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

6 )

(b + c)²	a²	a²
b²	(c + a)²	b²	= k(abc)(a + b + c)³, then k = ….
c²	c²	(a + b)²

-2

View Solution

Answer :

Let a = 1, b = – 1 and c = 2 then

1	1	1
1	9	1	= k(– 2)(2)³
4	4	0

∴ 1(0 – 4) – 1(0 – 4) + 1(4 – 36) = – 16k

∴ – 4 + 4 – 32 = – 16k

∴ k = 2.

7 )

[(a² + b²) / c]	c	c
a	[(b² + c²) / a]	a	= kabc, then k = ….
b	b	[(c² + a²) / b]

-2

View Solution

Answer :

Let a = 1, b = 2 and c = – 1 then given expression is

– 5	– 1	– 1
1	5	1	= k(– 2)
2	2	1

∴ – 5(5 – 2) + 1(1 – 2) – 1(2 – 10) = – 2k

∴ – 15 – 1 + 8 = – 2k

∴ – 8 = – 2k ⇒ k = 4.

8 )

a	a²	1 + a³
b	b²	1 + b³	= 0
c	c²	1 + c³

and the vectors (1, a, a²), (1, b, b²), (1, c, c²) are non-coplanar then abc = …..

-1

View Solution

Answer :

	a	a²	1 + a³		a	a²	1		a	a²	a³
D =	b	b²	1 + b³	=	b	b²	1	+	b	b²	b³
	c	c²	1 + c³		c	c²	1		c	c²	c³

	1	a	a²		1	a	a²
∴ D =	1	b	b²	+	1	b	b²	(abc)
	1	c	c²		1	c	c²

	1	a	a²
∴ D = (1 + abc)	1	b	b²
	1	c	c²

given D = 0 as

1	a	a²
1	b	b²	≠ 0
1	c	c²

hence 1 + abc = 0

i.e. abc = – 1.

9 )

√11 + √3	√20	√5
√15 + √22	√25	√10	= ……
3 + √55	√15	√25

5(5âˆš3 â€“ 3âˆš2)

5(3âˆš2 + 5âˆš3)

â€“ 5(5âˆš3 + 3âˆš2)

5(3âˆš2 â€“ 5âˆš3)

View Solution

Answer :

	√11 + √3	√20	√5
D =	√15 + √22	√25	√10
	3 + √55	√15	√25

c₂(1 / √5), c₃(1 / √5)

	√11 + √3	√2	√1
∴ D =	√15 + √22	√5	√2	(√5 × √5)
	3 + √55	√3	√5

	√3	2	1		√11	2	1
∴ D =	√15	√5	√2	+	√22	√5	√2	× 5
	3	√3	√5		√55	√3	√5

	√3	2	1		1	2	1
∴ D = 5 ×	√15	√5	√2	+ √11	√2	√5	√2
	3	√3	√5		√5	√3	√5

Here value of D₂ = 0 as two columns are same

∴ D = 5[√3(5 – √6) – 2{√75 – √18} + 1{√45 – √45}]

= 5[√3(5 – √6) – 2√3(5 – √6)]

= 5√3(5 – √6)(– 1)

= – 25√3 + 5√18

= 5[3√2 – 5√3].

10 )

If 2s = a + b + c and

	a²	(s – a)²	(s – a)²
A =	(s – b)²	b²	(s – b)²	then det A = ….
	(s – c)²	(s – c)²	c²

2s2 (s â€“ a)(s â€“ b)(s â€“ c)

2s3 (s â€“ a)(s â€“ b)(s â€“ c)

2s (s â€“ a)2(s â€“ b)2(s â€“ c)2

2s2 (s â€“ a)2(s â€“ b)2(s â€“ c)2

View Solution

Answer :

given 2s = a + b +c (1)

when s = 0 then

	a²	a²	a²
A =	b²	b²	b²	= 0
	c²	c²	c²

∴ s² is factor of det A as 3 rows are in identical.

Let s = a then

	a²	0	0
A =	(a – b)²	b²	(a – b)²
	(a – c)²	(a – c)²	c²

from (1)

2a – a = b + c

∴ a = b + c

	a²	0	0
∴ A =	c²	b²	c²
	b²	b²	c²

= a²(b²c² – b²c²) – 0 + 0 = 0

∴ (s – a) is factor of determinant A.

also (s – b) and (s – c) will be factor of A.

but det A is sixth degree polynomial

∴ The sixth factor is of the form k(a + b + c)

∴ det A = k(a + b + c)s³ (s – a)(s – b)(s – c) (2)

Let a = b = 0, c = 2 ⇒ s = 1

	0	1	1
∴ A =	1	0	1
	1	1	4

= 0 – 1(4 – 1) + 1(1 – 0)

= – 3 + 1 = – 2

from (2)

A = k(2)(– 1) = – 2k

hence

– 2k = – 2 ⇒ k = 1

∴ (2) becomes, det A = 2s³(s – a)(s – b)(s – c).

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