# Cls Jeead-15-16 Xii Mat Target-5 Set-2 Chapter-4

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Chapter

4

Determinants Solutions SECTION - A Objective Type Questions

1.

⎡ cos  sin  ⎤ Let A = ⎢ ⎥ , then |2A| is equal to ⎣  sin  cos  ⎦

(1) 4cos2

(2)

1

(3)

2

(4)

4

(4)

2

Sol. Answer (4) |A| = cos2 + sin2 = 1, |2A| = 22.|A| = 4

2.

2

1

If  is non-real complex cube root of unity, then 

2

1 is equal to

2

1

(1) 0

(2)

1

(3)

3

1 A  2

 3.

2

2 1 

1

1    2

2

   2  1 2 1 2

  1 

0 

2

 0 2 1

1

0 

0

1

If A is any of square matrix of order n, then A(adjA) is equal to (1) 1

(2)

|A|In

(3)

0

(4)

|A|n

(3)

(x – y)(y – z)(z – x)

(4)

xy + yz + zx

Sol. Answer (2) Clearly option (2) is correct. 1 x

4.

Value of determinant 1 y 1 z

(1) 0

x2 y 2 is equal to z2

(2)

xyz

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214

Determinants

Solution of Assignment (Set-2)

1 x

x2

1 y

y2

1 z

z2

R1  R1 – R3, R2  R2 – R3

0 xy

x2  y 2

  0 y z

y 2  z2

1

z2

z

= (x – y)(y – z)(z – x) 1 x

5.

The value of the determinant 1 y 1 z

x3 y 3 is equal to z3

(1) (x – y)(y – z)(z – x)

(2)

(x – y)(y – z)(z – x)(x + y + z)

(3) (x + y + z)

(4)

(x – y)(y – z)(z – x)(xy + yz + zx)

(3)

4

1 x

x3

1 y

y3

1 z

z3

R1  R1 – R2, R2  R2 – R3

0 x  y ( x  y )( x 2  xy  y 2 )   0 y z 1

( y  z )( y 2  yz  z 2 )

z

z3

= (x – y)(y – z)(z – x)(x + y + z) 6.

If A is 3 × 3 matrix and |A| = 4, then |A–1| is equal to (1)

1 4

(2)

1 16

(4)

2

(4)

81

Sol. Answer (1) A 1 

7.

1 1  A 4

If A is a square matrix of order 3, |A| = 3, then |adj adjA| is equal to (1) 35

(2)

37

(3)

9

Sol. Answer (4) |adjA| = 32 = 9 |adj(adjA)| = 92 = 81 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

8.

Determinants

215

If A, B and C are three square matrices of the same order such that A = B + C, then det A is equal to (1) det A + det B

(2)

det B

(3)

det (A)

(4)

det (B + C)

a1

9.

b1

If = a2 a3

b2 b3

2a1  3b1  4c1

c1

c2 , then value of 2a2  3b2  4c 2 c3 2a3  3b3  4c3

(1) 2

(2)

4

b1

c1

b2 b3

c 2 is equal to c3 (3)

(4)

2

(3)

HP

(4)

AGP

(4)

4

Sol. Answer (4) 10. If the system of equations x + 4ay + az = 0 x + 3by + bz = 0 x + 2cy + cz = 0 have a non-zero solution, then a, b, c ( 0) are in (1) AP

(2)

GP

Sol. Answer (3) 1 4a a 1 3b b  0 1 2c

c

R2  R2 – R1, R3  R3 – R1 1 4a a 0 3b  4a b  a  0 0 2c  4a c  a

 ab + bc = 2ac 

2 1 1   b a c

 a, b, c are in HP

11. Let

ax3

+

bx2

3x x 1 x 1 + cx + d = x  3 2 x x  2 , then the value of d is x  3 x  4 5x

(1) 5

(2)

0

(3)

–6

Sol. Answer (3) Put x = 0 0 d  3

1 1 0 2

3 4

0

 d = –6 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

216

Determinants

Solution of Assignment (Set-2)

12. Let x + y + z = 6, 4x + y – z = 0, 3x + 2y – 4z = –5. The value of  for which given system of equations does not have a unique solution is (1) 0

(2)

1

(3)

2

(4)

(1) 0

(2)

1

(3) 2sinB tanAcosC

(4)

2sinAsinBsinC

3

Sol. Answer (4) To not have unique solution, 1 1 1 4    0 3 2 4

1(2 – 0) – 1( – 3) + 1(2 –2) = 0 –3=0 =3 13. If A + B + C = , then value of

sin( A  B  C )

sin B

 sin B cos( A  B )

0  tan A

cos C tan A is 0



sin   sin B

sin B 0

cos C tan A

cos(   C )  tan A 0    sin B  cos C

0

sin B 0

cos C tan A

 tan A

0

= 0 (Skew symmetric matrix of odd order)

 14. The value of  lying between 0 and and satisfying the equation 2

(1)

5 3 , 24 24

(2)

7 5 , 24 24

(3)

1  sin2 

cos 2 

4 sin 4

sin2 

1  cos 2 

4 sin 4

2

sin 

7 11 , 24 24

2

cos 

= 0 are

1  4 sin 4

(4)

 11 , 24 24

Sol. Answer (3) R2  R2 – R1, R3  R3 – R1 1  sin2  cos2  4 sin 4 

1 1

1 0

0 1

0

C1  C1 + C2 + C3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

217

2  4 sin 4 cos 2  4 sin 4 0 1 0 0 0

0

1

 4sin4 = –2  sin4 =   

1 ⎛ ⎞  sin ⎜  ⎟ 2 ⎝ 6⎠

7 11 , 24 24

x

6

15. The number of real roots of the equation 2 3 x 3 2 x (1) 0

(2)

1

1 x  3 = 0 is x2 (3)

2

(4)

3

(3)

628

(4)

–4

(3)

x = –7

(4)

x = –9

6 3 x

1 x 3  0

3

2x

x2

 –5 (x – 2)(x + 3)(x – 1) = 0  x = 2, –3, 1

4 4 0 c = 0, then a + b + c is equal to 16. If a b  4 a b c4 (1) 41

(2)

116

Sol. Answer (4) C1  C1 + C2 + C3 0 4 abc 4 b4 abc 4

b

0 c

0

c4

 (a + b + c + 4) = 0  a + b + c = –4

x 2 3 1 17. The equation 4 x  2 10 4 = 0 is satisfied by 2x  1 5 1 (1) x = –2

(2)

x = –5

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218

Determinants

Solution of Assignment (Set-2)

Sol. Answer (3) x 2 3 1 4 x  2 10 4  0 2x  1

5

1

(x – 2)(10 – 20) – (4x – 2)(3 – 5) + (2x – 1)(12 – 10) = 0 = –10(x – 2) + 2(4x – 2) + 2(2x – 1) = 0 = –10x + 20 + 8x – 4 + 4x – 2 = 0  2x + 14 = 0  x = –7

18.

x

4 y z

y z

4 z  x is equal to 4 xy

(1) 4

(2)

x+y+z

(3)

xyz

(4)

0

(1) a + b + c + abc

(2)

a2 + b2 + c2 + ab + bc + ca

(3) 3abc – a3 – b3 – c3

(4)

a3 + b3 + c3 – 3abc

Sol. Answer (4) R1  R1 – R2, R2  R2 – R1 xy 0 y x y z 0 zy z

4 xy

= –4((x – y)(z – y) – (y – z)(y – x)) = 0

19.

a b c b c a is equal to c a b

Sol. Answer (3) C1  C1 + C2 + C3 abc b c   abc c a abc a b 1 b c  (a  b  c ) 1 c a 1 a b 0 bc c a  (a  b  c ) 0 c  a a  b 1

a

b

= (a + b + c)(ab + bc + ca – a2 – b2 – c2) = –(a3 + b3 + c3 – 3abc) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

x 1

(1) 1, 2

1

x  1 1 = 0 are 1 x 1

1 1

20. The roots of the equation

1

219

(2)

–1, 2

(3)

1, –2

(4)

–1, –2

Sol. Answer (2) C1  C1 + C2 + C3 x 1 1 1 x 1



1

1 1 x 1

1

x 1 1  x 1 x 1 x 1

1 1 x 1

1

1 1  ( x  1) 1 x  1 1

1 1 x 1

1

0 x  2 0  ( x  1) 0 x  2 2  x 1

x 1

1

= –(x + 1)(x – 2)2 Roots are –1, 2, 2

21. The value of the determinant (1) 1

1 logb a is equal to loga b 1

(2)

logab

(3)

logba

(4)

0

0

(3)

3

(4)

a+b+c

= 1.1 – logba × logab =1–1=0

22.

