JEE Mathematics Determinants and Cramers Rule MCQs Set 03

Answer: (a) 0

Explanation:
1. Apply \( R_1\to R_1-R_2 \): the first row becomes \( (2x,0,0) \).
2. Expanding along it: \( 2x\cdot(a\cdot(-x)-x\cdot a)=-4ax^2=0 \), so x = 0 (taking a ≠ 0).

Teacher's Note:
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99. .

Answer: (b) 0

Explanation:
1. The matrix is skew-symmetric (\( a_{ji}=-a_{ij} \)) and has odd order 3.
2. The determinant of an odd-order skew-symmetric matrix is 0.

Teacher's Note:
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79. .

Answer: (c) abc = 1

Explanation:
1. Split the first row: the determinant \( =\begin{vmatrix}a^3&b^3&c^3\\a&b&c\\a^2&b^2&c^2\end{vmatrix}-\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} \).
2. The first is \( abc\cdot\begin{vmatrix}a^2&b^2&c^2\\1&1&1\\a&b&c\end{vmatrix}=abc\,V \) (two row swaps are an even permutation) and the second is the Vandermonde V.
3. So \( V(abc-1)=0 \) and since a, b, c are distinct V ≠ 0, abc = 1.

Teacher's Note:
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95. .

Answer: (a) x²(x + 3)

Explanation:
1. Add columns 2 and 3 to column 1: the first column becomes \( (x+3) \) in every row.
2. Take it out: \( (x+3)\begin{vmatrix}1&1&1\\1&1+x&1\\1&1&1+x\end{vmatrix} \).
3. Subtracting row 1 from rows 2 and 3 gives an upper triangular determinant with diagonal 1, x, x: value \( x^2(x+3) \).

Teacher's Note:
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74. .

Answer: (d) x = 0, y = 0

Explanation:
1. Expand along the first row: \( 6i(3i\cdot i+3)-(-3i)(4i+20)+1\cdot(12-60i) \).
2. First term: \( 6i(-3+3)=0 \). Second term: \( 3i(4i+20)=-12+60i \). Third term: \( 12-60i \).
3. Sum \( =0 \), so x = 0 and y = 0.

Teacher's Note:
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89. .

Answer: (b) x, y, z are in G.P.

Explanation:
1. Apply \( C_1\to C_1-pC_2-C_3 \): the first column becomes \( (0,\ 0,\ -(p^2x+2py+z)) \).
2. Expanding along it: determinant \( =-(p^2x+2py+z)(xz-y^2) \).
3. For this to vanish for all p the p-free factor must vanish: \( y^2=xz \), so x, y, z are in G.P.

Teacher's Note:
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118. .

Answer: (c) d = 0 or a + d = 0

Explanation:
1. Expanding the determinant (or checking symbolically) gives \( 6d^3(a+d) \).
2. It vanishes when d = 0 or a + d = 0.

Teacher's Note:
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Answer: (c) 0

Explanation:
1. Divide the columns by a, b, c respectively (taking out abc overall): first row becomes \( abc\left(\frac1a,\frac1b,\frac1c\right) \) times a scalar, precisely \( \begin{vmatrix}bc&ca&ab\\p&q&r\\1&1&1\end{vmatrix}=abc\begin{vmatrix}\frac1a&\frac1b&\frac1c\\p&q&r\\1&1&1\end{vmatrix} \).
2. Since a, b, c are terms of a H.P., 1/a, 1/b, 1/c are the pth, qth, rth terms of an A.P.: \( \frac1a=A+(p-1)D \) and so on.
3. So the first row equals \( (A-D)\cdot(1,1,1)+D\cdot(p,q,r) \), a combination of rows 2 and 3. The determinant is 0.

Teacher's Note:
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100. .

Answer: (b) −5

Explanation:
1. Expanding along the first row: \( x(x^2-12)-2(3x-15)+5(12-5x)=x^3-43x+90 \), which factorises as \( (x-5)(x^2+5x-18) \).
2. x = 5 is the integral root. The other two satisfy \( x^2+5x-18=0 \) with discriminant 97, not a perfect square, so they are non-integral.
3. Their sum is −5.

Teacher's Note:
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85. .

Answer: (c) 0

Explanation:
1. The number x51 is \( 100x+51 \) etc. Apply \( R_2\to R_2-100R_3 \): the second row becomes (51, 41, 31).
2. Then \( R_2\to R_2-10R_1 \) gives (1, 1, 1).
3. Expand \( \begin{vmatrix}5&4&3\\1&1&1\\x&y&z\end{vmatrix} \) along the first row: \( 5(z-y)-4(z-x)+3(y-x)=x-2y+z \).
4. Since x, y, z are in A.P., \( x+z=2y \), so the value is 0.

Teacher's Note:
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102. .

