1

Q.1 If α, β, γ are the roots of x3 + px

2 + q = 0, where q ≠ 0, then

∆ =

βαγ

αγβ

γβα

/1/1/1

/1/1/1

/1/1/1

equals-

(A) –p/q (B) 1/q (C) p2/q (D) None of these

Sol.[D] We have βγ + γα + αβ = 0

We can write ∆ as

∆ = 333

1

γβαγαβγαβ

βγαβγα

αβγαβγ

= 333

1

γβαγαβγγα+βγ+αβ

βγαββγ+αβ+γα

αβγααβ+γα+βγ

[using C1 → C1 + C2 + C3]

=333

1

γβα

γαβγ

βγαβ

αβγα

0

0

0

= 0 [all zero property].

Q.2 If α, β, γ are different from 1 and are the roots of ax3 + bx

2

+ cx + d = 0 and (β – γ) (γ – α) (α – β) =2

25, then the determinant ∆ =

222

111

γβα

γβαγ−

γ

β−

β

α−

α

equals:

(A) a2

d25 (B)

a

d25 (C)

dcba

d25

+++

− (D) None of these

Sol.[D] Taking α, β, γ common from C1, C2, C3 respectively, we get

∆ = αβγ

γβα

γ−β−α−111

1

1

1

1

1

1

= αβγ

α−γα−βα

α−−

γ−α−−

β−α−001

1

1

1

1

1

1

1

1

1

1

2

[using C2 → C2 – C1 and C3 → C3 – C1]

= )1)(1)(1(

))(()1(–

γ−β−α−

α−γα−βαβγ

11

11 β−γ−

= )1)(1)(1(

))(()–(

γ−β−α−

α−γγ−ββααβγ

As α, β, γ are the roots of

ax3 + bx

2 + cx + d = 0,

ax3 + bx

2 + cx + d = a(x – α) (x – β) (x – γ)

and αβγ = –d/a

Thus, ∆ = a/)dcba(

)2/25)(a/d(

+++

−= –

)dcba(2

d25

+++.

Q.3 The value of θ for which the system of equations

(sin 3θ) x – 2y + 3z = 0

(cos 2θ) x + 8y – 7z = 0

2x + 14y – 11z = 0

has a non-trivial solution is -

(A) nπ (B) nπ + (–1)n π/3 (C) nπ + (–1)

n π/8 (D) None of these

Sol.[A] The system of equations has a non-trivial solution if and only if

11142

782cos

323sin

−

−θ

−θ

= 0

Applying R2 → R2 + 4R1, R3 → R3 + 7R1, we get

1003sin72

503sin42cos

323sin

θ+

θ+θ

−θ

= 0

Expanding along C2, we get

2(cos 2θ + 4 sin 3θ) – (2 + 7 sin 3θ) = 0

⇒ 2 – 2 cos 2θ – sin 3θ = 0

⇒ 4 sin2 θ – (3 sin θ – 4 sin

3 θ) = 0

⇒ sin θ (4 sin2 θ + 4 sin θ – 3) = 0

⇒ sin θ (2 sin θ – 1) (2 sin θ + 3) = 0

⇒ sin θ = 0 or sin θ = 1/2.

[∵ sin θ = –3/2 is non possible]

3

∴ For, θ = nπ the system of equations has a non-trivial solution.

Q.4 If ∆ =

xsin41xcosxsin

xsin4xcos1xsin

xcos4xcosxsin1

222

222

222

+

+

+

then the maximum

value of ∆ is -

(A) 4 (B) 6 (C) 8 (D) 10

Sol.[B] Apply C1 + C2 and then apply R2 – R1 and R3 – R1 and expand

∆ = 2 + 4 sin 2x

The maximum value of sin 2x is 1 and hence of ∆ is 6.

Q.5 If a ≠ p, b ≠ q, c ≠ r and

rba

cqa

cbp

= 0, then ap

p

−+

bq

q

−+

cr

r

−is equal to-

(A) 0 (B) 1 (C) –1 (D) 2

Sol.[D] Apply R1 – R2, R2 – R3

∆ =

rba

rcbq0

0qbap

−−

−−

= 0

= (p – a) [(q – b) r – b (c – r)] + a (b – q) (c – r) = 0

Dividing by (p – a) (q – b) (c – r), we get

rc

r

−–

bq

b

−–

ap

a

−= 0

or ap

a

−+

bq

b

−+

cr

r

−= 0

or ap

a

−+ 1 +

bq

b

−+ 1 +

cr

r

−= 1 + 1

or ap

p

−+

bq

q

−+

cr

r

−= 2.