11  a c 1  bc 11  a d 1  bd is equal to 11  a e 1  be

(1) 1

(2)

Sol. Answer (2) 0 0 11  a

c  d b(c  d ) d  e b(d  e ) e

= (c – d)(d – e)

1  be 0 0

1 1

b b

11  a e 1  be

0

(as R1  R2 )

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220

Determinants

a 2

ab

23. If ab

b

2

ac

bc

Solution of Assignment (Set-2)

ac bc = ka 2b 2c 2, then k is equal to c 2

(1) 2

(2)

4

(3)

–4

(4)

8

(4)

–100

Sol. Answer (2) Put a = b = c = 1 1 1 1 1

1 1 k

1

1

1

k = –1(0) – 1(–2) + 1(2) = 4

1 24. Let f(x) =

x 1

x

2x x( x  1) ( x  1)x , then f(100) is equal to 3 x( x  1) x( x  1)( x  2) ( x  1)x ( x  1)

(1) 0

(2)

1

(3)

100

f ( x )  x ( x  1)

1 x 1

1 x

3 x ( x  1) ( x  1)( x  2) x ( x  1)

C1  C1 – C2, C2  C2 – C3 0 x 1

x ( x  1)

0 1

1 x

2( x  1)( x  1) 2( x  1) x ( x  1)

= x(x + 1)(–2(x2 – 1) + 2(x2 – 1)) = 0  f(100) = 0 25. If the system of the equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then the possible value of k are (1) –1, 2

(2)

1, 2

(3)

0, 1

(4)

–1, 1

Sol. Answer (4) To have non-zero solution 1  k 1 k 1 1  0 1

1

1

1(1 + 1) + k(–k + 1) – 1(k + 1) = 0 2 – k2 + k – k – 1 = 0 –k2 + 1 = 0 k = ±1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

sc a 26. If s = (a + b + c), then value of c sa c

(1) 2s 2

(2)

2s 3

a

b b

221

is

sb

(3)

s3

(4)

3s 3

Sol. Answer (2) sc a  c sa c

a

b b sb

C1  C1 + C2 + C3 sabc a   sabc sa sabc

a

b b sb

1 a b ( s  a  b  c ) 1 s  a b = 1 a sb 0 s 0 (2 s ) 0 s  s = 1 a sb

= (2s)(s2 – c) = 2s3

sin x cos x cos x   27. The number of distinct real roots of cos x sin x cos x = 0 in the interval   x  is 4 4 cos x cos x sin x (1) 0

(2)

2

(3)

1

(4)

3

Sol. Answer (3) sin x cos x cos x   cos x sin x cos x cos x cos x sin x

C1  C1 + C2 + C3 sin x  2cos x cos x cos x  sin x  2cos x sin x cos x sin x  2cos x cos x

sin x

1 cos x cos x  (sin x  2cos x ) 1 sin x cos x 1 cos x sin x Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

222

Determinants

Solution of Assignment (Set-2)

0 cos x  sin x 0  (sin x  2cos x ) 0 sin x  cos x cos x  sin x 1

cos x

sin x

0  (sin x  2cos x )(cos x  sin x ) 0 2

1 1

0 1

1 cos x sin x

= (sinx + 2cosx)(cosx – sinx)2 = 0  tanx = –2 or sinx = cosx x=

 is only solution 4

 Number of solution = one

28.

(a x  a  x )2

(a x  a  x )2 1

( b x  b  x )2

(b x  b  x )2 1 is equal to

(c x  c  x ) 2

(c x  c  x )2 1

(1) 0

(2)

2abc

(3)

a 2b 2c 2

(4)

abc

(3)

 and 

(4)

Neither  nor 

Sol. Answer (1) C1  C1 – C2

4 (a x  a  x )2 1 4 ( b x  b  x )2 1 4 (c x  c  x )2 1 = 0 (as C1 and C3 are proportional) 29. The determinant D=

cos(  )  sin(  ) cos 2 sin  cos  sin  is independent of  cos 

(1) 

sin 

cos 

(2)

Sol. Answer (1) D = cos( + )(cos cos – sinsin) + sin( + )(sin cos + cos sin) + cos2(sin2 + cos2) = cos( + ) cos( + ) + sin( + ) sin( + ) + cos2.1 = cos2 1 4 30. The roots of the equation 1 2

20 5  0 are

1 2x 5 x 2

(1) –1, –2

(2)

–1, 2

(3)

1, –2

(4)

1, 2

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Solution of Assignment (Set-2)

Determinants

223

4

20

  1 2

5

1 2x 5 x 2

0

6

15

 0 2  2 x 5(1  x 2 ) 1

5x 2

2x

= 30(1 – x2) + 30(1 + x) = 0  1 – x2 + 1 + x = 0  x2 – x – 2 = 0  x = 2, –1

1 a b 31. If a, b, c are sides of the ABC, 1 c a  0 , then value of sin2A + sin2B + sin2C is equal to 1 b c

(1)

4 9

(2)

3 2

(3)

9 4

(4)

3 3 2

Sol. Answer (3) 1 a b 1 c a 0 1 b c

0 ac ba 0 c b ac  0 1

b

c

 (a – c)2 – (b – a)(c – b) = 0  a2 + c2 – 2ac – (bc – b2 – ac + ab)  a2 + b2 + c2 – ab – bc – ca = 0  a=b=c  sin2A + sin2B + sin2C =

3 3 3 9   = 4 4 4 4

a a2 32. The parameter, on which the value of the determinant cos( p  d )x cos px cos( p  d )x does not depend sin( p  d )x sin px sin( p  d )x 1

upon, is (1) a

(2)

p

(3)

d

(4)

x

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224

Determinants

Solution of Assignment (Set-2)

Sol. Answer (2) 1 a a2 cos( p  d )x cos px cos( p  d )x sin( p  d )x

sin px

sin( p  d )x

= (cospx .sin(p + d)x – sinpx . cos(p + d)x) – a(cos(p – d)x . sin(p + d)x – sin(p – d)x . cos (p + d)x) 4 + a2(cos(p – d)x . sinpx –sin(p – d)x . cospx) = sindx – asin2dx + a2sindx ⎡ 11 33. If A = ⎢ ⎢⎣ – 13 ⎡17 (1) ⎢ ⎢⎣13

– 7⎤ ⎥ 11 ⎥⎦

⎡ 11 ⎢ ⎢⎣ – 13

(2)

7⎤ ⎥ 17 ⎥⎦

(3)

⎡ – 17 ⎢ ⎢⎣ 13

(3)

(3)

7 ⎤ ⎥ – 11⎥⎦

(4)

⎡ – 11 ⎢ ⎢⎣ – 13

1

(4)

–1

(4)

A2

7⎤ ⎥ 17 ⎥⎦

Sol. Answer (2) For a square matrix of order 2 × 2, adj (adj A) = A ⎡1 34. If A2 = 8A + kI where A = ⎢ ⎣ 1

(1) 7

0⎤ ⎥ , then k is 7⎦

(2)

–7

Sol. Answer (2) ⎡ 1 0⎤ ⎡ 1 0⎤ ⎡ 1 0 ⎤ A2  ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 1 7 ⎦ ⎣ 1 7 ⎦ ⎣ 8 49 ⎦ 0 ⎤ ⎡ k 0 ⎤ ⎡8  k ⎡8 8 A  kI  ⎢ ⎥  ⎢ 0 k ⎥  ⎢ 8  8 56 ⎣ ⎦ ⎣ ⎦ ⎣

0 ⎤ 56  k ⎥⎦

Clearly, k = –7

⎡sin   cos  ⎢ 35. If A = ⎢cos  sin  ⎢⎣ 0 0 (1) –AT

0⎤ ⎥ 0⎥ , then A–1 is equal to 1⎥⎦ (2)

A

Sol. Answer (3) |A| = sin2 + cos2 = 1 ( 0 ). 1 Hence A–1 exists. Since, A 

1 adj( A) and for the given question |A| = 1, hence A–1 = adj(A) |A|

36. If A is a matrix of order 3 and |A| = 2, then |adj A| is (1) 1

(2)

2

(3)

8

(4)

4

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Solution of Assignment (Set-2)

Determinants

225

Sol. Answer (4) det (adj A) = (det A)n – 1, where n is the order of matrix A, therefore |adj A| = |A|3–1 = 22 = 4 5

C0 5 37. The value of the determinant C1 5

(1) 0

(2)

C2

5 5 5

C3 14 C 4 1 is C5

1

– (6!)

(3)

80

(4)

–576

(4)

x=1

C0

5

C1 5 C2

5

5

C3 14

C4 5 C5

1 10 14 1  5 5 1 1 10 1 1

16 16 16  5 10

5 1

R1  R1  R2  R3

1 1

1 1 1  16 5 5 1 10 1 1 0 0 1  16 0 4 1 9 0 1

C1  C1  C2 , C2  C2  C3

= 16(0 – 36) = – 16 × 36

10 4 3 4 x 5 3 38. If 1  17 7 4 ,  2  7 x  12 4 such that 1 + 2 = 0, then 4 5 7  5 x 1 7 (1) x = 5

(2)

x has no real value

(3)

x=0

10 4 3 1  17 7 4 4 5 7 4 10 3   7 17 4 5 4 7 If 1 + 2 = 0 then 1 = – 2

…(i) …(ii)

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226

Determinants

Solution of Assignment (Set-2)

By equation (i) and (ii)

x 5

4

3

4

10 3

7 x  12 4  7 17 4 5 x  1 7 5 4 7  x=5

x 39. The value of y

z

x 2  yz 1 y 2  zx 1 is z 2  xy 1

(1) 1

(2)

–1

(3)

0

(4)

–xyz

Sol. Answer (3) x  y z x  y z

x 2  yz 1 y 2  zx 1 can be resolved as sum by two determinants as, z 2  xy 1  yz 1 zx 1  xy 1

x2 1 x y2 1  y z2 1 z

= 1 + 2 (say)

x where,  2  y z

 yz 1 zx 1  xy 1

x2 1 y2  xyz 2 z

 xyz x  xyz y  xyz

z

(multiplying R1 by x, R2 by y and R3 by z)

x2 1 x

 xyz 2 y 1 y (taking common –xyz from C2) xyz 2 z 1 z x2

x 1

2 = ( 1) y z2

y 1 z 1

x    y ( 1)( 1) = z

x2 1 y2 1 z2 1

= –1   = 1 + 2 = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

i m 1 i m4

im

m 5

40. The value of i i m 6

Determinants

227

i m2 i m  3 , when i   1 , is

i m 7

i m 8

(1) 1 if m is a multiple of 4

(2)

