datavorous/entrance-exam-dataset huggingface dataset:
Let $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Assertion (A) : The identity
$|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2=2|\overrightarrow{\mathbf{a}}|^2$ holds for
$\vec{a}$.
Reason (R) : $\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}=a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}$,
$\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}=a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}, \overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}=a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}$
Which of the following is correct?

Options:
A Both $(\mathrm{A})$ and $(\mathrm{R})$ are true and $(\mathrm{R})$ is the correct reason for (A)
B Both $(A)$ and $(R)$ are true but $(R)$ is not the correct reason for (A)
C (A) is true, (R) is false
D (A) is false, (R) is true

The correct answer is:
Both $(\mathrm{A})$ and $(\mathrm{R})$ are true and $(\mathrm{R})$ is the correct reason for (A)

Solution:
Given,
$\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Now, $|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2=(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}) \cdot(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}})$
$\begin{aligned} & =\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \cdot\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \\ & =a_3^2+a_2^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2 & =\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \cdot\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \\ & =a_1^2+a_3^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 & =\left(a_1 \hat{\mathbf{i}}+a_1 \hat{\mathbf{j}}\right) \cdot\left(a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}\right) \\ & =a_2{ }^2+a_1{ }^2\end{aligned}$
Now,
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2 & +|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 \\ & =2\left(a_1^2+a_2^2+a_3^2\right) \\ & =2|\overrightarrow{\mathbf{a}}|^2\end{aligned}$
Hence, both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is correct reason for $\mathrm{A}$.

Topic: Mathematics, Vector Algebra, JEE Main
Let A=a355-1323-4 and B=b144c1-31d. If the trace of A is -4 and AB=-1017-3102528-83 then a+b+c+d=

Options:
A 7
B -1
C 3
D 1

The correct answer is:
3

Solution:
Given:

A=a355-1323-4

B=b144c1-31d

Since, the trace of A is -4, so

a-1-4=-4

⇒a=1

Also,

AB=-1017-3102528-83

⇒1355-1323-4b144c1-31d=-1017-3102528-83

⇒b-36+3c7+5d5b-138-c19+3d2b+243c-211-4d=-1017-3102528-83

∴b-3=-1⇒b=2

6+3c=0⇒c=-2

7+5d=17⇒d=2

Hence, a+b+c+d=1+2-2+2=3

Topic: Mathematics, Matrices, JEE Main
Let Aa, b, B3, 4 and −6, −8 respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P2a+3,7b+5 from the line 2x+3y−4=0 measured parallel to the line x−2y−1=0 is

Options:
A 1557
B 1756
C 1757
D 517

The correct answer is:
1757

Solution:
Given: Aa,b, B3,4, C−6,−8 denotes centroid, circumcentre and orthocentre respectively.

We know that, the line joining orthocentre and circumcentre is divided by centroid in the ratio 2:1, where distance from orthocentre being the larger part.

⇒a=2×3+1-62+1, b=2×4+1-82+1

⇒a=0, b=0

⇒P≡2×0+3, 7×0+5

⇒P≡3, 5

Distance from P measured along x−2y−1=0

⇒x=3+rcosθ, y=5+rsinθ

Where, tanθ=12

So, the distance of point from 2x+3y−4=0 will be,

23+rcosθ+35+rsinθ−4=0

⇒r2cosθ+3sinθ=−17

⇒r=−172·25+3·15

⇒r=−1757

⇒r=1757

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let A=abca âa,b,c,â0,1,2. If A is a singular matrix, then the number of possible matrices A are

Options:
A 18
B 27
C 7
D 3

The correct answer is:
7

Solution:
A=a2-bc=0 (for a singular matrix)
âa2=bc
7 cases a=0,b=0,c=0,1,2a=0,b=1,2,c=0a=1,b=1,c=1a=2b=2c=2
Hence, total number of possible matrices =7

Topic: Mathematics, Determinants, JEE Main
Let A=abcd and B=αβ≠00 such that AB=B and a+d=2021, then the value of ad-bc is equal to ______ .