Q.6 The value of θ lying between θ = 0 and π/2 and satisfying

the equation

θ+θθ

θθ+θ

θθθ+

4sin41cossin

4sin4cos1sin

4sin4cossin1

22

22

22

= 0 are -

4

(A) 7π/24 (B) 5π/24 (C) 11π/24 (D) π/24

Sol.[A,C]

Apply C1 + C2

∆ =

θ+θ

θθ+

θθ

4sin41cos1

4sin4cos12

4sin4cos2

2

2

2

= 0

Now applying R2 – R1 and R3 – R1

∆ =

101

010

4sin4cos2 2

−

θθ

= 0

or 2 + 4 sin 4θ = 0 by expansion with R2

or sin 4θ = –1/2 = – sin(π/6) = sin (–π/6)

∴ 4θ = nπ + (–1)n

π−

6 ∴ θ =

24

)1(n6 n−−π

The value of θ lying between 0 and π/2 are 7π/24 and 11π/24 for n = 1 and 2.

Q.7 The values of λ for which the system of equations

x + y – 3 = 0

(1 + λ)x + (2 + λ)y – 8 = 0

x – (1 + λ)y + (2 + λ) = 0

is consistent are -

(A) –5/3, 1 (B) 2/3, –3 (C) –1/3, –3 (D) 0, 0 1

Sol.[A] The given system of equations will be consistent if

∆ =

λ+λ+−

−λ+λ+

−

2)1(1

821

311

= 0

Applying C2 → C2 – C1 and C3 → C3 + 3 C1 ,

we get ∆ =

λ+λ−

λ+−λ+

5–21

3511

001

= 0

⇒ (5 + λ) + (2 + λ) (3λ – 5) = 0

5

⇒ 5 + λ + 6λ – 10 + 3λ2 – 5λ = 0

⇒ 3λ2 + 2λ – 5 = 0 ⇒ (3λ + 5) (λ – 1) = 0

⇒ λ = –5/3 or λ = 1.

Q.8 Let m be a positive integer and ∆r

=

)1m(sin)m(sin)m(sin

1m21m

1C1r2

2222

m2r

m

+

+−

−

= (0 ≤ r ≤ m) then the value of ∑=

∆m

0r

r is given

by -

(A) 0 (B) m2 – 1 (C) 2

m (D) 2

m sin

2 (2

m)

Sol.[A] ∑=

−m

0r

)1r2( = sum of (m + 1) terms of an A.P.

whose first term is –1, d = 2 ∴ S = m2 – 1

∑=

m

0r

rm

C = sum of binomial coefficient = 2m

∑1= m + 1 ∴ R1 and R2 will be identical and hence ∆ = 0.

Q.9 If the three digit numbers A28, 3B9 and 62C when A, B

and C are integers between 0 and 9 which are divisible by λ, then ∆ =

2B2

C98

63A

is divisible by -

(A) λ (B) λ2 (C) 2λ (D) None

Sol.[A] A28 = 100A + 2 . 10 + 8.

Apply R2 + 10R3 + 100R1 etc.

∆ =

2B2

kkk

63A

321 λλλ = λ∆1

∴ ∆ is divisible by λ.

Q.10 If ∆ =

0ycos.xsinysin.xsin

xsinysin.xcosycos.xcos

xcosysin.xsinycos.xsin

−

− , then ∆ is independent of -

(A) x (B) y (C) constant (D) None of these

Sol.[B] Take sin x, cos x and sin x common from R1, R2 and R3 respectively

6

∴ ∆ = sin2 x . cos x

0ycosysin

xtanysinycos

xcotysinycos

−

−

Make two zeros by R1 – R2

= sin2 x cos x

0ycosysin

xtanysinycos

xtanxcot00

−

−

+

= sin2 x . cos x .

xcos.xsin

xcosxsin 22 +

[cos2 y + sin

2 y]

= sin x, which is independent of y.

Q.11 f(n), g(n), and h(n) are second degree polynomials in n having n + 2 as a

common factor then value of ∑=

∆n

1r

r is, where ∆r =

)n(h)2n)(1n(n)1rn(r

)n(g62.32

)n(f)2n(n61r21n1r

+++−

−

+++−

(A) 6

)2n()1n()n(2n ++ (B)

12

)2n()1n(n ++

(C) 3

)1n()12( n +− (D) None of these

Sol.[D] We have to find the sum of n determinants whose second and third columns

are the same for all determinants. Hence, we use the sum of determinants.

∑=

∆n

1r

r = ∆1 + ∆2 + … + ∆n

= +

)n(h)2n()1n(n)11n.(1

)n(g62.32

)n(f)2n(n611.21n0

++−+

−

+++ +

)n(h)2n()1n(n)21n(2

)n(g62.32

)n(f)2n(n612.21n1

++−+

−

+++ …

to n determinants

=

)n(h)2n()1n(n)n1n(n...)21n(2)11n.(1

)n(g62.32...222

)n(f)2n(n6)1...11()n...321(21n1n210

++−+++−++−+

−++++

++++++++++− + ….