0 for all real m

(3) – i if m is a multiple of 3

(4)

2

(3)

2

m 5

= i i m 6

i m 1 i m4 i m 7

i m2 i m 3 i m 8

1

i

i2

 i m .i m  3 .i m  6 i 2 1

i i

1 i2

=0

(∵ R1 & R3 are identical)

41. The value of the determinant

1

– (25  1)2

210 – 1

210 – 1

– (25 – 1)2

1

1

25 – 1

25  1

5

2 –1 1 –

(1) 0

(2)

25  1 1

is

(210 – 1)2 1

(4)

4

Sol. Answer (4) Taking 25 + 1 = a and 25 – 1 = b, then 210 – 1 = (25 + 1) (25 – 1) = ab, therefore the given determinant equals.

a 2

ab

 ab

b 2

1 b

1 a

1 b 1 a 1  2 2 a b

Multiplying R1 with b, R2 with a and R3 with a2b2 ba 2 1   3 3 ba 2 a b ba 2

ab 2 ab 2 ab 2

1 1 1

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228

Determinants

0 1  3 3 ba 2 a b ba 2

Solution of Assignment (Set-2)

0

2

ab ab 2

2

ab 2 ab 2

2 ba 2 . a b3 ba 2

1 1 2 (a 3 b 3 ) 3 1 1 a b

3

R1  R1  R2

1 1

3

=4 cos 2 x sin 2 x cos 4 x 42. If the determinant sin 2 x cos 2 x cos 2 x is expanded in powers of sin x then the constant term in the cos 4 x cos 2 x cos 2 x

expansion is (1) 1

(2)

2

(3)

–1

(4)

0

Sol. Answer (3) For constant term, we can substitute x = 0.

1 0 1 0 1 1  The constant term = 1 1 1 = –1

43. If  are non-real numbers satisfying

(1) 0

(2)

x3

 1    1 is equal to – 1 = 0 then the value of   1 

3

(3)

3 + 1

(4)

4

Sol. Answer (2) x3 – 1 = 0  x = 1, , 2   = ,  = 2  1 +  +  = 0 and  = 1

 1    1     1    1         1       1   1    1 

C1  C1  C2  C3

1    1  1 1 1 

(∵ 1      0)

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Solution of Assignment (Set-2)

Determinants

    0  1  0 1     1

229

R2  R2  R1 R3  R3  R1

= [( +  – )( +  – ) – (1 – )(1 – )] =  [2 – ( – )2 – (1 –  –  + )] = [2 – ( + )2 + 4 – (1 – ( + ) + )] = [2 – (–1)2 + 4 – (1 – (–1) + 1)] = (2 – 1 + 4 – 3) = 3

n 2 U n if U = n 44. The value of n n 1 3 n N

(1) 0

(2)

1 5 2N  1 2N  1 is 3N 2 3N 1

(3)

–1

(4)

2

∑n

1

5

n 1

N

∑U n 1

N

n

=

∑n

2

2N  1 2N  1

n 1 N

∑n

3

3N 2

3N

n 1

N (N  1) 1 5 2 N (N  1)(2N  1)  2N  1 2N  1 6 N 2 (N  1)2 3N 2 3N 4

0 1 5 N (N  1)  0 2N  1 2N  1 12 0 3N 2 3N

6 1 5 N (N  1)  2(2N  1) 2N  1 2N  1 12 3N (N  1) 3N 2 3N

C1  C1  C2 – C3

=0 45. If a > 0, b > 0, c > 0 are respectively the pth, qth, rth terms of a G.P., then the value of the determinant

log a p 1 log b q 1 log c r 1

(1) 1

is

(2)

–1

(3)

abc pqr

(4)

0

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230

Determinants

Solution of Assignment (Set-2)

Sol. Answer (4) Let, a = ARp – 1, b = ARq – 1, c = ARr – 1

log A  ( p  1)log R

log a p 1

 log A  (q  1)log R q 1 log A  (r  1)log R r 1

log b q 1 log c r 1

p 1

( p  1)log R

p 1

 (q  1)log R q 1 (r  1)log R r 1

C1  C1  (log A)C3

0 p 1  0 q 1 0 r 1

C1  C1  (C2  C3 )log R

=0

a b c 46. If a, b, c are positive and not equal then value of b c a c a b (1) 1

(2)

–3

(3)

may be

2

(4)

4

212 – 211

(4)

5 × 211

a b c b c a = – (a3 + b3 + c3 – 3abc) c a b = – (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 

(a  b  c ) {(a – b)2 + (b – c)2 + (c – a)2} < 0 2

Hence possible value = –3 47. The greatest value of x satisfying the equation 211 – x

– 212

211

– 212

211 – x

211

11

11

2

12

2

–2

(1) 3 × 211

0

is

–x

(2)

211 + 213

(3)

Sol. Answer (1) Let 211 = a, then the given equation reduces to

ax

2a

2a a

ax a 0 a 2a  x

a

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Solution of Assignment (Set-2)

Determinants

231

Applying C1  C1 + C2 + C3

x

2a

a

x a  x a 0 x a 2a  x 1

2a

a

x 1 ax a 0 1 a 2a  x

x 0 3a  x 0 0 0 3a 3a  x

1

2a

a

R2  R2  R1 R3  R3  R1

 x[(3a – x) (–3a – x) – 0] = 0  x(x – 3a) (x + 3a) = 0  Greatest value of x satisfying the given equation is x = 3a = 3 × 211 48. The system of equations ax + 4y + z = 0, bx + 3y + z = 0, cx + 2y + z = 0 has non-trival solution if a, b, c are in (1) AP

(2)

GP

(3)

HP

(4)

None of these

(4)

A2 = I

Sol. Answer (1) For non-trivial solutions,

a 4 1 b 3 1  0 ⇒ a  2b  c  0 c 2 1  a, b, c are in A.P.

⎡ 0 0  1⎤ ⎥ ⎢ 49. Let A = ⎢ 0  1 0 ⎥ . The only correct statement about the matrix A is ⎢⎣ 1 0 0 ⎥⎦ (1) A is a zero matrix

(2)

A = (–1)I3

(3)

A–1 doesn’t exist

Sol. Answer (4) Clearly, options (1), (2) & (3) are false. Option (4) can be easily verified as

⎡ 0 0 –1⎤ ⎡ 0 0 –1⎤ ⎡ 1 0 0 ⎤ ⎢ 0 –1 0 ⎥ ⎢ 0 –1 0 ⎥  ⎢0 1 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ –1 0 0 ⎥⎦ ⎢⎣ –1 0 0 ⎥⎦ ⎢⎣0 0 1⎥⎦ 50. The largest value of a third order determinant whose elements are equal to 1 or 0 is (1) 0

(2)

2

(3)

4

(4)

6

Sol. Answer (2) 0 1 1 1 0 1 2 1 1 0

Option (2) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

232

Determinants

xk

x k +2

k 51. If y zk

y k +2

Solution of Assignment (Set-2)

x k +3

⎧ 1 1 1⎫ y k +3 =(x  y )(y  z )(z  x ) ⎨ + + ⎬ , ⎩x y z⎭ z k +3

z k +2

then (1) k = –2

(2)

k = –1

(3)

k = 0

(4)

k = 1

Sol. Answer (2) xK yK zK

xK 2 y K 2 zK  2

x K 3 K K K y K  3  x .y .z ( x  y )( y  z )( z  x ){ xy  yz  zx } K 3 z ⎧ 1 1 1⎫  ( xyz )K 1( x  y )( y  z )( z  x ) ⎨   ⎬ ⎩x y z⎭

 k+ 1 = 0  k = –1 Option (2) is correct. 52. If all elements of a third order determinant are equal to 1 or –1, then deteminant itself is (1) An odd integer

(2)

An even integer

(3) An imaginary number

(4)

A real number

Sol. Answer (2) Let third order determinant be a11 a12   a21 a22 a31 a32

a13 a23 a33

a31 a21 when aij = 1 on –1 then both a and a equals ±1. 11 11 ⎛ a21 ⎞ ⎛ a31 ⎞ ⎟ R1 and R3  R3  ⎜ ⎟ R1 . a ⎝ 11 ⎠ ⎝ a11 ⎠

Apply R2  R2  ⎜

a11 a12 a13 D  0 2 s where , , ,  are all 1 or –1. 0 2 2

Expand along C1   = 4a11{ – }. which is an even integer hence option (2) is correct. 53. If A is a 3×3 matrix and det (3A) = k{det(A)}, then k is (1) 9

(2)

6

(3)

1

(4)

27

Sol. Answer (4) det (3A) = k{det(A)}  33det(A) = k{det(A)}  k = 27, option (4) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

233

⎡ x ⎤ ⎡ 1⎤ 54. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A ⎢ y ⎥  ⎢ 0 ⎥ has exactly ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 0 ⎥⎦ two distinct solutions, is (1) 0

[IIT-JEE 2010] (2)

29 – 1

(3)

168

(4)