Options:

The correct answer is:
2020

Solution:
Given A=abcd, B=αβ

And, AB=B

⇒(A-I)B=O

⇒(A-I)B=0

⇒(A-I)×B=0

⇒|A-I|=0, since B≠0

⇒(a-1)bc(d-1)=0

⇒a-1d-1-bc=0

⇒ad-a-d+1-bc=0

⇒ad-bc=a+d-1

⇒ad-bc=2021-1=2020.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let AA denote the set of all 4-digit natural numbers with no digit being 0. Let B⊂A consist of all numbers x such that no permutation of the digits of x gives a number that is divisible by 4. Then the probability of drawing a number from B with all even digits is

Options:
A 6251641
B 16641
C 161641
D 10001641

The correct answer is:
161641

Solution:
All even digit numbers in B=2,4,6,8 fav : forming a 4 digit nos with all digit is even and not divisible by 4

2222, 6666, 2266,

4!2!2!=6 cases

2226, 2666

4!3!=4 cases 4!3!=4 cases

Total =1+1+6+4+4=16 cases

Total : forming a 4-digit no. from 1,2,…,9 but not divisible by 4

C-1 all 4 digit are add =54=625

C-2 all 4 digit are even =16

C-3 one even and 3 odd

you can not take 2 or 6 as one of even digit
12,16,32,36,… all are divisible by 4

But you can take 4 or 8 as one of even digit ...........

C12×C14×53=1000

Take one even digit out of 4 and 8 select one place for 4 and 8out of 4-places

Total =1000+625+16=1641

P=fav. total=161641

Topic: Mathematics, Probability, JEE Main
Let a→=aii^+a2j^+a3k^ and b→=b1i^+b2j^+b3k^ be two vectors such that |a→|=1;a→·b→=2 and |b→|=4. If c→=2(a→×b→)-3b→, then the angle between b→ and c→ is equal to :

Options:
A cos-123
B cos-1-13
C cos-1-32
D cos-123

The correct answer is:
cos-1-32

Solution:
Given |a→|=1,|b→|=4,a→·b→=2

Also, c→=2(a→×b→)-3b→ ...i

Applying dot product with a→ on both sides of equation i

⇒c→·a→=-6 ...ii

as (a→×b→)·a→=(a→×b→)·b→=0

Again, applying dot product with b→ on both sides of equation i

⇒b→·c→=-48 ...iii

Now, using equation i

⇒c→2=4|a→×b→|2+9|b→|2-12a→×b→·b→

We know that, a→×b→2+a→.b→2=a→2b→2

⇒|c→|2=4|a→|2| b→|2-(a→·b→)2+9|b→|2

⇒|c→|2=4(1)(4)2-(4)+9(16)

⇒|c→|2=4×12+144

⇒|c→|2=48+144

⇒|c→|2=192

Now, cosθ=b→·c→|b→||c→|

⇒cosθ=-48192×4

⇒cosθ=-4883.4

⇒cosθ=-323

⇒cosθ=-32

⇒θ=cos-1-32

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=aii^+a2j^+a3k^ and b→=b1i^+b2j^+b3k^ be two vectors such that |a→|=1;a→·b→=2 and |b→|=4. If c→=2(a→×b→)-3b→, then the angle between b→ and c→ is equal to :

Options:
A cos-123
B cos-1-13
C cos-1-32
D cos-123

The correct answer is:
cos-1-32

Solution:
Given |a→|=1,|b→|=4,a→·b→=2

Also, c→=2(a→×b→)-3b→ ...i

Applying dot product with a→ on both sides of equation i

⇒c→·a→=-6 ...ii

as (a→×b→)·a→=(a→×b→)·b→=0

Again, applying dot product with b→ on both sides of equation i

⇒b→·c→=-48 ...iii

Now, using equation i

⇒c→2=4|a→×b→|2+9|b→|2-12a→×b→·b→

We know that, a→×b→2+a→.b→2=a→2b→2

⇒|c→|2=4|a→|2| b→|2-(a→·b→)2+9|b→|2

⇒|c→|2=4(1)(4)2-(4)+9(16)