7

=

)n(h)2n()1n(n)n...21()n...21)(1n(

)n(g)22(312

12

)n(f)2n(n6n2

)1n(n.2

222

1nn

+++++−++++

−−

−

+++

+

=

)n(h6

)2n()1n(n.6

6

)1n2)(1n(n

2

)1n(n)1n(

)n(g)12(612

)n(f)2n(n6)2n(nnn

++++−

++

−−

++

= 6.

)n(h6

)2n()1n(n)2n(

6

)1n(n

)n(g1212

)n(f)2n(n)2n(nnn

+++

+−−

++

= 6 × 0 (∵ C1 ≡ C2)

= 0

∴∑=

∆n

1r

r is independent of n.

Q.12 If ∆r =

n3n3n3)1r(

2n4n2)1r(

6n1r

223

22

−−

−−

−

then ∑=

∆n

1r

r equals -

(A) n2(n + 1) (B) n(n + 1)

2 (C)

12

1n(n

3 + 1) (D) None of these

Sol.[D] Using the sum property of determinants, we get

∑

∑

∑

∑

=

=

=

=

−−

−−

−

=∆n

1r

23n

1r

3

2n

1r

2

n

1r

r

n3n3n3)1r(

2n4n2)1r(

6n)1r(

But ∑=

−n

1r

)1r( =2

1 n (n – 1),

∑=

−n

1r

2)1r( = 6

1n (n – 1) (2n – 1) and

8

∑=

−n

1r

3)1r( =4

1n

2 (n – 1)

2. Thus,

∑=

∆n

1r

r =

n3n3n3)1n(n4

1

2n4n2)1n2)(1n(n6

1

6n)1n(n2

1

2322

2

−−

−−−

−

Taking 12

1n (n – 1) common from C1 and n from C2, we get

∑=

∆n

1r

r =12

1n (n – 1) (n)

n3n3n3)1n(n3

2n4n2)1n2(2

616

22 −−

−−

= 12

1 n (n – 1) (n) (0) = 0 [∵ C1 and C3 are identical].

Q.13 If the system of equations λx1 + x2 + x3 = 1, x1 + λx2 + x3 = 1, x1 + x2 + λx3 =

1 is consistent, then λ can be-

(A) 5 (B) –2/3 (C) –3 (D) None of these

Sol.[D] Let ∆ =

λ

λ

λ

11

11

11

=

λ+λ

λ+λ

+λ

12

12

112

[C1 → C1 + C2 + C3]

= (λ + 2)

λ

λ

11

11

111

= (λ + 2)

101

011

001

−λ

−λ = (λ + 2) (λ – 1)2

[using C2 → C2 – C1 and C3 → C3 – C1]

If ∆ = 0, then λ = –2 or λ = 1.

But when λ = 1, the system of equation becomes x1 + x2 + x3 = 1 which has

infinite number of solutions. When λ = – 2, by adding three equations, we

obtain 0 = 3 and thus, the system of equations is inconsistent.

Q.14 If α be a repeated roots of the quadratic equation f(x) = 0 and A(x), B(x), C(x)

are polynomials of degree 3, 4, 5 respectively then determinant

)(C)(B)(A

)(C)(B)(A

)x(C)x(B)x(A

α′α′α′

ααα is divisible by (where A′(α) =α=

xdx

dA, etc) -

9

(A) f(x) (x – α)3 (B) (f(x))

2 (C) f(x) (D) None of these

Sol.[C] We know that α is a root of f(x) = 0. This means (x – α) is a factor of f(x). If α

is a repeated root of f(x) = 0 then f(x) has a repeated factor (x – α), i.e., (x –

α)2 is a factor of f(x). But here f(x) is quadratic

∴ f(x) = λ(x – α)2 …(1)

Where λ is a constant.

Let ∆(x) =

)(C)(B)(A

)(C)(B)(A

)x(C)x(B)x(A

α′α′α′

ααα

Which is of the degree 5 at most and 3 at least.

Clearly, ∆(α) = 0 …(2)

Differentiating ∆(x) w.r.t. x,

∆′(x) =

)(C)(B)(A

)(C)(B)(A

)x(C)x(B)x(A

α′α′α′

ααα

′′′

+

)(C)(B)(A

000

)x(C)x(B)x(A

α′α′α′

+

000

)(C)(B)(A

)x(C)x(B)x(A

ααα

because derivatives of constants = 0

=

)(C)(B)(A

)(C)(B)(A

)x(C)x(B)x(A

α′α′α′

ααα

′′′

∴ ∆′(α) =

)(C)(B)(A

)(C)(B)(A

)(C)(B)(A

α′α′α′

ααα

α′α′α′

= 0 …(3)

because R1 ≡ R3.