2

⎡x The equation A ⎢ y ⎢ ⎢⎣ z

⎤ ⎡1⎤ ⎥  ⎢ 0 ⎥ has two distinct solutions. It should be noted here that the given equation is linear ⎥ ⎢ ⎥ ⎥⎦ ⎢⎣ 0 ⎥⎦

equation in 3 variables, which may have no solution, or unique solution or infinitely many solutions. Hence there does not exist any matrix A such that the given equation has exactly two solutions and consequently number of 3 × 3 matrices is 0. 55. Let   1 be a cube root of unity and S be the set of all non-singular matrices of the form ⎡1 ⎢ ⎢ ⎢ 2 ⎣

b⎤ ⎥ c⎥ 1⎥⎦

a 1 

where each of a, b and c is either  or 2. Then the number of distinct matrices in the set S is [IIT-JEE 2011] (1) 2

(2)

6

(3)

4

(4)

8

Sol. Answer (3) We have, ⎡1 ⎢ M= ⎢ ⎢ 2 ⎣

b⎤ ⎥ c⎥ 1⎥⎦

a 1 

 |M|=

0 

a  2 1

b  c 1 c

2

1

= –(a – 2 – b – c – ) + 2(ac – 2c – b – c – 1) = –(a + c) + ac2 + 1 a + c  1, ac  1 Since a, b, c are  or 2  a=c If a =  c =   Number of ways of selecting a, b, c = 1 × 1 × 2 = 2 If a = 2, then number of ways = 1 × 1 × 1 = 2 Total number of distinct matrices in the given set S = 4. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

234

Determinants

Solution of Assignment (Set-2)

56. Let P = [aij] be a 3 × 3 matrix and let Q = [bij], where bij = 2i+jaij for 1  i, j  3. If the determinant of P is 2, then the determinant of the matrix Q is [IIT-JEE 2012] (1) 210

(2)

211

⎡ a11 a12 Given, P  [aij ]33  ⎢a21 a22 ⎢ ⎢⎣a31 a32

a13 ⎤ a33 ⎥⎥ a33 ⎥⎦

(3)

212

(4)

213

i j Now, Q  2 aij  [bij ]

⎡ b11 b12  Q  ⎢b21 b22 ⎢ ⎢⎣ b31 b32

b13 ⎤ b23 ⎥⎥ b33 ⎥⎦

⎡ 22 a11 23 a12 ⎢ 3 ⎢ 2 a21 24 a22 = ⎢ 4 5 ⎢⎣ 2 a31 2 a32

a11 a12 a21 a22 a31 a32

12

So, Q  2

24 a13 ⎤ ⎥ 25 a23 ⎥ ⎥ 26 a33 ⎥⎦

a13 a23 a33

= 213. (as |P| = 2)

SECTION - B Objective Type Questions (More than one options are correct) 1.

The values of  for which the system of equations x + y + z = 1, x + 2y + 4z = , x + 4y + 10z = 2 is consistent, are given by (1) 1, 2

(2)

–1, 2

(3)

1, –2

(4)

–1, –2

Sol. Answer (1) Clearly, the determinant

1 1

1

1 0 0

1 2 4  1 1 3 0 1 4 10 1 3 9

(C2  C2 – C1, C3  C3 – C1)

therefore, the given equations are consistent, if each of the three determinants

1  2

1

1

1

1

1

1 1

1

2 4 , 1  4 , 1 2  are zero. 4 10 1  2 10 1 4  2

Solving,  = 1, 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

2.

Determinants

235

The system of equations x – y cos  + z cos 2 = 0 – x cos  + y – z cos  = 0 x cos 2 – y cos  + z = 0 has non-trivial solution for  equals  3

(1)

(2)

 6

(3)

2 3

(4)

 12

(4)

Sol. Answer (1, 2, 3, 4) For non-trivial solution

1

 cos  cos 2

 cos  1  cos   0 cos 2  cos  1 on expansion, the determinant equals (1 – cos2) + cos(–cos + coscos2) + cos2(cos2 – cos2) = 1 – 2cos2 + 2cos2cos2 – cos22 = 1 – 2cos2 (1 – cos2) – cos22 = 1 – 4cos2sin2 – cos22 = 1 – 4cos2 (1 – cos2) – (2cos2 – 1)2 = 1 – (4cos2 – 4cos4) – (4cos4 – 4cos2 + 1) = 0, for all values of . 3.

If f(x) and g(x) are functions such that f(x + y) = f(x) g(y) + g(x) f(y), then

f (  ) g (  ) f (   ) f () g () f (  ) is independent of f (  ) g (  ) f (   ) (1) 

(2)

(3)

Sol. Answer (1, 2, 3, 4)

f (  ) g (  ) f (   ) f () g () f (  ) f (  ) g (  ) f (   ) f ( ) g ( ) f ( )g ()  g ( )f () g () f ()g ()  g ()f () g (  ) f (  )g ()  g (  )f ()

= f () f ()

f ( ) g ( ) 0 = f () g () 0 f () g() 0

C3  C3 – g()C1 – f()C2

=0 Hence, independent of , ,  and . Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

236 4.

Determinants

Solution of Assignment (Set-2)

The digits A, B, C are such that the three digit numbers A88, 6B8, 86C are divisible by 72, then the determinant

A 6 8 B 8

8 6 is divisible by

8 C

(1) 72

(2)

144

(3)

288

(4)

1216

Sol. Answer (1, 2, 3) 3 digit numbers divisible by 72, include 288, 648 and 864. Hence, the only possibility is A = 2, B = 4, C = 4

2 6 8 

  8 4 6  288 8 8 4

  is divisible by 72, 144 and 288.

5.

a b a  b If b c b  c  0 then a  b b  c 0 (1) a, b, c are in A.P.

(2)

a, b, c are in G.P.

(3)  is a root of ax2 + 2bx + c = 0

(4)

( x   ) is a factor of ax2 + 2bx + c = 0

a b a  b b c b  c  0 a  b b  c 0  R3  R3 – (R1 + R2)

a b

b c

a  b b  c

0 0

0

2

(a  b  b  c )

 – (a2 + 2b + c) (b2 – ac) = 0 b2 = ac, a2 + 2b + c = 0 Hence option (2, 3, 4) are true

6.

1

bc

a(b  c )

The value of the determinant 1

ca

b(a  c ) doesn’t depend on

1

ab

c (a  b )

(1) a

(2)

b

(3)

c

(4)

a+b+c

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Solution of Assignment (Set-2)

Determinants

237

1 bc a(b  c ) 

 1 ca b(a  c ) 1 ab c(a  b ) 0 c ( b  a ) c (a  b )  0 a(c  b ) a(b  c ) 1 ab c (a  b )

 

(R1  R1 – R2, R2  R1 – R3)

c ( b  a ) c (a  b ) a(c  b ) a(b  c )

 ac (a  b )(b  c )

1 1 1 1

=0   does not depend on a, b, c.

7.

⎡ 1 0 0⎤ ⎡1 0 0⎤ ⎢ ⎥ Let A  ⎢0 1 1⎥ and I  ⎢0 1 0 ⎥ , then ⎢ ⎥ ⎢⎣0 0 1⎥⎦ ⎢⎣0 2 4 ⎥⎦ 1 2 ( A  6 A  11I ) 6

(1) A3 – 6A2 + 11A – 6I = 0

(2)

A 1 

(3) A2 is non-singular

(4)

A is singular

1  0 0 A  I  0 1   1 0 0 2 4    (1 – ) {(1 – ) (4 – ) + 2} = 0  (1 – ) (2 – 5 + 4 + 2) = 0 ( – 1) (2 – 5 + 6) = 0 3 – 52 + 6 + 5 – 2 – 6 = 0 3 – 62 + 11 – 6 = 0 Hence the characteristics equation of the matrix is A3 – 6A2 + 11A – 6 = 0  A–1 =

1 2 (A – 6A + 11I) 6

|A2|  0 |A|  0 Hence options (1, 2, 3) are correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

238 8.

Determinants

Solution of Assignment (Set-2)

A is a matrix of order 3 × 3. If A = A and five entries in the matrix are of one kind and remaining four are of another kind, then the maximum number of such matrices is greater than or equal to (1) 9

(2)

10

(3)

11

(4)

8

Sol. Answer (1, 2, 3, 4) Since, the matrix is symmetric, it can be of the following type

⎡a x y ⎤ A  ⎢⎢ x b z ⎥⎥ ⎢⎣ y z c ⎥⎦ Case (i) : Let a = b = c = x  y = z In this case total matrices = 3C1 = 3 Case (ii) : Let a = x = y  b = c = z In this case total matrices = 3C2 × 3C1 = 9 Hence, maximum number of such matrices = 3 + 9 = 12

9.

⎡ 1 4 4⎤ ⎢ ⎥ If the adjoint of a 3 × 3 matrix P is ⎢2 1 7 ⎥ , then the possible value(s) of the determinant of P is (are) ⎢⎣ 1 1 3 ⎥⎦ [IIT-JEE 2012] (1) –2

(2)

–1

(3)

1

(4)

2

⎡ 1 4 4⎤ adj P  ⎢⎢ 2 1 7 ⎥⎥ ⎢⎣ 1 1 3 ⎥⎦ |adj P| = 1(3 – 7) – 4(6 – 7) + 4(2 – 1) = –4 + 4 + 4 = 4  |adj P| = |P|2  |P|2 = 4  |P| = ±2

SECTION - C Linked Comprehension Type Questions Comprehension-I Consider a matrix A = [aij]n × n. Form the matrix A – I,  being a number, real or complex. ⎡a11 –  ⎢ ⎢ a21 A – I = ⎢ ... ⎢ ⎢ a ⎣ n1

a12

...

a22 – 

...