⇒|c→|2=4×12+144

⇒|c→|2=48+144

⇒|c→|2=192

Now, cosθ=b→·c→|b→||c→|

⇒cosθ=-48192×4

⇒cosθ=-4883.4

⇒cosθ=-323

⇒cosθ=-32

⇒θ=cos-1-32

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let A=aij1≤i,j¯3 be a 3×3 invertible matrix where each aii is a real number. Denote the inverse of the matrix A by A-1. If ∑itl3aij=1
for 1≤i≤3, then

Options:
A sum of the diagonal entries of A is 1
B sum of each row of A-1 is 1
C sum of each row and each column of A-1 is 1
D sum of the diagonal entries of A-1 is 1

The correct answer is:
sum of each row of A-1 is 1

Solution:
Sum of elements in each row of A is 1.

So, A111=111
⇒A-1 A111=A-1111
⇒111=A-1111
⇒ sum of elements in each row of A-1 is 1

Topic: Mathematics, Matrices, JEE Main
Let A=aij2×2, where a≠ij0 for all i,j and A2=I, Let a be the sum of all diagonal elements of A and b=A Then 3a2+4b2 is equal to

Options:
A 4
B 14
C 7
D 3

The correct answer is:
4

Solution:
Let,

A=pqrs

A2=pqrspqrs=p2+qrpq+qsrp+rsqr+s2

Now given, A2=I

⇒p2+qrpq+qsrp+rsqr+s2=1001

Now on comparing both side we get,

⇒p2+qr=1, q(p+s)=0

And r(p+s)=0, qr+s2=1

Now on solving above relation we get,

q≠0⇒p+s=0⇒a=0

And b=|A|=ps-qr=-p2-qr=-1(∵s=-p)

∴3a2+4b2=4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let A=aij3Ã3 and B=bij3Ã3, where bij=aij+aji2 âi,j. Number of such matrices A whose elements are selected from the set 0,1,2,3 such that A=B, are

Options:
A 29
B 212
C 26
D 28

The correct answer is:
212

Solution:
A=a11a12a13a21a22a23a31a32a33 and B=b11b12b13b21b22b23b31b32b33
∵A=B⇒aij=bij=aij+aji2
⇒aij=aji
⇒A is a symmetric matrix

Hence, the number of such matrix=4×4×4×4×4×4=46=212

Topic: Mathematics, Matrices, JEE Main
Let A=aij3Ã3 be a matrix such that aij=i+2j2 where i,jâ1,3 and i,jâN . If Cij be a cofactor of aij, then the value of a11C21+a12C22+a13C23+a21C31+a22C32+a33C33+a31C11+a32C12+a33C13 is equal to

Options:

The correct answer is:
0

Solution:
A=325272234527292
∵ Sum of the product of elements of any row to the corresponding cofactor of any other row is zero.
a11C21+a12C22+a13C23=0
a21C31+a22C32+a23C33=0
a32C12+a31C11+a33C13=0

Topic: Mathematics, Matrices, JEE Main
Let A=aij3Ã3 be a matrix, where aij=x iâ j1i=jâi,jâN & i,jâ¤3. If Cij be the cofactor of aij and C12+C23+C32=6, then the number of value(s) of x âxâR is(are)

Options:

The correct answer is:
2

Solution:
A=1xxx1xxx1
C12=x2-x
C23=x2-x
C32=x2-x
C12+C23+C32=6
3x2-x=6
x2-x-2=0
x=2,-1

Topic: Mathematics, Matrices, JEE Main
Let A=aij3×3 be a scalar matrix whose elements are the roots of the equation x9-15x8+75x7-125x6=0. If A·adjA=k, then the value of k is equal to

Options:
A 512
B 59
C 312
D 39

The correct answer is:
59

Solution:
x9-15x8+75x7-125x6=0⇒x6(x-5)3=0
⇒x=0,0,0,0,0,0,5,5,5
So, A=500050005 and A=53
A·adjA=AI=A3=533=59