We know that

φ(α) = 0 ⇒ (x – α) is a factor of φ (x)

and φ′(α) = 0 ⇒ (x – α)2 is a factor of φ(x)

∴ from (2) and (3), ∆(x) has a factor (x – α)2.

∴ ∆(x) = (x – α)2 . F(x)

= λ

1λ(x – α)

2 . F(x) =

λ

1f(x) . F(x), using (1)

∴ ∆(x) is divisible by f(x).

10

Q.15 If an = ∫π

−

−2/

0dx

x2cos1

nx2cos1 then the value of determinant

987

654

321

aaa

aaa

aaa

is -

(A) 1 (B) –1 (C) 2 (D) None of these

Sol.[D] Here an + an+2

= dxx2cos1

nx2cos12/

0∫π

−

−+ dx

x2cos1

x)2n(2cos12/

0∫π

−

+−

= dxx2cos1

}x)2n(2cosnx2{cos22/

0∫π

−

++−

= ∫π

−

+−2/

0dx

x2cos1

x2cos.x)1n(2cos22

Also 2.an+1 = 2 ∫π

−

+−2/

0dx

x2cos1

x)1n(2cos1

∴ an + an+2 – 2an+1 = { }

∫π

−

++−+−2/

0dx

x2cos1

x)1n(2cos1x2cos.x)1n(2cos12

= ∫π

−

−+2/

0dx

x2cos1

}x2cos1{.x)1n(2cos2

= ∫π

+2/

0dxx)1n(2cos2

∴ an + an+2 – 2an+1 = 2 2/

0)1n(2

x)1n(2sinπ

+

+ =

1n

1

+[0 – 0] = 0, …(1)

for all n.

∴ an+1 = 2

aa 2nn ++

∴ an + 1 is the AM between an, an+2.

Now, ∆ =

987

654

321

aaa

aaa

aaa

= 2

1

987

654

321

aa2a

aa2a

aa2a

= 2

1

99787

66454

33121

a)aa(a2a

a)aa(a2a

a)aa(a2a

+−

+−

+−

, C2 → C2, – C1 – C3

= 2

1

97

64

31

a0a

a0a

a0a

, using (1)

∴ ∆ = 0.

11

Q.1. Let A and B be two 2 × 2 matrices. Consider the statements

(i) AB = O ⇒ A = O or B = O (ii) AB = I2 ⇒ A = B–1

(iii) (A + B)2 = A

2 + 2AB + B

2

Then

(A) (i) is false, (ii) and (iii) are true (B) (i) and (iii) are false, (ii) is true

(C) (i) and (ii) are false, (iii) is true (D) (ii) and (iii) are false, (i) is true

Sol.[B] (i) is false,

If A =

−10

10 and B =

00

11, then AB =

00

00 = O

Thus, AB = 0 ⇒ A = O or B = O

(iii) is false since matrix multiplication is not commutative.

(ii) is true as product AB is an identity matrix, if B is inverse of the matrix A.

Q.2. Let E(α) =

ααα

ααα2

2

sinsincos

sincoscos. If α and β differs by an odd multiple of π/2,

then E(α) E(β) is a -

(A) null matrix (B) unit matrix

(C) diagonal matrix (D) orthogonal matrix

Sol.[A] We have

E(α) E(β) =

ααα

ααα2

2

sinsincos

sincoscos

βββ

βββ2

2

sinsincos

sincoscos

=

β−αβαβ−αβα

β−αβαβ−αβα

)cos(sinsin)cos(cossin

)cos(sincos)cos(coscos

As α and β differ by an odd multiple of π/2, α – β = (2n + 1) π/2 for some As

α and β integer n. Thus, cos [(2n + 1)π/2] = 0

∴ E (α) E (β) = O.

Q.3. The inverse of a skew symmetric matrix (if it exists) is -

(A) a symmetric matrix (B) a skew symmetric matrix

(C) diagonal matrix (D) None of these

12

Sol.[B] We have A′ = – A

Now, AA–1

= A–1

A = In

⇒ (AA–1

)′ = (A–1

A)′ = (In)′

⇒ (A–1

)′ A′ = A′(A–1)′ = In

⇒ (A–1

)′ (–A) = (–A) (A–1

)′ = In

Thus, (A–1

)′ = –(A–1

) [inverse of a matrix is unique].

Q.4. Let A, B, C be three square matrices of the same order, such that whenever

AB = AC then B = C, if A is -

(A) singular (B) non-singular

(C) symmetric (D) skew-symmetric

Sol.[B] If A is non-singular, A–1

exists.

