Q: If ω=1 is the complex cube root of unity and matrix H=[ω00ω], then H70 is equal to

Exp. (a) H=[ω00ω]∴ H2=[ω00ω][ω00ω] ⇒ Hk=[ωk00ωk]Then, Hk+1=[ωk+100ωk+1]H70=[ω7000ω70][ω00ω] =H

Q: The number of 3×3 non-singular matrices, with four entries as 1 and all other entries as 0 , is (a) less than 4 (b) 5 (c) 6 (d) atleast 7

Exp. (d)Consider ⎝⎛1∗∗∗1∗∗∗1⎠⎞. By placing 1 in anyone of the 6∗ position and 0 elsewhere, we get 6 non-singular matrices.Similarly, ⎝⎛∗∗1∗1∗1∗∗⎠⎞ gives atleast one non-singular matrix.Hence, we get atleast 7 non-singular matrix.

Q: Let A be 2×2 matrix with real entries. Let I be the 2×2 identity matrix. Denote by tr(A), the sum of diagonal entries of A. Assume that A2=I. Statement I. If A=I and A=−I, then det(A)=−1. Statement II. If A=I and A=−I, then tr(A)=0. (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

Exp. (d) Let A=[acbd] ∴[acbd]2=[1001][∵A2=I] ⇒[a2+bcac+cdab+bdbc+d2]=[1001] ⇒b(a+d)=0, c(a+d)=0 anda2+bc=1, bc+d2=1 ⇒a=1,d=−1,b=c=0 If A=[100−1], then A2=[100−1][100−1]=[1001]=IA=I,A=−I det(A)=−1 (Statement I is true) Statement II, tr(A)=1−1=0, Statement II is false.

Q: Let A be a square matrix all of whose entries are integers. Then, which one of the following is true? (a) If det(A)=±1, then A−1 need not exist (b) If det(A)=±1, then A−1 exists but all its entries are not necessarily integers (c) If det(A)=±1, then A−1 exists and all its entries are non-integers (d) If det(A)=±1, then A−1 exists and all its entries are integers

Exp. (d) As det(A)=±1,A−1 exists. And A−1=det(A)1(adjA)=±(adjA) All entries in adj (A) are integers. Hence, A−1 has integer entries.

Q: If D=∣∣11111+x1111+y∣∣ for x=0,y=0, then D is

Given that, D=∣∣11111+x1111+y∣∣Applying C2→C2−C1 and C3→C3−C1,=∣∣1110x000y∣∣=xyHence, D is divisible by both x and y.

Q: Let A=[1324] and B=[a00b];a,b∈N. Then, (a) there exists more than one but finite number of B 's such that AB=BA (b) there exists exactly one B such that AB=BA (c) there exist infinitely many B 's such that AB=BA (d) there cannot exist any B such that AB=BA

Exp. (c) Given that, A=[1324] and B=[a00b] Now, AB=[1324][a00b]=[a3a2b4b] and BA=[a00b][1324]=[a3b2a4b] If AB=BA, then a=b. Hence, AB=BA is possible for infinitely many values of B′ s.

Question 1

If ω=1 is the complex cube root of unity and matrix H=[ω0​0ω​], then H70 is equal to

Accepted Answer

Exp. (a) H=[ω0​0ω​]∴ H2=[ω0​0ω​][ω0​0ω​]  ⇒ Hk=[ωk0​0ωk​]Then, Hk+1=[ωk+10​0ωk+1​]H70=[ω700​0ω70​][ω0​0ω​] =H

Question 2

Consider the system of linear equations
x1​+2x2​+x3​=3,
2x1​+3x2​+x3​=3
and 3x1​+5x2​+2x3​=1
The system has
(a) an infinite number of solutions
(b) exactly 3 solutions
(c) a unique solution
(d) no solution

Accepted Answer

Exp. (d)
The given system of linear equations can be put in the matrix form as
⎣⎡​123​235​112​⎦⎤​⎣⎡​x1​x2​x3​​⎦⎤​=⎣⎡​331​⎦⎤​
⇒⎣⎡​100​2−1−1​1−1−1​⎦⎤​⎣⎡​x1​x2​x3​​⎦⎤​=⎣⎡​3−3−8​⎦⎤​[ by R2​→R2​−2R1​R3​→R3​−3R1​​]
⇒⎣⎡​100​210​110​⎦⎤​⎣⎡​x1​x2​x3​​⎦⎤​=⎣⎡​335​⎦⎤​
Clearly, the given system of equations has no solution.
Alternate Solution
Subtracting the addition of first two equations from
third equation, we get
0=−5, which is an absurd result.
Hence, the given system of equations has no solution.