Topic: Mathematics, Determinants, JEE Main
Let A=aij3×3 be a square matrix such that AAT=4I, A<0 . If a11+4a12a13a21a22+4a23a31a32a33+4=5λA+I. Then, λ is equal to

Options:
A 45
B -45
C 85
D -85

The correct answer is:
-85

Solution:
AAT=4I⇒AAT=4I⇒A2=64
⇒A=-8 or 8
As, A<0⇒A=-8
a11+4a12a13a21a22+4a23a31a32a33+4=A+4I=A+AAT
=AI+AT
⇒-8A+I=5λA+I
⇒λ=-85

Topic: Mathematics, Determinants, JEE Main
Let A=aij5×5 is a matrix such that aij=3∀i=j0∀i≠j. If adj(adjA)3=3λ, then λ is equal to (where, adj(M) represents the adjoint matrix of matrix M)

Options:

The correct answer is:
150

Solution:
A=3000003000003000003000003=A=35
adjadjA3=A4235=A1635=38035=375=3150
⇒λ=150

Topic: Mathematics, Determinants, JEE Main
Let A=aij and B=bij be two 3×3 real matrices such that bij=3i+j-2aij , where i,j=1,2,3 . If the determinant of B is 81 , then determinant of A i s

Options:
A 13
B 3
C 181
D 19

The correct answer is:
19

Solution:
B=b11b12b13b21b22b23b31b32b33=30a1131a1232a1331a2132a2233a2332a3133a3234a33

⇒81=33∙3∙32A⇒34=36A⇒A=19

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let A=aij be a 3Ã3 matrix where aij=ij-ji+2ijxi<j1i>j0i=j,
then the minimum value of A is equal to (where x is a real number)

Options:
A 14
B -833
C 7
D -433

The correct answer is:
-433

Solution:
A=a11a12a13a21a22a23a31a32a33
a11=0,a12=3x,a13=4x,a21=1
a22=0,a23=11x,a31=1,a32=1,a33=0
A=03x4x1011x110⇒A=33x2+4x
Minimum value of A=-433

Topic: Mathematics, Determinants, JEE Main
Let A=aij be a real matrix of order 3×3, such that ai1+ai2+ai3=1, for i=1,2,3 . Then, the sum of all the entries of the matrix A3 is equal to:

Options:
A 2
B 1
C 3
D 9

The correct answer is:
3

Solution:
Given A=a11a12a13a21a22a23a31a32a33

Let, X=111

Then, we have AX=a11a12a13a21a22a23a31a32a33111

⇒a11+a12+a13a21+a22+a23a31+a32+a33

Given, ai1+ai2+ai3=1, for i=1, 2, 3

⇒a11+a12+a13a21+a22+a23a31+a32+a33=111

⇒AX=X ...i

Pre multiply by A, we get

A2X=AX ...ii

From i & ii, we get A2X=X

Again, pre multiply by A, we get

A3X=AX ...iii

From i & iii, we get A3X=X

Let A3=x1x2x3y1y2y3z1z2z3

⇒A3111=x1+x2+x3y1+y2+y3z1+z2+z3=111

⇒x1+x2+x3=1, y1+y2+y3=1 & z1+z2+z3=1

⇒x1+x2+x3+y1+y2+y3+z1+z2+z3=3.

Thus, the sum of all the elements is=3.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let A=aij be a square matrix of order 3 and B=bij be a matrix such that bij=2i-jaij for 1â¤i,jâ¤3, âi,jâN. If the determinant of A is same as its order, then the value of BT-1 is

Options:
A 13
B 3
C 9
D 127

The correct answer is:
13

Solution:
∵A=3
B=a1112a1214a132a21a2212a234a312a32a33
B=8a1112a1214a13a2112a2214a23a3112a3214a33=a11a12a13a21a22a23a31a32a33=A=3
BT-1=1BT=1B=13

Topic: Mathematics, Determinants, JEE Main
Let A=aij be a square matrix of order 3 such that aij=2j-i, for all i, j=1,2,3. Then, the matrix A2+A3+…+A10 is equal to