Thus, AB = AC ⇒ A–1

(AB) = A–1

(AC)

⇒ (A–1

A) B = (A–1

A) C ⇒ IB = IC

Q.5. If A =

rqp

zyx

cba

, B =

−

−−

−

zcr

xap

ybq

then -

(A) |A| = |B| (B) |A| = – |B|

(C) |A| = 2|B| (D) A is invertible if and only if B is invertible

Sol.[B, D]

We have

|B| =

zcr

xap

ybq

−

−−

−

= (–1)

zcr

xap

ybq

−

−

−

=

zcr

xap

ybq

= –

zcr

ybq

xap

=

zrc

yqb

xpa

= –

rzc

qyb

pxa

= – |A| [using reflection property]

∴From here, we get |A| ≠ 0 if and only if |B| ≠ 0.

⇒ B = C.

13

Q.6. If A and B are two square matrices such that B = –A–1

BA, then (A + B)2 is

equal to -

(A) 0 (B) A2 + B

2 (C) A

2 + 2AB + B

2 (D) A + B

Sol.[B] As B = –A–1

BA, we get AB = –BA or AB + BA = O

Now, (A + B)2 = (A + B) (A + B) = A

2 + BA + AB + B

2

= A2 + O + B

2 = A

2 + B

2

Q.7. If a, b and c are all different from zero such that 0c

1

b

1

a

1=++ , then the matrix

A=

+

+

+

c111

1b11

11a1

is-

(A) symmetric (B) non-singular

(C) can be written as sum of a symmetric and a skew symmetric matrix

(D) none of these

Sol.[B, C]

We have

|A| = abc

c/11b/1a/1

c/1b/11a/1

c/1b/1a/11

+

+

+

= abc

c/11b/1c/1b/1a/11

c/1b/11c/1b/1a/11

c/1b/1c/1b/1a/11

++++

++++

+++

[C1 → C1 + C2 + C3]

= abc

c/11b/11

c/1b/111

c/1b/11

+

+

=++ 0

c

1

b

1

a

1∵

= abc

100

010

c/1b/11

= abc ≠ 0 [R2 → R2 – R1, R3 → R3 – R1]

∴ A is non-singular.

Also, every matrix can be written as sum of a symmetric and a skew

symmetric matrix.

14

Q.8. If A is non-singular matrix of order 3 × 3, then adj (adj A) is equal to -

(A) |A| A (B) |A|2 A (C) |A|

–1 A (D) None of these

Sol.[A] As A is a non-singular matrix A–1

= |A|

1 (adj A)

⇒ adj A = |A| A–1

= B(say).

Now,

adj (adj A) = adj (B) = |B|B–1

= ||A|A–1

| (|A| A–1

)–1

= |A|3 |A

–1| |A|

–1 (A

–1)–1

[using scalar multiple property of determinants]

= |A|3

|A|

1·

|A|

1A = |A| A.

Q.9. The value of a for which the system of equations

x + y + z = 0

ax + (a + 1)y + (a + 2)z = 0

a3x + (a + 1)

3y + (a + 2)

3z = 0

has a non-zero solution is -

(A) 1 (B) 0 (C) –1 (D) None of these

Sol.[C] The system of equations will have a non-zero solution if and only if

∆ = 333 )2a()1a(a

2a1aa

111

++

++ = 0

Using C3 → C3 – C2, C2 → C2 – C1, we get

∆ =

7a9a31a3a3a

11a

001

223 ++++

= 0

⇒ 6a + 6 = 0 or a + 1 = 0 or a = –1.

15

Q.10. Let A =

− 51

32if A

–1 = xA + yI2 then x and y are respectively -

(A) 13

1,

13

7− (B)

13

7,

13

1− (C)

13

1−,13

7 (D)

13

7−,13

1

Sol.[C] A–1

= 13

1

−

21

35= x

− 51

32+ y

10

01

⇒ 13

5 = 2x + y,

13

3− = 3x

⇒ 13

1 = –x,

13

2 = 5x + y

x = 13

1−, y =

13

7.

Q.11. If A =

1x3

321

210

and A–1

=

−

−

−

2/12/32/5

y34

2/12/12/1

, then -

(A) x = 1, y = –1 (B) x = –1, y = 1

(C) x = 2, y = –1/2 (D) x = 1/2, y = 1/2

Sol.[A] We have

100

010

001

= AA–1

=

1x3

321

210

−

−

−

2/12/32/5

y34

2/12/12/1

=

+−−

+

+

xy2)1x(3)x1(4

)1y(210

1y01

1 – x = 0, x – 1 = 0, y + 1 = 0, y + 1 = 0, 2 + xy = 1

x = 1, y = –1.

16

Q.12. If A is a 3 × 3 skew-symmetric matrix, then |A| is given by -

(A) 0 (B) –1 (C) 1 (D) None of these

Sol.[A] Let A =

−−

−

0cb

c0a

ba0

be the given skew symmetric matrix.

We have |A| =

0cb

c0a

ba0

−−

− =

0cb

c0a

ba0

−

−−

[∵ |A| = |A′|]

= (–1)3

0cb

c0a

ba0

−−

− = – |A|

⇒ 2 |A| = 0 ⇒ |A| = 0.