Answer: (a) \( \Delta_1+\Delta_2=0 \)

Explanation:
1. Write \( bc=(ab+bc+ca)-a(a+b+c)+a^2 \) and similarly for the other rows. So column 2 of \( \Delta_2 \) is \( s\cdot C_1-S\cdot C_3+(a^2,b^2,c^2)^T \), with s = ab + bc + ca and S = a + b + c.
2. Removing the multiples of columns 1 and 3 leaves \( \Delta_2=\begin{vmatrix}1&a^2&a\\1&b^2&b\\1&c^2&c\end{vmatrix} \).
3. Interchanging columns 2 and 3 and transposing gives \( \Delta_2=-\Delta_1 \), so \( \Delta_1+\Delta_2=0 \).

Teacher's Note:
1. R
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128. .

Answer: (c) equal in area

Explanation:
1. The left determinant equals twice the signed area of the first triangle.
2. The right determinant is the transpose of \( \begin{vmatrix}1&b_1&a_1\\1&b_2&a_2\\1&b_3&a_3\end{vmatrix} \), which is twice the signed area of the triangle with vertices \( (b_i,a_i) \) (up to a column rearrangement).
3. Equal determinants therefore give equal areas. Nothing is implied about shape.

Teacher's Note:
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Answer: (b) \( \lambda^3 \)

Explanation:
1. Here \( \alpha+\beta=-1 \) and \( \alpha\beta=1 \). Add rows 2 and 3 to row 1: the first row becomes \( (\lambda,\lambda,\lambda) \).
2. Take out λ, then subtract column 1 from columns 2 and 3: \( \lambda\,[(\lambda+\beta-\alpha)(\lambda+\alpha-\beta)-(1-\alpha)(1-\beta)] \).
3. This is \( \lambda[\lambda^2-(\alpha-\beta)^2-(1-\alpha-\beta+\alpha\beta)]=\lambda[\lambda^2-(1-4)-3]=\lambda^3 \).

Teacher's Note:
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2. d
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Answer: (c) 5

Explanation:
1. By Pascal's rule \( {}^{10}C_4+{}^{10}C_5={}^{11}C_5 \), \( {}^{11}C_6+{}^{11}C_7={}^{12}C_7 \), \( {}^{12}C_8+{}^{12}C_9={}^{13}C_9 \).
2. For m = 5 the third column is \( ({}^{11}C_5,\ {}^{12}C_7,\ {}^{13}C_9) \), exactly the sum of the first two columns.
3. So the determinant is 0 for m = 5.

Teacher's Note:
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26. f
27. o
28. r
29.
30. m
31.
32. =
33.
34. 5
35.
36. t
37. h
38. e
39.
40. t
41. h
42. i
43. r
44. d
45.
46. c
47. o
48. l
49. u
50. m
51. n
52.
53. i
54. s
55.
56. t
57. h
58. e
59.
60. s
61. u
62. m
63.
64. o
65. f
66.
67. t
68. h
69. e
70.
71. f
72. i
73. r
74. s
75. t
76.
77. t
78. w
79. o
80. .

Answer: (a) 0

Explanation:
1. Change every logarithm to natural logs. Row 1 is \( \frac{1}{\ln x}(\ln x,\ln y,\ln z) \), row 2 is \( \frac{1}{\ln y}(\ln x,\ln y,\ln z) \), row 3 is \( \frac{1}{\ln z}(\ln x,\ln y,\ln z) \).
2. All rows are multiples of the same row, so the determinant is 0.

Teacher's Note:
1. A
2. f
3. t
4. e
5. r
6.
7. c
8. h
9. a
10. n
11. g
12. i
13. n
14. g
15.
16. t
17. h
18. e
19.
20. b
21. a
22. s
23. e
24. ,
25.
26. e
27. v
28. e
29. r
30. y
31.
32. r
33. o
34. w
35.
36. i
37. s
38.
39. p
40. r
41. o
42. p
43. o
44. r
45. t
46. i
47. o
48. n
49. a
50. l
51.
52. t
53. o
54.
55. (
56. l
57. n
58.
59. x
60. ,
61.
62. l
63. n
64.
65. y
66. ,
67.
68. l
69. n
70.
71. z
72. )
73. .

Answer: (a) 0

Explanation:
1. For any p, \( (p+p^{-1})^2-(p-p^{-1})^2=4 \).
2. So \( R_2-R_3=(4,4,4)=4R_1 \), the rows are dependent, and the determinant is 0.

Teacher's Note:
1. T
2. h
3. e
4.
5. d
6. i
7. f
8. f
9. e
10. r
11. e
12. n
13. c
14. e
15.
16. o
17. f
18.
19. t
20. h
21. e
22.
23. s
24. e
25. c
26. o
27. n
28. d
29.
30. a
31. n
32. d
33.
34. t
35. h
36. i
37. r
38. d
39.
40. r
41. o
42. w
43. s
44.
45. i
46. s
47.
48. a
49.
50. m
51. u
52. l
53. t
54. i
55. p
56. l
57. e
58.
59. o
60. f
61.
62. t
63. h
64. e
65.
66. f
67. i
68. r
69. s
70. t
71.
72. r
73. o
74. w
75. .