...

...

an 2

...

a1n

⎤ ⎥ a2 n ⎥ ⎥ ... ⎥ ann –  ⎥⎦

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Solution of Assignment (Set-2)

Determinants

239

Then det(A – I) = (–1)n[n + b1n – 1 + b2n – 2 + ... + bn]. An important theorem tells us that the matrix A satisfies the equation xn + b1xn – 1 + b2xn – 2 + ... + bn = 0. This equation

⎡1 is called the characteristic equation of A. For all the questions on this passage, take A = ⎢ ⎢⎣2 1.

4⎤ ⎥ 3 ⎥⎦

The matrix A satisfies the matrix equation (1) A2 – 4A – 5I = 0

(2)

A2 – 4A – 5 = 0

(3) A2 + 4A – 5I = 0

(4)

A2 + 4A – 5 = 0

(3)

1 ( A – 5I ) 4

Sol. Answer (1) | A  I | 0 ⇒

1  4 0 2 3

 (1 – ) (3 – ) – 8 = 0  2 – 4l – 5 = 0  A2 – 4A – 5I = 0 2.

Which of the following is inverse of A? (1)

1 ( A – 4I ) 5

1 ( A  4I ) 5

(2)

(4)

1 ( A  5I ) 4

Sol. Answer (1) A2 – 4A – 5I = 0  A–1(A2 – 4A – 5I) = 0  A – 4I – 5A–1 = 0  5A–1 = A – 4I  3.

A 1 

1 ( A  4I ) 5

The matrix A5 – 4A4 – 7A3 + 11A2 – A – 10I, when expressed as a linear polynomial in A , is (1) A + 5I

(2)

A – 5I

(3)

–A + 5I

(4)

–A – 5I

Sol. Answer (1) A5 – 4A4 – 7A3 + 11A2 – A – 10I = A3(A2 – 4A – 5) – 2A3 + 11A2 – A – 10I = – 2A(A2 – 4A – 5I) + 3A2 – 11A – 10I = 3(4A + 5I) – 11A – 10I = A + 5I Comprehension-II Matrix theory can be applied to investigate the conditions under which a given system of linear equations possesses unique, infinite or no solutions. Consider the system of 3 non-homogeneous linear equations in 3 unknowns x+y+z=6 x + 2y + 3z = 10 x + 2y + z =  and answer the questions that follow. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

240 1.

Determinants

Solution of Assignment (Set-2)

The system possesses a unique solution if (1)   3

(2)

  3,   10

(3)

  10

(4)

 = 10

Sol. Answer (1) For unique solution

1 1 1 1 2 3 0 1 2  1 0 

0 (C2  C2 – C1, C3  C3 – C2)

1 1 1 0 1 1 2

 ( – 2) – 1  0  3 2.

The system possesses no solutions if (1)   3,  = 10

(2)

 = 3,   10

(3)

 = 10

(4)

 = 3,   10

(3)

 = 3,  = 10

(4)

  3,   10

(3)

8A

(4)

16A

Sol. Answer (2) For no solution

1 1 1 1 2 3  0 and   10 1 2  i.e.,  = 3 and  10. The last two equations must not be identical. 3.

The system possesses infinite solutions if (1)  = 3,   10

(2)

  3,  = 10

Sol. Answer (3) For infinite solutions

1 1 1 1 2 3  0 and  = 10 1 2  i.e.,  = 3 and  = 10 The last two equations must be identical. Comprehension-III A and B are two matrices of same order 3 × 3, where

⎡1 2 3⎤ ⎢ ⎥ A = ⎢2 3 4 ⎥ , B = ⎢5 6 8 ⎥ ⎣ ⎦ 1.

⎡3 2 5 ⎤ ⎢ ⎥ ⎢2 3 8 ⎥ ⎢7 2 9 ⎥ ⎣ ⎦

(2)

4A

Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

2.

Determinants

241

The value of |adj(AB)| is (1) 24

(2)

242

(3)

243

(4)

65

29

(3)

1

(4)

219

(2)

Sol. Answer (3) Comprehension-IV Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. [IIT-JEE 2009] 1.

The number of matrices in A is (1) 12

(2)

6

(3)

9

(4)

3

Sol. Answer (1) The matrix is of order 3 × 3 where there are 9 elements. This has to be symmetric matrix with five of the entries as 1 and four of the entries as zero. For the matrix to be symmetric (i) either all the three elements along principal diagonal should be (1) or exactly one element along principal diagonal should be 1. The possible matrixes are as follows

⎡ 1 1 0⎤ ⎡ 1 0 1⎤ ⎡ 1 0 0⎤ ⎢ ⎥ ⎢ ⎥ A1  ⎢ 1 1 0 ⎥ , A2  ⎢0 1 0 ⎥ , A3  ⎢⎢0 1 1⎥⎥ ⎢⎣0 0 1⎥⎦ ⎢⎣ 1 0 1⎥⎦ ⎢⎣0 1 1⎥⎦ ⎡1 1 1⎤ ⎡ 1 1 0⎤ ⎡ 1 0 1⎤ ⎢ ⎥ ⎢ ⎥ B1  ⎢1 0 0 ⎥ , B2  ⎢ 1 0 1⎥ , B3  ⎢⎢0 0 1⎥⎥ ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣ 1 1 0 ⎥⎦ ⎡ 0 1 1⎤ ⎡0 1 0 ⎤ ⎡0 0 1⎤ ⎢ ⎥ ⎢ ⎥ C1  ⎢ 1 1 0 ⎥ , C2  ⎢ 1 1 1⎥ , C3  ⎢⎢0 1 1⎥⎥ ⎢⎣ 1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣ 1 1 0 ⎥⎦ ⎡ 0 1 1⎤ ⎡0 1 0 ⎤ ⎡0 0 1⎤ ⎢ ⎥ ⎢ ⎥ D1  ⎢ 1 0 0 ⎥ , D2  ⎢ 1 0 1⎥ , D3  ⎢⎢0 0 1⎥⎥ ⎣⎢ 1 0 1⎥⎦ ⎣⎢0 1 1⎦⎥ ⎣⎢ 1 1 1⎦⎥ There are 12 such matrices. 2.

The number of matrices A in A for which the system of linear equations

⎡ x ⎤ ⎡ 1⎤ A ⎢⎢ y ⎥⎥  ⎢⎢0 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣0 ⎥⎦ has a unique solution, is (1) Less than 4

(2)

At least 4 but less than 7

(3) At least 7 but less than 10

(4)

At least 10

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242

Determinants

Solution of Assignment (Set-2)

Sol. Answer (2) The determinant corresponding to matrices B2, B3, C1, C3, D1 and D2 are non-zero. Therefore in these three cases, the given linear equations will have unique solution. Number of required matrices in this case is 6. 3.

The number of matrices A in A for which the system of linear equations

⎡ x ⎤ ⎡ 1⎤ A ⎢⎢ y ⎥⎥  ⎢⎢0 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣0 ⎥⎦ is inconsistent, is (1) 0

(2)

More than 2

(3)

2

(4)

1

⎡1 1 0⎤ (i) If we take (i) A1  ⎢ 1 1 0 ⎥ ⎢ ⎥ ⎣⎢0 0 1⎥⎦ then |A1| = 0

1 1 0 x  0 1 0  1  0 , hence in this case system is inconsistent. 0 0 1 1 0 1 (ii) If we take A2  0 1 0 ⇒ | A2 |  0

1 0 1

1 0 1 x  0 1 0  1  0 , hence system is inconsistent. 0 0 1 1 0 0 (iii) If we take A3  0 1 1 ⇒ | A3 |  0 0 1 1 1 0 0 x  0 1 1  0 , Similarly y = A3 = 0. In this case we get infinite solution. 0 1 1

1 1 1 (iv) If we take B1  1 0 0 ⇒ | B1 |  0 . In this case x = y = z = 0 1 0 0 Hence infinite solution consistent system. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

243

Comprehension-V Let a, b and c be three real numbers satisfying

[a

1.

⎡1 c ] ⎢⎢8 ⎢⎣7

b

9

7⎤ 7 ⎥⎥  [0 7 ⎥⎦

2 3

0

0]

...(E)

[IIT-JEE 2011]

If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (1) 0

(2)

12

(3)

7

(4)

6

Sol. Answer (4) From the given condition, a + 8b + 7c = 0 9a + 2b + 3c = 0 7a + 7b + 7c = 0 a+b+c=0 Also, 2a + b + c = 1  a=1  7a + b + c = 6a + (a + b + c) = 6

2.

Let  be a solution of x3 – 1 = 0 with Im() > 0. If a = 2 with b and c satisfying (E), then the value of

3 a

1 b

3 c

is equal to (1) –2

(2)

2

(3)

3

(4)

–3

Sol. Answer (1) When a = 2  b + c = –2 Also, 8b + 7c = –2  b = 12 and c = –14 Thus,

3 a

1 b

3 c

=

3 2

1 12

3 14

= 1 + 3 + 32 =1–3 = –2 3.