Q.13. If the system of equations x + 2y – 3z = 2, (k + 3)z = 3, (2k + 1)y + z = 2 is

inconsistent, then value of k can be -

(A) –3 (B) –1/2 (C) 1 (D) 2

Sol.[A,B]

For the given system to be inconsistent we must have

11k20

3k00

321

+

+

−

= 0

⇒ – (k + 3) (2k + 1) = 0 ⇒ k = –3, –1/2.

For k = –3, the second becomes 0z = 3 which is impossible.

For k = –1/2, the second equation gives z = 6/5 and the third equation gives

z = 2. A contradiction.

Q.12 For each real x such that –1 < x < 1 .

Let A (x) denote the matrix (1 – x)–1/2

−

−

1x

x1 and A(x) A(y) = k A(z)

where x, y ∈ R, –1 < x, y < 1 and z = xy1

yx

+

+ then k is -

(A) xy1

1

+ (B)

xy1

x

+ (C) xy1+ (D)

x

xy1+

17

Sol.[C] A(x) A(y) = ( ) 1/21 x

−− (1 – y)

–1/2

−

−

1x

x1

−

−

1y

y1

= (1 – x – y + xy)–1/2

++−

+−+

xy1)yx(

)yx(xy1

= )xy()yx(1

xy1

++−

+

+

+−+

+−

1xy1

)yx(xy1

)yx(1

=

xy1

yx1

xy1

+

+−

+

+

+−+

+−

1xy1

)yx(xy1

)yx(1

= xy1+ A(z)

⇒ k = xy1+ .

Q.15. If the matrix

γβ−α

γ−βα

γβ20

is orthogonal, then -

(A) α = ±2

1 (B) β = ±

6

1 (C) γ = ±

3

1 (D) all of these

Sol.[D] Let A =

γβ−α

γ−βα

γβ20

, A′ =

γγ−γ

β−ββ

αα

2

0

Since A is orthogonal, ∴ AA′ = I

⇒

γβ−α

γ−βα

γβ20

γγ−γ

β−ββ

αα

2

0

=

100

010

001

⇒

γ+β+αγ−β−αγ+β−

γ−β−αγ+β+αγ−β

γ+β−γ−βγ+β

22222222

22222222

222222

2

2

224

=

100

010

001

18

Equation the corresponding elements, we have

=γ−β

=γ+β

02

1422

22

⇒ β = ± 6

1, γ = ±

3

1

α2 + β2

+ γ2 = 1 ⇒ α2

+6

1+

3

1= 1 ⇒ a = ±

2

1.

Hence (D) is correct answer.

Q.16. The rank of the matrix

+−

−−

−

1a21

4a42

521

is -

(A) 2 if a = 6 (B) 2 if a = 1 (C) 1 if a = 2 (D) 1 if a = –6

Sol.[B,D]

Let A =

+−

−−

−

1a21

4a42

521

~

+

+

−

6a00

6a00

521

[R2 → R2 + 2R1, R3 → R3 + R1]

Clearly rank of A is 1 if a = –6.

Also, for a = 1, |A| =

221

342

521

−

−−

−

= 0

and 34

52

−−= –6 + 20 = 14 ≠ 0

∴ rank of A is 2 if a = 1.

Hence (b), (d) are correct answer.

Q.17. If [x] stands for the greatest integer less or equal to x, then in order than the set

of equations x – 3y = 4 ; 5x + y = 2 ;

[2π]x – [e]y = [2a] may be consistent, then ‘a’ should lie in -

(A)

2

7,3 (B)

3

7,3 (C)

3

7,3 (D) None of these

Sol.[A] On solving x – 3y = 4 and 5x + y = 2, we get x = 8

5, y = –

8

9. As [2π] = 6 and

[e] = 2.

19

So that the three equations are consistence if

[2π].8

5 + [e]

8

9= [2a] or [2a] = 6

This gives 6 ≤ 2a < 7 or 3 ≤ a < 2

7.

Hence (A) is correct answer.

Q.18. If matrix A =

bac

acb

cba

where a, b, c are real positive numbers, abc = 1 and AT

A = 1, then the value of a3 + b

3 + c

3 is -

(A) 3 (B) 2 (C) 4 (D) 1

Sol.[C] AT A = I ⇒ |A

T A| = |I| ⇒ |A|

2 = 1

⇒ (a3 + b

3 + c

3 – 3abc)

2 = 1

⇒ a3 + b

3 + c

3 – 3abc = 1

(since a, b, c are positive real number ⇒ a3 + b

3 + c

3 ≥ 3abc from AM ≥ GM)

⇒ a3 + b

3 + c

3 = 4.

Hence (C) is correct answer.