Options:
A 310-12A
B 310+12A
C 310+32A
D 310-32A

The correct answer is:
310-32A

Solution:
Given, A=aij3×3 where aij=2j-i we get,

A=124121214121

A2=124121214121124121214121

⇒A2=3612323634323=3A ⇒A3=3A2=9A

Now, A2+A3+A4+⋯A10

=3A+32A+33A+⋯39A

=3A39-13-1=A310-32

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Let $a, a r, a r^2$, $\qquad$ be an infinite G.P. If $\sum_{n=0}^{\infty} a r^n=57$ and $\sum_{n=0}^{\infty} a^3 r^{3 n}=9747$, then $a+18 r$ is equal to

Options:
A 46
B 38
C 31
D 27

The correct answer is:
31

Solution:
$\begin{aligned}
& \sum_{n=0}^{\infty} \operatorname{ar}^{\mathrm{n}}=57 \\
& \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 \\
& \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots(I)\\
& \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 \\
& \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots . \ldots=9746 \\
& \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots(II)\\
& \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\frac{\mathrm{a}^3}{(1-\mathrm{r})^3}}{\frac{\mathrm{a}^3}{1-\mathrm{r}^3}}=\frac{57^3}{9717}=19 \\
& \mathrm{a}=19 \\
& \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31 \\
&
\end{aligned}$
On solving, $\mathrm{r}=\frac{2}{3}$ and $\mathrm{r}=\frac{3}{2}$ (rejected $)$
$\begin{aligned} & a=19 \\ & \therefore a+18 r=19+18 \times \frac{2}{3}=31\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let A and B are 3×3 matrices with real number entries, where A is symmetric, B is skew-symmetric and A+BA-B=A-BA+B. If ABT=-1kAB, then the sum of all possible integral value of k in 2,10 is equal to (where AT represent transpose of matrix A)

Options:
A 20
B 24
C 28
D 45

The correct answer is:
24

Solution:
AT=A,BT=-B
A+BA-B=A-BA+B
⇒A2-AB+BA-B2=A2+AB-BA-B2
⇒AB=BA……i
Now, ABT=BTAT=-BA=-AB=-1kAB
⇒k is an odd number
⇒k=3,5,7,9
Sum =24

Topic: Mathematics, Matrices, JEE Main
Let \(A\) and \(B\) are orthogonal matrices and \(\operatorname{det} A+\operatorname{det} B=0\). Then

Options:
A \(A+B\) is singular
B \(A+B\) is non-singular
C \(A+B\) is orthogonal
D \(A+B\) is skew symmetric

The correct answer is:
\(A+B\) is singular

Solution:
Hint: \(A A^{\top}=I\) and \(B B^{\top}=I\)
\(\Rightarrow \operatorname{det}\left(A A^{\top}\right)=1\) and \(\operatorname{det}\left(B B^{\top}\right)=1\)
\(\operatorname{det}(A)=-\operatorname{det}(B)\) (Given) \( \Rightarrow(\operatorname{det}(A))^2=1 \Rightarrow(\operatorname{det}(B))^2=1\)
\(\therefore \operatorname{det}(A+B)=\operatorname{det}\left(A\left(B^{\top}+A^{\top}\right) B\right)\)
\(=-\operatorname{det}\left(B^{\top}+A^{\top}\right) \quad[\because(\operatorname{det}(A))(\operatorname{det}(B))=-1]\) as \(\operatorname{det}(A)=-\operatorname{det}(B)\)
\(=-\operatorname{det}(B+A)^{\top}\)
\(=-\operatorname{det}(A+B)^{\top}\)
\(=-\operatorname{det}(A+B)\)
\(\Rightarrow 2 \operatorname{det}(A+B)=0\)
\(\Rightarrow \operatorname{det}(A+B)=0\)
\(\therefore(A+B)\) is singular