Let b = 6, with a and c satisfying (E). If  and  are the roots of the quadratic equation ax2 + bx + c = 0, then 

⎛ 1 1⎞ ⎜  ⎟  ⎠ n 0 ⎝

(1) 6

n

(2)

7

(3)

6 7

(4)

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244

Determinants

Solution of Assignment (Set-2)

Sol. Answer (2) For b = 6  a + c = –6 a + 7c = –48 a + c = –6 when a = 1, c = –7  and  are roots of the given equation ax2 + bx + c = 0, hence  +  = –6,  = –7

1 1  6       7 n

2

⎛ 1 1⎞ 6 ⎛6⎞ ⎜  ⎟  1   ⎜ ⎟  ......  7 ⎝7⎠  ⎠ n 0 ⎝

1 1

6 7

7

SECTION - D Assertion-Reason Type Questions

1.

⎡0  h  g ⎤ Consider the matrix A  ⎢⎢ h 0  f ⎥⎥ ⎢⎣g f 0 ⎥⎦ STATEMENT-1 : det A = 0. and STATEMENT-2 : The value of the determinant of a skew symmetric matrix of odd order is always zero.

⎡ 0 h g ⎤ | A | ⎢⎢ h 0 f ⎥⎥ ⎢⎣g f 0 ⎥⎦ 0  h

h g

f h g 0 g

0 f

= 0 + h (0 + gf) – g(hf – 0) =0

2.

2 1 3 2 0 2 Consider the determinants   1 1 1 ,    2  1 1 1 1 1

4

1

3

STATEMENT-1 :  = 2. and STATEMENT-2 : The determinant formed by the cofactors of the elements of a determinant of order 3 is equal in value to the square of the original determinant. Sol. Answer (1) Clearly, ' is formed by the cofactor of   ' = 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

3.

Determinants

245

⎡a 0 0 0 ⎤ ⎢ ⎥ STATEMENT-1 : Matrix ⎢0 b 0 0 ⎥ is a diagonal matrix. ⎢0 0 c 0 ⎥ ⎣ ⎦

and STATEMENT-2 : A = [aij] is a square matrix such that aij = 0  i  j, then A is called diagonal matrix. Sol. Answer (4) A is not diagonal matrix ∵ A is not square matrix Option (4) is correct.

4.

⎡1 1 1⎤ ⎢ ⎥ STATEMENT-1 : Inverse of A= ⎢1 2 3 ⎥ does not exist. ⎢1 4 7 ⎥ ⎣ ⎦

and STATEMENT-2 : If any matrix is singular, then its inverse does not exist. Sol. Answer (1) ∵ |A| = 0, A–1 does not exists. Option (1) correct. 5.

STATEMENT-1 : The system of equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 3y – 4z = 0, possesses a non-trival solution, then value of k is

31 . 2

and STATEMENT-2 : Three linear equations in x, y, z can never, hence exactly two solutions. Sol. Answer (4)

1 k 3   3 k 2  0 2 3 4

 20k + 33 – 22k = 0

 k

33 2

Option (4) is correct.

(1+x )21 (1+x )22 6.

STATEMENT-1 : f (x )= (1+x )31 (1+x )32

(1+x )41 (1+x )42

(1+x )23 (1+x )33 , then coefficient of x in f(x) is zero. (1+x )43

and STATEMENT-2 : If F(x) = A0 + A1x + ...... + Anxn, then A1 = F(0), dash denotes differential coefficient. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

246

Determinants

Solution of Assignment (Set-2)

Sol. Answer (1) f'(x) = a1 + 2a2x + or f'(0) = a1 

21 22 23 1 1 1 1 1 1 a1  1 1 1  31 32 33  1 1 1  0 1 1 1 1 1 1 41 42 43

Option (1) is correct. 7.

Consider the system of equations x – 2y + 3z = –1 –x + y – 2z = k x – 3y + 4z = 1 STATEMENT-1 : The system of equations has no solution for k  3. and

1

3

STATEMENT-2 : The determinant  1  2

1

4

1 k  0, for k  3. 1

[IIT-JEE 2008]

Sol. Answer (1) The given system of equations can be expressed as

⎡ 1 2 3 ⎤ ⎡ x ⎤ ⎡ 1⎤ ⎢ 1 3 4 ⎥ ⎢ y ⎥  ⎢ 1 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 1 2⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ k ⎥⎦ Applying R2  R2  R1, R3  R1  R3

⎡ 1 2 3 ⎤ ~ ⎢⎢0 1 1⎥⎥ ⎣⎢0 1 1⎥⎦

⎡ x ⎤ ⎡ 1 ⎤ ⎢y ⎥  ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ z ⎦⎥ ⎢⎣k  1⎥⎦

⎡ 1 2 3 ⎤ ~ ⎢⎢0 1 1⎥⎥ ⎢⎣0 0 0 ⎥⎦

⎡ x ⎤ ⎡ 1 ⎤ ⎢y ⎥  ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ by R3  R3 – R2 ⎢⎣ z ⎥⎦ ⎢⎣k  3 ⎥⎦

When k  3 , the given system of equations has no solution.  Statement 1 is true. Clearly statement–2 is also true as it is rearrangement of rows and columns of

⎡ 1 2 3 ⎤ ⎢ 1 3 4 ⎥ ⎢ ⎥ ⎢⎣ 1 1 2 ⎥⎦ Hence option (1) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

247

SECTION - E Matrix-Match Type Questions 1.

Let f(x) denote the determinant x2 f (x)  x 2  1

2x

1  x2

x 1

1

–1

x –1

x

on expansion f(x) is seen to be a 4th degree polynomial given by f(x) = a0x4 + a1x3 + a2x2 + a3x + a4. Using differentiation of determinant or otherwise, match the values of the quantities on the left to those on the right. Column-I

Column-II

(A) a4

(p)

–1

(B) a3

(q)

1

(C) a0

(r)

–3

(D) a12 + a1 + 1

(s)

3

Sol. Answer A(p), B(q), C(r), D(s)

0 0 1 If x = 0, a4  1 1 1  1 0 1 1 f(x) = x2[(x2 – 1) – 1] – 2x [(x2 + 1)(x – 1) – x] + (1 + x2) [(1 + x2)(–1) – x(x + 1)] = x2(x2 – 2) – 2x(x3 – x2 – 1) – (1 + x2)2 – x(1 + x + x2 + x3) = (x4 – 2x2) – (2x4 – 2x3 – 2x) – (1 + 2x2 + x4) – (x + x2 + x3 + x4) = – 3x4 + x3 – 5x2 + x – 1  a4 = –1, a3 = 1, a0 = –3, a1 = 1

 a12 + a1 + 1 = 3 2.

If A is a non-singular matrix of order n × n, then match the following Column-I

Column-II

(p)

(q)

(r)

(s) (t)

n –1

Sol. Answer A(q, s), B(t), C(r), D(q, s) We know that (adj A)(A) = A(adj A) = |A|In and adj(adj A) = |A|n – 2A; |A|  0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

248 3.

Determinants

Solution of Assignment (Set-2)

The entries in a 3 × 3 determinant are either 1 or –1, then match the following Column-I

Column-II

(A) Total number of such determinants

(p)

4

(B) The maximum value of such a determinant

(q)

3

(C) The maximum value of trace of such determinant

(r)

512

(D) The minimum value of such determinant

(s)

Zero

(t)

–4

(u)

–3

Sol. Answer A(r), B(p), C(q), D(t) (A) 29 = 512 (B) For maximum value the determinant will be

1

1 1

1 1 1 = 1(1 + 1) +1(1 + 1) + 0 = 4 1 1 1 (C) Maximum trace = 1 + 1 + 1 = 3 (D) Minimum value = –4

SECTION - F Integer Answer Type Questions 1.

Let |A| = |aij|3 × 3  0. Each element aij is multiplied by ki–j. Let |B| be the resulting determinant, where k1|A| + k2|B| = 0, then k1 + k2 is

a11 a12 a13 A  a21 a22 a23 a31 a32 a33 a11 B  k a21 k 2 a31

k 1a12 k 2a13 a22 k a32

k 1 a23  a33

k 2a11 ka12 a13

1 2 k a21 ka22 a23  A k3 2 k a31 ka32 a33

 |A| = |B|  |A| – |B| = 0 Comparing it with k1|A| + k2|B| = 0 We get k1 + k2 = 0 x 2  6x  5

2.

If

2x  6 2 2 x 2  5 x  9 4 x  5 4  Ax 3  Bx 2  Cx  D , then A + B + 2C is equal to 6 x 2  4 x  6 12 x  4 12

Sol. Answer (0) As the value of the determinant is zero  A=B=C=D=0  A + B + 2C + 3D = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

3.

Determinants

249

Let A = [aij]n × n, n is odd. Then det((A – AT)2009) is equal to

Sol. Answer (0) ∵ A – AT is skew symmetric.  (A – AT)2009 is also skew symmetric  det (A – AT)2009 = 0, as determinant value of every skew symmetric matrix of odd order is 0.

1 4.

3cosθ

If  = sinθ

1

1

1 3cosθ , then the

sinθ

1

1 [|maximum value of  – minimum value of |3] is equal 1000

to_________. Sol. Answer (1) ∵ 

= 1{1 – 3sincos} – 3cos{sin – 3cos} + 1{sin2 – 1} = 1 – 6sincos + 9cos2 + sin2 – 1

  

= (sin – 3sin)2

 10  sin   3 cos   10

 0  (sin – 3cos)2  10  0    10 

1 [|maximum value of  – minimum value of |]3. 100

= 5.