Q.19. Let p be the sum of all possible determinants of order 2 having 0, 1, 2 and 3 as

their four elements. Then the common root α of the equations

x2 + ax + [m + 1] = 0

x2 + bx + [m + 4] = 0

x2 – cx + [m + 15] = 0

Such that α > p, where a + b + c = 0 and m =∞→n

limn

1∑

= +

n2

1r22 rn

r. (where [.]

denotes the greatest integer function) -

(A) ± 3 (B) 3 (C) –3 (D) None of these

20

Sol.[B] Let α be the common root, then

α2 + aα + [m + 1] = 0 …(1)

α2 + αb + [m + 4] = 0 …(2)

α2 – cα + [m + 15] = 0 …(3)

Apply (1) + (2) – (3)

α2 + [m] – 10 = 0 …(4)

but m = ∞→n

limn

1∑

= +

n2

1r22 rn

r=

∞→nlim

n

1∑

=

+

n2

1r2

n

r1

n/r

= ∫+

2

02x1

xdx =

2

0

2x1

+ = 5 – 1

Now [m] = [ 5 – 1] = 1

Now from (4), α2 + 1 – 10 = 0 ⇒ α = ± 3 …(5)

Now number of determinants of order 2 having 0, 1, 2, 3 = 4! = 24.

Let ∆1 = 43

21

aa

aabe one such determinant and there exists another determinant.

∆2 = 21

43

aa

aa(obtained on interchanging R1 and R2)

such that ∆1 + ∆2 = 0.

p = sum of all the 24 determinants = 0

since α > p ⇒ α > 0

from (5), α = 3.

Hence (B) is the correct answer.

21

Q20. For k =50

1, the value of a, b, c such that PP′ = I, where P =

−

−−

ck53/2

bk43/1

ak33/2

is-

(A) ±25

16, ±

25

13, ∓

23

1 (B) ∓

23

1, ±

25

13, ±

25

16

(C) ±25

13, ±

25

16, ∓

23

1 (D) None of these

Sol.[C] For PP′ = 1,

−

−−

ck53/2

bk43/1

ak33/2

−−

−

cba

k5k4k3

3/23/13/2

=

100

010

001

.

Performing matrix multiplication, we have

9

4+ 9k

2 + a

2 = 1,

9

1 + 16k

2 + b

2 = 1,

9

4 + 25k

2 + c

2 = 1

⇒ a2 =

450

169, b

2 =

450

256, c

2 =

450

25.

Also 9

4– 15k

2 + ac = 0, –

9

2 + 20k

2 + bc = 0, –

9

2 – 12k

2 + ab = 0

⇒ ab = 450

208, bc = –

450

80, ac =

450

65−.

Hence a = ± 25

13, b = ±

25

16, c ∓

23

1.

Hence (C) is correct answer.

Q.21. If B is an idempotent matrix, and A = I – B, then -

(A) A2 = A (B) A

2 = I (C) AB = 0 (D) BA = 0

Sol.[A,C,D]

22

Since B is an idempotent matrix, ∴ B2 = B.

Now, A2 = (I – B)

2 = (I – B) (I – B)

= I – IB – BI + B2 = I – B – B + B

2 = I – 2B + B

2

= I – 2B + B = I – B = A.

∴ A is idempotent.

Again AB = (I – B) B = IB – B2 = B – B

2 = B

2 – B

2 = 0.

Similarly, BA = (I – B) = BI – B2 = B – B = 0.

Hence (A), (C), (D) are correct answers.

Q.22. If the matrix

γβ−α

γ−βα

γβ20

is orthogonal, then -

(A) α = ±2

1 (B) β = ±

6

1 (C) γ = ±

3

1 (D) all of these

Sol.[D] Let A =

γβ−α

γ−βα

γβ20

, A′ =

γγ−γ

β−ββ

αα

2

0

Since A is orthogonal, ∴ AA′ = I

⇒

γβ−α

γ−βα

γβ20

γγ−γ

β−ββ

αα

2

0

=

100

010

001

⇒

γ+β+αγ−β−αγ+β−

γ−β−αγ+β+αγ−β

γ+β−γ−βγ+β

22222222

22222222

222222

2

2

224

=

100

010

001

Equation the corresponding elements, we have

23

=γ−β

=γ+β

02

1422

22

⇒ β = ± 6

1, γ = ±

3

1

α2 + β2

+ γ2 = 1 ⇒ α2

+6

1+

3

1= 1 ⇒ a = ±

2

1.

Hence (D) is correct answer.

Q.23. If A is non-singular matrix of order 3 × 3, then adj (adj A) is equal to -

(A) |A| A (B) |A|2 A (C) |A|

–1 A (D) None of these

Sol.[A] As A is a non-singular matrix A–1

= |A|

1 (adj A)

⇒ adj A = |A| A–1

= B(say).

