ω)2	1	a^ω	a^2ω	+	1	a^2ω	a^ω
	1	a^(ω)2	a^2(ω)2		1	a^2(ω)2	a^(ω)2

∵ ω + ω² + 1 = 0

⇒ a ∙ a^ω ∙ a^(ω)2 = a^[(ω)2+ω+1] = a° = 1

	2	a + a²	a +a²
∴ D =	2	a^ω + a^2ω	a^2ω + a^ω
	2	a^(ω)2 + a^2(ω)2	a^2(ω)2 + a^(ω)2

∵ Two columns are same

∴ D = 0.

14 )

If a₁, a₂, a₃,…. are in GP, then

log a_n	log a_n+1	log a_n+2
log a_n+3	log a_n+4	log a_n+5	= ……
log a_n+6	log a_n+7	log a_n+8

-1

View Solution

Answer :

in G.P. n^th term a_n = a₁r^n–1

∴ log a_n = log a₁ + (n – 1) log r (1)

	log a_n	log a_n+1	log a_n+2
∴ D =	log a_n+3	log a_n+4	log a_n+5
	log a_n+6	log a_n+7	log a_n+8

from (1) log a_n+3 = log a₁ + (n + 2)log r

log a_n+4 = log a₁ + (n + 3)log r

Similarly putting other value D becomes

	log a₁ + (n – 1)log r	log a₁ + (n)log r	log a₁ + (n + 1)log r
D =	log a₁ + (n + 2)log r	log a₁ + (n + 3)log r	log a₁ + (n + 4)log r
	log a₁ + (n + 5)log r	log a₁ + (n + 6)log r	log a₁ + (n + 7)log r

	log a₁	log a₁	log a₁		(n – 1)log r	n log r	(n + 1)log r
=	log a₁	log a₁	log a₁	+	(n + 2)log r	(n + 3)log r	(n + 4)log r
	log a₁	log a₁	log a₁		(n + 5)log r	(n + 6)log r	(n + 7)log r

	n – 1	n	n + 1
= 0 + (log r)³	n + 2	n + 3	n + 4
	n + 5	n + 6	n + 7

for n = 0,

	– 1	0	1
D = (log r)³	2	3	4
	5	6	7

= (log r)³ [(– 1)(21 – 24) – 0 + 1(12 – 15)]

= (log r)³ [3 – 3]

= 0.

15 )

P =	(√3 / 2)	(1/2)	, A =	1	1
P =	– (1/2)	(√3 / 2)	, A =	0	1

and Q = PAP^T, Then P^T Q²⁰¹³P = …

Choose correct answer from below options

(a)	1	2013	(b)	4 + 2013√3	6039
	0	1		2012	4 – 2013√3

(c) (1/4)	2 + √3	1	(d) (1/4)	2012	2 + √3
	– 1	2 – √3		2 + √3	2012

View Solution

Answer :

P =	(√3 / 2)	(1/2)
P =	– (1/2)	(√3 / 2)

P^T =	(√3 / 2)	– (1/2)
P^T =	(1/2)	(√3 / 2)

P ∙ P^T	1	0	= I
	0	1

∴ P is orthogonal matrix.

∵ Q = PAP^T

Q²⁰¹³ = (PAP^T) (PAP^T)……2013 times

= IA²⁰¹³ …… P ∙ P^T = I

given expression = P^T ∙ Q²⁰¹³ ∙ P = Q²⁰¹³ = A²⁰¹³

∴ A² =	1	1		1	1	=	1	2
∴ A² =	0	1		0	1	=	0	1

A³ =	1	2		1	1	=	1	3
A³ =	0	1		0	1	=	0	1

Similarly we can write

A²⁰¹³ =	1	2013
A²⁰¹³ =	0	1

∴ P^TQ²⁰¹³P = A²⁰¹³ =	1	2013
∴ P^TQ²⁰¹³P = A²⁰¹³ =	0	1

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