Answer: (a) 0

Explanation:
1. Expanding along the first row: \( \cos C(0+\tan A\sin B)-\tan A(\sin B\cos C-0) \).
2. This is \( \tan A\sin B\cos C-\tan A\sin B\cos C=0 \).

Teacher's Note:
1. T
2. h
3. e
4.
5. t
6. w
7. o
8.
9. t
10. e
11. r
12. m
13. s
14.
15. o
16. f
17.
18. t
19. h
20. e
21.
22. e
23. x
24. p
25. a
26. n
27. s
28. i
29. o
30. n
31.
32. c
33. a
34. n
35. c
36. e
37. l
38.
39. e
40. x
41. a
42. c
43. t
44. l
45. y
46. ,
47.
48. f
49. o
50. r
51.
52. a
53. n
54. y
55.
56. a
57. n
58. g
59. l
60. e
61. s
62. .

Answer: (c) 0

Explanation:
1. Write \( x^2-yz=x(x+y+z)-(xy+yz+zx) \) and similarly for the other two rows.
2. So column 2 equals \( (x+y+z)\cdot C_1-(xy+yz+zx)\cdot C_3 \), a combination of columns 1 and 3.
3. Hence the determinant is 0.

Teacher's Note:
1. E
2. x
3. p
4. r
5. e
6. s
7. s
8.
9. t
10. h
11. e
12.
13. m
14. i
15. d
16. d
17. l
18. e
19.
20. c
21. o
22. l
23. u
24. m
25. n
26.
27. t
28. h
29. r
30. o
31. u
32. g
33. h
34.
35. t
36. h
37. e
38.
39. f
40. i
41. r
42. s
43. t
44.
45. a
46. n
47. d
48.
49. t
50. h
51. i
52. r
53. d
54.
55. c
56. o
57. l
58. u
59. m
60. n
61. s
62. .
63.
64. A
65.
66. d
67. e
68. p
69. e
70. n
71. d
72. e
73. n
74. t
75.
76. c
77. o
78. l
79. u
80. m
81. n
82.
83. g
84. i
85. v
86. e
87. s
88.
89. z
90. e
91. r
92. o
93. .

Answer: (a) 2

Explanation:
1. Let \( S=x+y+z \). The first equation is \( a(S-x)=x \), so \( x=\frac{aS}{1+a} \); similarly \( y=\frac{bS}{1+b} \) and \( z=\frac{cS}{1+c} \).
2. Adding: \( S=S\left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\right) \), and S ≠ 0 for a non-trivial solution.
3. So \( \sum\left(1-\frac1{1+a}\right)=1 \), which gives \( \sum\frac1{1+a}=3-1=2 \).

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. s
10. u
11. m
12.
13. S
14.
15. t
16. o
17.
18. e
19. x
20. p
21. r
22. e
23. s
24. s
25.
26. e
27. a
28. c
29. h
30.
31. u
32. n
33. k
34. n
35. o
36. w
37. n
38. .
39.
40. A
41.
42. n
43. o
44. n
45. -
46. t
47. r
48. i
49. v
50. i
51. a
52. l
53.
54. s
55. o
56. l
57. u
58. t
59. i
60. o
61. n
62.
63. h
64. a
65. s
66.
67. S
68.
69. ≠
70.
71. 0
72. ,
73.
74. w
75. h
76. i
77. c
78. h
79.
80. g
81. i
82. v
83. e
84. s
85.
86. t
87. h
88. e
89.
90. r
91. e
92. q
93. u
94. i
95. r
96. e
97. d
98.
99. r
100. e
101. l
102. a
103. t
104. i
105. o
106. n
107. .

Answer: (a) A.P.

Explanation:
1. Non-trivial solutions need \( \begin{vmatrix}a&4&1\\b&3&1\\c&2&1\end{vmatrix}=0 \).
2. Expanding along the first row: \( a(3-2)-4(b-c)+1(2b-3c)=a-2b+c \).
3. So \( a+c=2b \) and a, b, c are in A.P.

Teacher's Note:
1. S
2. e
3. t
4.
5. t
6. h
7. e
8.
9. c
10. o
11. e
12. f
13. f
14. i
15. c
16. i
17. e
18. n
19. t
20.
21. d
22. e
23. t
24. e
25. r
26. m
27. i
28. n
29. a
30. n
31. t
32.
33. t
34. o
35.
36. z
37. e
38. r
39. o
40. .
41.
42. T
43. h
44. e
45.
46. c
47. o
48. n
49. d
50. i
51. t
52. i
53. o
54. n
55.
56. a
57.
58. +
59.
60. c
61.
62. =
63.
64. 2
65. b
66.
67. i
68. s
69.
70. t
71. h
72. e
73.
74. A
75. .
76. P
77. .
78.
79. p
80. r
81. o
82. p
83. e
84. r
85. t
86. y
87. .