1 1000 [| (10  0) |3 ]  1 1000 1000

Given 2x + 4y + z = 1, x + 2y + z = 2, x + y – z = 3, then one of the value of  such that the given system of equations has no solution, is_______.

Sol. Answer (0) Since, given system has no solution,   = 0 and any one amongst x, y, z is non-zero. 2 4 1 Let  2 1 1 1 

=0

 42 – 3 = 0  (4 – 3) = 0    0,

3 4

1 4 1 and  x  2 2 1  0 3 1 

6.

Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations: 3x – y – z = 0 –3x + z = 0 –3x + 2y + z = 0 Then the number of such points for which x2 + y2 + z2  100 is

[IIT-JEE 2009]

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250

Determinants

Solution of Assignment (Set-2)

Sol. Answer (7) We have 3x – y – z = 0

...(i)

–3x + z = 0

...(ii)

–3x + 2y + z = 0

...(iii)

Applying (i) and (ii) we get  y=0

...(iv)

Also 3x = z Points satisfying x2 + y2 + z2  100 with integral coordinates can be (0, 0, 0), (1, 0, 3), (2, 0, 6), (3, 0, 9), (–1, 0, –3), (–2, 0, –6), (–3, 0, –9) Hence 7 such points exist.

7.

Let  be the complex number cos z 1

2

z  2

1

1

z

2

2 2  i sin . Then the number of distinct complex numbers z satisfying 3 3

 0 is equal to

[IIT-JEE 2010]

Sol. Answer (1) ⎡1 ⎢ Let A  ⎢  ⎢ 2 ⎣⎢

 2 1

2 ⎤ ⎥ 1⎥ ⎥  ⎦⎥

⎡0 0 0 ⎤ A  ⎢⎢0 0 0 ⎥⎥ , Tr ( A)  0, | A |  0 ⎢⎣0 0 0 ⎥⎦ 2

 A3 = 0

z 1

2

z  2

1

2

1

z

 | A  zI |  0

 z3 = 0  z = 0, the number of z satisfying the given equation is 1. 8.

The number of all possible values of , where 0 <  < , for which the system of equations (y + z)cos3 = (xyz)sin3

x sin3 

2cos3 2sin3  y z

(xyz)sin3 = (y + 2z)cos3 + ysin3 have a solution (x0, y0, z0) with y0z0  0, is

[IIT-JEE 2010]

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Solution of Assignment (Set-2)

Determinants

251

Sol. Answer (3) Given equation can be written as

x sin3 

cos3 cos3  0 y z

...(i)

x sin3 

2cos3 2sin3  0 y z

...(ii)

x sin3 

2 1 cos3  (cos3  sin3)  0 y z

...(iii)

Equations (ii) and (iii) imply 2sin3 = cos3 + sin3  sin3 = cos3

9.

tan3  1 ⇒ 3 

or



 5 9 , , 4 4 4

 5 9 , , 12 12 12

Let k be a positive real number and let ⎡ 2k  1 2 k ⎢ A⎢2 k 1 ⎢ ⎣⎢ 2 k 2k

⎡ 0 2 k⎤ ⎥ ⎢ 2k ⎥ and B  ⎢1  2k ⎥ ⎢ 1 ⎦⎥ ⎢⎣  k

2k  1 0 2 k

k ⎤ ⎥ 2 k ⎥. ⎥ 0 ⎦⎥

If det (adj A) + det(adj B) = 106, then [k] is equal to

[IIT-JEE 2010]

[Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k] Sol. Answer (4) |A| = (2k + 1)3, |B| = 0  det (adj A)  det (adj B) = (2k + 1)6 = 106  k 

9 . 2

 [k] = 4 10. Let M be a 3 × 3 matrix satisfying

⎡0 ⎤ ⎡ 1⎤ ⎡1⎤ ⎡1⎤ ⎡1⎤ ⎡ 0 ⎤ M ⎢⎢ 1⎥⎥  ⎢⎢ 2 ⎥⎥ , M ⎢⎢ 1⎥⎥  ⎢⎢ 1 ⎥⎥ , and M ⎢⎢1⎥⎥  ⎢⎢ 0 ⎥⎥ ⎢⎣0 ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣1⎥⎦ ⎢⎣12⎥⎦ Then the sum of the diagonal entries of M is

[IIT-JEE 2011]

⎡ a11 Let M  ⎢a21 ⎢ ⎢⎣a31

a12 a22 a32

a13 ⎤ a23 ⎥⎥ be the given matrix. a33 ⎥⎦

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252

Determinants

Solution of Assignment (Set-2)

Using the given conditions, we have

⎡ a11 ⎢a ⎢ 21 ⎢⎣a31

a12 a22 a32

a13 ⎤ ⎡0 ⎤ ⎡ 1⎤ a23 ⎥⎥ ⎢⎢ 1⎥⎥  ⎢⎢ 2 ⎥⎥ a33 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 3 ⎥⎦

 a12 = –1 a22 = 2 a32 = 3

⎡1⎤ ⎡1⎤ Also, M ⎢ 1⎥  ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦  a11 – a12 = 1 a21 – a22 = 1 a31 – a32 = –1 Using above equations, we shall get a11 = 0

⎡1⎤ ⎡ 0 ⎤ Moreover, M ⎢1⎥  ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣1⎥⎦ ⎢⎣12⎥⎦  a11 + a12 + a13 = 0 a21 + a22 + a23 = 0 a31 + a32 + a33 = 12 Using above results, we get a33 = 7 Finally, the sum of elements of leading diagonals = a11 + a22 + a33 =0+2+7 =9

SECTION - G Multiple True-False Type Question 1.

⎡ 1 1⎤ The matrices which commute with A = ⎢ ⎥ in case of multiplication ⎣0 1⎦

STATEMENT-1 : Are always singular. STATEMENT-2 : Are always non-singular. STATEMENT-3 : Are always symmetric. (1) F F F

(2) T T F

(3)

TTT

(4)

TFT

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Solution of Assignment (Set-2)

Determinants

253

Sol. Answer (1) Let B  ⎡ x y ⎤ commute with A ⎢a b ⎥ ⎣ ⎦  AB = BA ⎡x  a y  b⎤ ⎡x x  y ⎤  i.e., ⎢ a b ⎥⎦ ⎢⎣ a a  b ⎥⎦ ⎣

 x + a = x, y + b = x + y, b = a + b  a = 0 and b = x 

⎡x y ⎤ B⎢ ⎥ ; option (1) is correct. ⎣0 x ⎦

SECTION - H Aakash Challengers Questions 1.

If the system of equation ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 has non-trivial solution then

find the value of

bc  a 2 ca  b 2 ab  c 2

ca  b 2 ab  c 2 bc  a 2

ab  c 2 bc  a 2 ca  b 2

.

Sol. Answer (0) For non-trivial solution

a b c A b c a 0 c a b And the determinant bc  a 2

ca  b 2

ab  c 2

ca  b 2

ab  c 2

bc  a 2

2

2

2

ab  c

bc  a

ca  b

is determinant of cofactor matrix of A.

Hence its value = 0

2.

If a, b, c are real such that a2 + b2 + c2 = 1 then show that

ax  by  c bx  ay cx  a bx  ay ax  by  c cy  b 0 cx  a cy  b ax  by  c

will represent a straight line. Sol. Applying C1  aC1 + bC2 + cC3 (a 2  b 2  c 2 ) x

bx  ay

  (a 2  b 2  c 2 )y ax  by  c (a 2  b 2  c 2 )

cy  b

cx  a cy  b ax  by  c

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254

Determinants

Solution of Assignment (Set-2)

bx  ay

x

cx  a

  y ax  by  c

cy  b

cy  b

1

∵ a2  b2  c 2  1

ax  by  c

C2  C2 – bC1, C3  C3 – cC1

x

ay

a

  y ax  c 1

b ax  by

cy

R3  xR1 + yR2 + R3 x 

ay

a

ax  c b

y x2  y 2 1

0

0

Expand w.r.t. R3 

= (x2 + y2 + 1) [aby + a2x + ac] = a(x2 + y2 + 1) (ax + by + c)

 3.

= 0  ax + by + c = 0 represents a straight line.

Prove that

(i)

(ii)

1+ a

b

c

a

1+ b

c

a

b

1+ c

1+ a

1

1

1

1+ b

1

1

1

1+ c

Sol. (i) Let  =

=1+a+b+c

1 1 1⎞ ⎛ = abc ⎜ 1 + + + ⎟ , where abc  0 a b c⎠ ⎝

b c b c 1 a 1 a  b  c a c c (Performing C1  C1 + C2 + C3)  1 a  b  c 1  b 1 b a b b 1 c 1 a  b  c 1 c

1 b c c = (1  a  b  c ) 1 1  b 1 b 1 c

⎛ Taking 1  a  b  c common ⎞ ⎜ ⎟ ⎝ from the 1st column ⎠

1 b c = (1  a  b  c ) 0 1 0 (Performing R2  R2 – R1, R3  R3 – R1) 0 0 1

= (1 + a + b + c) (1.1 – 0) = 1 + a + b + c, expanding along first row. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Determinants

255

(ii) It follows as a corollary of (i)

1 a 1 1   abc 1 1 b 1 = 1 1 1 c

1 1 b 1 c

1 a

1 a 1 1 c

1 b

1 a 1 b 1

⎛ Taking a, b, c common from ⎞ ⎜ ⎟ ⎝ R1, R2 , R3 respectively ⎠

1 c

1 1 1⎞ ⎛ = abc ⎜ 1    ⎟ , from (i) part. a b c⎠ ⎝

If some of a, b, c, say a, is zero, we rewrite the left hand side as abc + ab + bc + ca, and the determinant then evaluates to bc. 4.