Topic: Mathematics, Determinants, JEE Main
Let A and B are square matrices of order 2 such that A+adjBT=3223 and AT-adjB=-2-1-1-2, then A2+2A3+3A4+5A5 is equal to (where MT and adj(M) represent the transpose matrix and adjoint matrix of matrix M respectively and I represents the identity matrix of order 2)

Options:
A 4A
B 7A
C 11A
D 10Ι

The correct answer is:
11A

Solution:
Let, X=3223, Y=-2-1-1-2
A+adjBT=X
AT-adjB=Y
A-adjBT=YT
2A=X+YT=1111⇒A=121111
2adjBT=X-YT=5335⇒adjBT=125335
A2=1411111111=142222=121111=A
A3=A4=A5=A

Topic: Mathematics, Matrices, JEE Main
Let A and B are square matrices of order 3. If A=4, B=6, B=A-2I and adjI-2A-1=k, then the value of k is equal to

Options:

The correct answer is:
2.25

Solution:
B=A-2I
Multiplying by A-1
A-1B=I-2A-1…….i
adjI-2A-1=I-2A-12=A-1B2=B2A2
=94=2.25

Topic: Mathematics, Determinants, JEE Main
Let A and B are square matrices of order 3 such that AB2=BA and BA2=AB. If AB3=A3Bm, then m is equal to

Options:
A 3
B 4
C 5
D 7

The correct answer is:
7

Solution:
AB3=ABABAB
=ABABAB
=AAB2AB2B=A2BBAB3
=A2BAB2B3
=A2BAB5=A2AB2B5
=A3B7

Topic: Mathematics, Matrices, JEE Main
Let \( A \) and \( B \) are two events such that \( P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} \) and \( P(\bar{A})=\frac{1}{4} \), where \( \bar{A} \) represents the complement
of event \( A \). Then, events \( A \) and \( B \) are

Options:
A mutually exclusive and independent.
B independent but not equally likely.
C equally likely but not independent.
D equally likely and mutually exclusive.

The correct answer is:
independent but not equally likely.

Solution:
Given, PA∪ B¯=16⇒1-PA∪B=16

⇒1-PA-PB+PA∩B=16

⇒P(A¯)-PB+PA∩B=16

⇒14-PB+14=16

⇒PB=13 and PA=1-14=34

As, PA∩B=PAPB

So, events A and B are independent events but they are not equally likely.

Topic: Mathematics, Probability, JEE Main
Let A and B are two independent events such that PB=12 and PAâ©B=110, then the value of 9PA-AâªB is

Options:

The correct answer is:
6

Solution:
PA∩B=PA·PB
⇒PA=15
Now, PA-A∪B=PA-∩A∪BPA∪B

⇒PA-A∪B=PB-PA∩B15+12-110=12-110610=23
So, 9PA-A∪B=6

Topic: Mathematics, Probability, JEE Main
Let A and B are two matrices of order 3×3, where A=-2 and B=2, then A-1adjB-1adj2A-1 is equal to

Options:
A 2
B -2
C 4
D 8

The correct answer is:
-2

Solution:
We know that,
A-1=1A
adjA=A2 (for 3×3 matrices)
A-1adjB-1adj2A-1=1A×B-12×2A-12
=1A×1B2×26×1A2
=26-25=-2

Topic: Mathematics, Determinants, JEE Main
Let a^ and b^ are two non-collinear unit vectors. If u→=a^-a^·b^b^ and v→=a^×b^, then |v→| is equal to

Options:
A u→
B u→+v→.a^
C 2v→
D v→+u→·a^+b^

The correct answer is:
u→

Solution:
We have, a^=b^=1

Given, v→=a^×b^=a^b^sinθn^=sinθn^

where, θ is the angle between a^ and b^.