Question 3

The number of 3×3 non-singular matrices, with four entries as 1 and all other entries as 0 , is
(a) less than 4
(b) 5
(c) 6
(d) atleast 7

Accepted Answer

Exp. (d)Consider ⎝⎛​1∗∗​∗1∗​∗∗1​⎠⎞​. By placing 1 in anyone of the 6∗ position and 0 elsewhere, we get 6 non-singular matrices.Similarly, ⎝⎛​∗∗1​∗1∗​1∗∗​⎠⎞​ gives atleast one non-singular matrix.Hence, we get atleast 7 non-singular matrix.

Question 4

Let A be 2×2 matrix.
Statement I. adj (adjA)=A
Statement II. ∣adjA∣=A
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
(c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I
(d) Statement I is true, Statement II is false

Accepted Answer

Exp. (c)

∣adjA∣=∣A∣n−1=∣A∣2−1=∣A∣
adj (adjA)=∣A∣n−2A=∣A∣0A=A
Hence, both the statements are true but Statement II is not a correct explanation of Statement I.

Question 5

Let A be 2×2 matrix with real entries. Let I be the 2×2 identity matrix. Denote by tr(A), the sum of diagonal entries of A. Assume that A2=I.
Statement I. If A=I and A=−I, then det(A)=−1.
Statement II. If A=I and A=−I, then tr(A)=0.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
(c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I
(d) Statement I is true, Statement II is false

Accepted Answer

Exp. (d)
Let  A=[ac​bd​]
∴[ac​bd​]2=[10​01​][∵A2=I]
⇒[a2+bcac+cd​ab+bdbc+d2​]=[10​01​]

⇒b(a+d)=0,   c(a+d)=0
anda2+bc=1,  bc+d2=1
⇒a=1,d=−1,b=c=0
If A=[10​0−1​], then
A2=[10​0−1​][10​0−1​]=[10​01​]=IA=I,A=−I
det(A)=−1 (Statement I is true)
Statement II, tr(A)=1−1=0, Statement II is false.

Question 6

Let A be a square matrix all of whose entries are integers. Then, which one of the following is true?
(a) If det(A)=±1, then A−1 need not exist
(b) If det(A)=±1, then A−1 exists but all its entries are not necessarily integers
(c) If det(A)=±1, then A−1 exists and all its entries are non-integers
(d) If det(A)=±1, then A−1 exists and all its entries are integers

Accepted Answer

Exp. (d)
As det(A)=±1,A−1 exists.
And A−1=det(A)1​(adjA)=±(adjA)
All entries in adj (A) are integers.
Hence, A−1 has integer entries.

Question 7

If D=∣∣​111​11+x1​111+y​∣∣​ for x=0,y=0, then D is

Accepted Answer

Given that, D=∣∣​111​11+x1​111+y​∣∣​Applying C2​→C2​−C1​ and C3​→C3​−C1​,=∣∣​111​0x0​00y​∣∣​=xyHence, D is divisible by both x and y.

Question 8

If A and B are square matrices of size n×n such that A2−B2=(A−B)(A+B), then which of the following will be always true?
(a) AB=BA
(b) Either of A or B is a zero matrix
(c) Either of A or B is an identity matrix
(d) A=B

Accepted Answer

Exp. (a)
 Since, A2−B2=(A−B)(A+B)∴A2−B2=A2−B2+AB−BA⇒AB=BA​

Question 9

Let A=[13​24​] and B=[a0​0b​];a,b∈N. Then,
(a) there exists more than one but finite number of B 's such that AB=BA
(b) there exists exactly one B such that AB=BA
(c) there exist infinitely many B 's such that AB=BA
(d) there cannot exist any B such that AB=BA

Accepted Answer

Exp. (c)
Given that, A=[13​24​]
and B=[a0​0b​]
Now, AB=[13​24​][a0​0b​]=[a3a​2b4b​]
and BA=[a0​0b​][13​24​]=[a3b​2a4b​]
If AB=BA, then a=b.
Hence, AB=BA is possible for infinitely many values of B′ s.