Now,

adj (adj A) = adj (B) = |B|B–1

= ||A|A–1

| (|A| A–1

)–1

= |A|3 |A

–1| |A|

–1 (A

–1)–1

[using scalar multiple property of determinants]

= |A|3

|A|

1·

|A|

1A

= |A| A.

Q.24. If the matrices A, B, (A + B) are non-singular, then [A (A + B)–1

B]–1

, is equal

to -

(A) A + B (B) A–1

+ B–1

(C) A (A + B)–1

(D) None of these

Sol.[B] We have

[A (A + B)–1

B]–1

= B–1

((A + B)–1

)–1

A–1

= B–1

(A + B)A–1

= (B–1

A + I) A–1

= B–1

I + IA–1

= B–1

+ A–1

.

Hence (B) is correct answer.

24

Q.25. If B is a non-singular matrix and A is a square matrix, then det (B–1

AB) is

equal to -

(A) det (A–1

) (B) det (B–1

) (C) det (A) (D) det (B)

Sol.[C] det (B–1

AB) = det (B–1

) det A det B

= det (B–1

).det B. det A = det (B–1

B).det A

= det (I).det A = 1.det A = det A.

Hence (C) is correct.

Q.26. If the matrix A =

+

+

+

cyba

cbya

cbay

has rank 3, then -

(A) y ≠ (a + b + c) (B) y ≠ 1

(C) y = 0 (D) y ≠ – (a + b + c) and y ≠ 0

Sol.[D] Here the rank of A is 3

Therefore, the minor of order 3 of A ≠ 0.

⇒

cyba

cbya

cbay

+

+

+

≠ 0

⇒ (y + a + b + c)

cyb1

cby1

cb1

+

+ ≠ 0

[Applying C1 → C1 + C2 + C3 and taking (y + a + b + c) common from C1]

⇒ (y + a + b + c)

y00

0y0

cb1

≠ 0

[Applying R2 → R2 – R1, R3 → R3 – R1]

⇒ (y + a + b + c) (y2) ≠ 0 [Expanding along C1]

⇒ y ≠ 0 and y ≠ –(a + b + c)

Hence (D) is correct answer.

25

Q.27. If Ak = 0, for some value of k, (I – A)

p = I + A + A

2 + …. A

k–1, thus p is (A is

nilpotent with index k).

(A) –1 (B) –2 (C) –3 (D) None of these

Sol.[A] Let B = I + A + A2 + … + A

k–1

Post multiply both sides by (I – A), so that

B(I – A) = (I + A + A2 + … + A

k–1) (I – A)

= I – A + A – A2 + A

2 – A

3 + … –A

k–1 + A

k–1 – A

k

= I – Ak = I, since A

k = 0

⇒ B = (I – A)–1

Hence (I – A)–1

= I + A + A2 + … + A

k–1.

Thus, p = –1.

Hence (A) is correct answer.

Q.28. The value(s) of m does the system of equations 3x + my = m and 2x – 5y = 20

has a solution satisfying the conditions

x > 0, y > 0

(A) m ∈ (0, ∞) (B) m ∈

∞

2

15–,– ∪ (30, ∞)

(C) m ∈

∞,

2

15– (D) None of these

Sol.[B] By using Cramer’s rule, the solution of the system is

x = ∆

∆ x , y = ∆

∆ y, where ∆ =

52

m3

−= –(15 + 2m)

∆x = 520

mm

− = –25m, ∆y =

202

m3 = 60 – 2m

⇒ x = )m215(–

m25

+

−=

2)m215(

)m215(m25

+

+> 0,

26

for m > 0, or m < 2

15– .

Also y = )m215(

m260

+−

−=

2)m215(

)m215)(30m(2

+

+−>0

for m > 30, or m < 2

15– .

⇒ x > 0, y > 0 for m > 30 or m < 2

15–

For m = 2

15– , the system has no solution.

Hence (B) is correct answer.

Q.29. The number of values of k for which the system of equations (k + 1) x + 8y =

4k, kx + ( k + 3) y = 3k – 1 has no solution is-

(A) 0 (B) 3 (C) 1 (D) infinite

Sol.[B] For the system of equations to have no solution, we must have

k

1k +=

3k

8

+≠

1k3

k4

−

⇒ (k + 1) (k + 3) = 8k and 8(3k – 1) ≠ 4k (k + 3)

But (k + 1) (k + 3) = 8k ⇒ k2 + 4k + 3 = 8k

or k2 – 4k + 3 = 0 or (k – 1) (k – 3) = 0

⇒ k = 1, 3.

For k = 1, 8(3k – 1) = 16 and 4k (k + 3) = 16

∴ 8(3k – 1) = 4k (k + 3) for k = 1

For k = 3, 8(3k – 1) = 64 and 4k (k + 3) = 72

i.e. 8(3k – 1) ≠ 4k (k + 3) for k = 3.

Thus, there is just one value for which the system of equations has no solution.

27

28