Show that

2bc – a 2

c2

b2

bc – a 2

ca – b 2

ab – c 2

c2

2ac – b 2

a2

= ca – b 2

ab – c 2

bc – a 2

b2

a2

2ab – c 2

ab – c 2

bc – a 2

ca – b 2

Sol. A direct evaluation of the determinants involved is not at all attractive. One is required to think out of the box to discover what may be the idea behind the solution. Write LHS as product of two determinants 2bc – a 2 c2 b2

c2 2ac – b 2 a2

b2 – a 2  bc  bc – ab  ab  c 2 a2  – ab  c 2  ba – b 2  ac  ca 2 2ab – c – ac  ca  b 2 – bc  a 2  bc

– ac  b 2  ca – bc  bc  a 2 – c 2  ab  ab

a b c –a c b = b c a – b a c (Row-by-row multiplication) c a b –c b a a b c a b c a b c = b c a b c a  b c a c a b c a b c a b

2bc – a 2 c2 i.e. b2

c2 2ca – b 2 a2

b2 a b c a2  b c a c a b 2ab – c 2

bc – a 2 2 Now how to relate the determinant ca – b ab – c 2

2

2

…(A)

ca – b 2 ab – c 2 bc – a 2

2

ab – c 2 a b c 2 bc – a to the determinant b c a ? c a b ca – b 2

a b c Consider  = b c a c a b

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256

Determinants

Solution of Assignment (Set-2)

bc – a 2 2 1 = ca – b ab – c 2

ca – b 2 ab – c 2 bc – a 2

ab – c 2 bc – a 2 ca – b 2

and we know that 1 = 2 bc – a 2 2 Thus ca – b ab – c 2

ca – b 2 ab – c 2 bc – a 2

ab – c 2 a b c bc – a 2  b c a c a b ca – b 2

2

…(B)

From (A) and (B) we have 2bc – a 2 c2 b2

5.

Let

c2 2ca – b 2 a2

b2 bc – a 2 2 a  ca – b 2 2 2ab – c ab – c 2

ca – b 2 ab – c 2 bc – a 2

ab – c 2 bc – a 2 ca – b 2

x 2 + 3x

x –1

x +3

x +1

–2x

x – 4 = ax 4 + bx 3 + cx 2 + dx + k

x–3

x+4

3x

where a, b, c, d, k are independent of x. Find the value of d. Sol. Rather than expand the determinant, we differentiate both sides w.r.t. x and then put x = 0 to obtain the value of d. x 2  3x (x) = x  1 x3

x –1 x 3 –2 x x – 4  ax 4  bx 3  cx 2  dx  k x  4 3x

Differentiating w.r.t. x 2x  3 1 1 x 2  3x 1 ´(x) = x  1 –2 x x – 4  x – 3 x  4 3x x–3

x – 1 x  3 x 2  3x –2 1  x 1 1 x  4 3x

x –1 x 3 –2x x – 4 1 3

= 4ax3 + 3bx2 + 2cx + d Set x = 0 on both sides to obtain

d=

3 1 1 0 –1 3 0 –1 3 1 0 –4  1 –2 1  1 0 –4 –3 4 0 –3 4 0 1 1 3

= {3(0 + 16) – 1(0 – 12) + 1(4 – 0)} + {0 + 1(0 + 3) + 3(4 – 6)} + {0 + 1(3 + 4) + 3(1 – 0)} = (48 + 12 + 4) + (3 – 6) + (7 + 3) = 64 – 3 + 10 = 71 Hence the value of d is 71. Remarks : (1) If the value of k(the constant term) is to be found out, just put x = 0 in the determinant. (2) If the value of b is to found out, then divide both sides by x4 and write the polynomial in t, where t=

1 . Now differentiate, as in the example above, w.r.t. t to find the desired coefficient. x

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Solution of Assignment (Set-2)

1

2

0

3 , value of co-factor to its minor of element –3 is

3

0

0

6.

In the determinant 1 2

Determinants

(1) –1

(2)

0

(3)

1

(4)

257

2

Sol. Answer (1) Ratio of cofactor to its minor of element –3, which is in 3rd row and 2nd column = (–1)3+2 = –1

7.

1

5

loge e

5

5 is equal to

log10 10 5

e

(1)

(2)

e

(3)

1

(4)

0

(3)

a+b+c

(4)

0

(4)

–1

1 5

5 51 1

 5 0

1 5

e

e

1 1 1 1

x + 2 x +3 x + a

8.

If a, b, c are in A.P., then x + 4 x +5 x + b is x + 6 x +7 x + c

(1) x – (a + b + c)

9x2 + a + b + c

(2)

Sol. Answer (4) Apply C2  C2 – C1 x 2 1 x a

x 2 1 x a

R R  R

2 2 1 x  4 1 x  b  R R  R 

2

0 ba

x6 1 x c

4

0 c a

3

3

1

x 2 1 x a



2

0 b  a  1[2(c  a )  4(b  a )]

4

0 c a

= 2{2b – c – a}  = 0

9.

{∵ 2b = a + c}

1+ax

1+bx

1+cx

If 1+a1x

1+b1x

1+c1x = A0 + A1x + A2x2 + A 3x3, then A0 is

1+a2 x 1+b2 x 1+c2 x

(1) abc

(2)

0

(3)

1

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258

Determinants

Solution of Assignment (Set-2)

Sol. Answer (2) 1 1 1

Put x = 0 on both sides  A0  1 1 1  0 1 1 1

x 3 7

10. If –9 is a root of equation 2 x

2 = 0, then other two roots are

7 6 x

(1) 2, 7

(2)

–2, 7

(3)

2, –7

(4)

–2, –7

–1

(4)

0

Sol. Answer (1) x 3 7 2 x

1 1 1

2  0 ⇒ ( x  9) 2 x

7 6 x

2 0

7 6 x

[R1  R1 + R2 + R3]  (x + 9){x2 – 12 – (2x – 14) + (12 – 7x)} = 0  (x + 9)[(x – 2)(x – 7)] = 0  x = 2, 7 x

11. The coefficient of x in f (x )= 1 x

1+ sin x

cos x

log(1+x )

2

2

1+x

2

where

0

–1 < x  1, is (1) 1

(2)

–2

(3)

Sol. Answer (2) f(x) = x{–2(1 + x2)} – (1 + sinx)(–2x2) + cosx{1 + x2 – x2log(1 + x)} = –2x – 2x3 + 2x2 + 2x2sinx + cosx{1 + x2 – x2log(1 + x)} Hence, –2 12. If adjB = A, |P| = |Q| = 1, then adj(Q–1BP–1) equals (1) PQ

(2)

QAP

(3)

PAQ

(4)

PA–1Q

(3)

Nilpotent

(4)

Involutary

⎡ ⎢ 13. The matrix A = ⎢ ⎢ ⎢ ⎣

(1) Unitary

1 2 1 2

P Q A.  PAQ | P | |Q |

1 ⎤ ⎥ 2⎥ is 1 ⎥ ⎥ 2⎦

(2)

Orthogonal

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Solution of Assignment (Set-2)

Determinants

259

⎡a 2 ⎤ 14. The matrix A = ⎢ ⎥ is singular if ⎣2 4 ⎦

(1) a  1

(2)

a=1

(3)

a=0

(4)

a = –1

k = –1

(3)

k±1

(4)

k = 0

Any odd integer

(4)

Any integer

Sol. Answer (2) A is singular  |A| = 0  4a – 4 = 0  a = 1

15.

⎡1 0 k ⎤ ⎢ ⎥ A = ⎢ 2 1 3 ⎥ is invertible for ⎢k 0 1⎥ ⎣ ⎦

(1) k = 1

(2)

Sol. Answer (3) A is invertible  |A|  0  K  ±1 16. Let a, b, c be such that (b + 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

c

c  1 c +1

(  1)

n +2

=0

a (  1) b (  1) c n +1

n

then value of n is (1) 0

(2)

Any even integer

(3)

Sol. Answer (3) a 1 a 1

a

a 1 b 1 c 1

b b  1 b  1  ( 1) a  1 b  1 c  1 n

c 1 c 1

c a

b

a

a 1 a 1

c

a 1 a 1

a

 b b  1 b  1  ( 1) b  1 b  1 b n

c

c 1 c 1

a

a 1 a 1

c 1 c 1 a 1

a

c a 1

 b b  1 b  1  ( 1)n 1 b  1 b b  1 {C2  C3} c

c 1 c 1

a n 2

 [1  ( 1)

c 1

c

c 1

a 1 a 1

] b b  1 b  1  any odd integer c

c 1 c 1

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

260

Determinants

Solution of Assignment (Set-2)

x2 1 + x3

x 17. If y

y2

1 + y 3 = 0 and x, y, z are all distinct, then xyz equals

z

z2

1 + z3

(1) –1

(2)

1

(3)

0

(4)

3

x2 1 x3

y

y 2 1 y 3  0

z

z2

1  z3

x

x2 1

y

y 2 1  xyz 1 y

y2  0

z

z2 1

z2

1 x

 (1  xyz ) 1 y 1 z

1 x 1 z

x2

x2 y 2  0  xyz = –1 z2

  

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456