Question 10

If A2−A+I=O, then the inverse of A is

(a) I−A
(b) A−I
(c) A
(d) A+I

Accepted Answer

Exp. (a)
∵ A2−A+I=O 
⇒  A−1A2−A−1A+A−1I=O
⇒ (A−1A)A−(A−1A)+A−1=O
⇒ A−I+A−1=O
⇒ A−1=I−A

Question 11

If a2+b2+c2=−2 and f(x)=∣∣​1+a2x(1+a2)x(1+a2)x​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​∣∣​, then f(x) is a polynomial of degree

Accepted Answer

Given that,f(x)=∣∣​1+a2x(1+a2)x(1+a2)x​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​∣∣​Applying C1​→C1​+C2​+C3​, we getf(x)=∣∣​1+a2x+x+b2x+x+c2xx+a2x+1+b2x+x+c2xx+a2x+x+b2x+1+c2x​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​∣∣​ =∣∣​1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​∣∣​, given =∣∣​111​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​∣∣​Applying [∵a2+b2+c2=−2, we getR1​→R1​−R3​,R2​→R2​−R3​Hence, =∣∣​001​01−x(1+b2)x​x−1x−11+c2x​∣∣​=∣∣​01−x​x−1x−1​∣∣​ is of degree =(x−1)2

Question 12

A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then
(a) m=n=68
(b) m+n=68
(c) m=n=78
(d) n=m−8

Accepted Answer

Exp. (c)
Since there are 8 males and 5 females. Out of these 13 members committee of 11 members is to be formed.

According to the question, m= number of ways when there is at least 6 males
=(8C6​×5C5​)+(8C7​×4​)+(8C8​×5C3​) =(28×1)+(8×5)+(1×10)
=28+40+10=78
and n= nurnber of ways wher there is al least 3 females
=(5C3​×8C8​)+(5C4​×8C7​)+(5C5​×8C6​)
=10×1+5×8+1×28=78
So, m=n=78

Question 13

The number of 6 digits numbers that can be formed using the digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is
(a) 60
(b) 72
(c) 48
(d) 36

Accepted Answer

Exp. (a)

Since, the sum of given digits
0+1+2+5+7+9=24
Let the six-digit number be abcdef and to be divisible by 11 , so the difference of sum of odd placed digits and sum of even placed digits should be either 0 or a multiple of 11 means ∣(a+c+e)−(b+d+f)∣ should be either 0 or a multiple of 11.
Hence, possible case is
a+c+e=12=b+d+f (only) 
Now, 
Case I
set {a,c,e}={0,5,7} and set {b,d,f}={1,2,9}
So, number of 6 -digits numbers
=(2×2!)×(3!)=24
[∵ a can be selected in ways only either 5 or 7].
Case II
Set {a,c,e}={1,2,9} and set {b,d,f}={0,5,7}
So, number of 6-digits numbers =3!×3!=36
So, total number of 6-digits numbers
=24+36=60

Question 14

Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is
(a) 180
(b) 210
(c) 170
(d) 190

Accepted Answer

Exp.(c)
It is given that, there are 20 pillars of the same height have been erected along the boundary of a circular stadium.
Now, the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then total number of beams = number of diagonals of 20 -sided polygon.
∵20C2​ is selection of any two vertices of 20 -sided polygon which included the sides as well.
So, required number of total beams =20C2​−20
[∵ the number of diagonals in a n-sided closed polygon =nC2​−n]
=220×19​−20
=190−20=170

Question 15

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is
(a) 220−1
(b) 221
(c) 220
(d) 220+1

Accepted Answer

Exp.(c)
Given that, out of 31 objects 10 are identical and remaining 21 are distinct, so in following ways, we can choose 10 objects.
0 identical +10 distincts, number of ways
=1×21C10​
1 identical +9 distincts, number of ways
=1×21C9​
2 identicals +8 distincts, number of ways
=1×21C8​
10 identicals +0 distinct, number of ways
=1×21C0​
So, total number of ways in which we can choose 10 objects is
21C10​+21C9​+21C8​+…+21C0​=x (let) ... (i) 
⇒21C11​+21C12​+21C13​+…+21C21​=x… (ii) 
[∵nCr​=nCn−r​]
{ On adding both Eqs. (i) and (ii), we get }
2x=21C0​+21C1​+21C2​+…+21C10​+21C11​+21C12​+…+21C21​
⇒2x=221⇒x=220

Question 16

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750 , then n is equal to
(a) 28
(b) 27
(c) 25
(d) 24

Accepted Answer

Exp. (c)
It is given that a group of students comprises of 5 boys and n girls. The number of ways, in which a team of 3 students can be selected from this group such that each team consists of at least one boy and at least one girls, is = (number of ways selecting one boy and 2 girls) + (number of ways selecting two boys and 1 girl)

=(5C1​×nC2​)(5C2​×nC1​)=1750
⇒ (5×2n(n−1)​)+ (25×4​×n)=1750
 ⇒n(n−1)+4n=52​×1750
⇒n2+3n=2×350
⇒n2+3n−700=0
⇒n2+28n−25n−700=0
⇒n(n+28)−25(n+28)=0
⇒n=25, ∵n∈N

Question 17

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is
(a) 350
(b) 500
(c) 200
(d) 300

Accepted Answer

Exp. (d)
Number of girls in the class =5 and number of boys in the class =7

Now, total ways of forming a team of 3 boys and 2 girls =7C3​⋅5C2​=350

But, if two specific boys are in team, then number of ways =5C1​⋅5C2​=50

Required ways, i.e. the ways in which two specific boys are not in the same team =350−50=300.

Alternate Method-
 Number of ways when A is selected and B is not =5C2​⋅5C2​=100
Number of ways when B is selected and A is not =5C2​⋅5C2​=100

Number of ways when both A and B are not selected =5C3​⋅5C2​=100
∴ Required ways =100+100+100=300.

Question 18

Consider three boxes, each containing 10 balls labelled 1,2,…,10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni​, the label of the ball drawn from the i th box, (i=1,2,3). Then, the number of ways in which the balls can be chosen such that n1​<n2​<n3​ is
(a) 82
(b) 120
(c) 240
(d) 164

Accepted Answer

Exp. (b)
Given there are three boxes, each containing 10 balls labelled 1, 2, 3, ..., 10 .

Now, one ball is randomly drawn from each boxes, and ni​ denote the label of the ball drawn from the ith box, (i=1,2,3).

Then, the number of ways in which the balls can be chosen such that n1​<n2​<n3​ is same as selection of 3 different numbers from numbers {1,2,3,…, 10}=10C3​=120

Question 19

There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is 
(a) 12
(b) 11
(c) 9
(d) 7

Accepted Answer

Exp. (a)
Since, there are m-men and 2-women and each participant plays two games with every other participant.
∴ Number of games played by the men between themselves =2×mC2​
and the number of games played between the men and the women =2×mC1​×2C1​
Now, according to the question,
2mC2​=2mC1​2C1​+84
⇒2!(m−2)!m!​=m×2+42
⇒m(m−1)=4m+84
⇒m2−m=4m+84
⇒m2−5m−84=0
⇒m2−12m+7m−84=0
⇒m(m−12)+7(m−12)=0
⇒m=12[∵m>0]

Question 20

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf, so that the dictionary is always in the middle.
The number of such arrangements is
(a) at least 1000
(b) less than 500
(c) at least 500 but less than 750
(d) at least 750 but less than 1000

Accepted Answer

Exp. (a)
Given 6 different novels and 3 different dictionaries.
Number of ways of selecting 4 novels from 6 novels is 6C4​=2!4!6!​=15
Number of ways of selecting 1 dictionary is from 3 dictionaries is 3C1​=1!2!3!​=3
∴ Total number of arrangement of 4 novels and 1 dictionary where dictionary is always in the middle, is
15×3×4!=45×24=1080

Filters

Filters

book

topic

difficulty

Solutions for Arihant JEE Main Chapterwise Solutions Mathematics (2019-2002)

Found the solution, but did not understand the concept?

Ask my first question