∴ v→=sinθ

Now, u→=a^-a^·b^b^=a^-a^b^cosθb^

u→=a^-cosθb^

∴ u→=a^2+-cosθb^2-2a^-cosθb^cosθ

⇒ u→=1+cos2θ-2cos2θ

⇒u→=1-cos2θ=sin2θ

∴ u→=sinθ

So, v→=u→

Topic: Mathematics, Vector Algebra, JEE Main
Let aâ and bâ are two non-collinear vector such that aâ=1. The angles of a triangle whose two sides are represented by vectors 3aâÃbâ and bâ-aââbâaâ are

Options:
A π2,π4,π4
B π2,π3,π6
C π2,5π12,π12
D π4,π3,5π12

The correct answer is:
π2,π3,π6

Solution:
Let, A→=3a→×b→
B→=b→-a→⋅b→a→
A→⋅B→=0⇒A→ is perpendicular to B→
A→=3a→×b→=3a→b→sin⁡θ=3b→sin⁡θ
B→2=b→-a→⋅b→a→2
=b→2+a→⋅b→2a→2-2a→⋅b→2
=b→2+b→2cos2⁡θ-2b→2cos2⁡θ
=b→2sin2⁡θ
⇒B→=b→sin⁡θ
A→B→=31⇒ other two angles are π6,π3

Topic: Mathematics, Vector Algebra, JEE Main
Let A and B are two non-singular matrices of order 3 such that A=3 and A-1B2+2AB=O, then the value of A4-2A2B is equal to (where O is the null matrix of order 3)

Options:
A 0
B 56
C 2356
D 3453

The correct answer is:
3453

Solution:
A-1B2+2AB=O
⇒A-1BBB-1+2ABB-1=O
⇒A-1B+2A=O
⇒AA-1B+2AA=O⇒B=-2A2
Hence, A4-2A2B=5A4=53A4=53.34

Topic: Mathematics, Determinants, JEE Main
Let A and B are two non-singular matrices of order 3 such that A+B=2I and A-1+B-1=3I, then AB is equal to (where, I is the identity matrix of order 3)

Options:
A A
B B
C 2I3
D 2I

The correct answer is:
2I3

Solution:
Given A+B=2I ... (i)

and A-1+B-1=3I ...(ii)
Multiplying by matrix A in eq. (ii), we get,
AA-1+AB-1=3AI
I+AB-1=3A
Now, multiplying by matrix B, we get,
IB+AB-1B=3AB
⇒B+A=3AB
⇒3AB=2I (from (i))

⇒AB=2I3

Topic: Mathematics, Matrices, JEE Main
Let A and B are two non-singular matrices such that AB=BA2, B4=I and Ak=I, then k can be equal to

Options:
A 5
B 10
C 15
D 16

The correct answer is:
15

Solution:
AB4=AB(B)(B)(B)
=BA2(B)(B)(B)
=BA(AB)BB
=BA(BA2)BB
=B(AB)(A2BB)
=BBA4BB
=B2BA8B=B4A16
⇒A=A16⇒A15=I

Topic: Mathematics, Matrices, JEE Main
Let a→ and b→ are two vectors such that a→=2, b→=3 and a→⋅b→=4. If c→=3a→×b→-4b→, then c→ is equal to

Options:

The correct answer is:
18

Solution:
c→2=3a→×b→-4b→2=9a→×b→2+16b→2-24a→×b→⋅b→
∵a→×b→2=a→2b→2-a→⋅b→2=4×9-16=20
and a→×b→⋅b→=0
Hence, c→2=9×20+16×9=324⇒c→=18

Topic: Mathematics, Vector Algebra, JEE Main
Let a→ and b→ are unit vectors such that a→+b→=32, then the value of 2a→+7b→⋅4a→+3b→+2020a→×b→ is equal to

Options:
A 1334
B 133
C 30
D 120

The correct answer is:
1334

Solution:
a→+b→2=94⇒1+1+2a→⋅b→=94⇒a→⋅b→=18
2a→+7b→⋅4a→+3b→+2020a→×b→
=8a→2+6a→⋅b→+0+28a→⋅b→+21b→2+0
=8+6×18+28×18+21=29+174=1334

Topic: Mathematics, Vector Algebra, JEE Main
Let A and B be 3×3 real matrices such that A is a symmetric matrix and B is a skew-symmetric matrix. Then the system of linear equations A2 B2-B2 A2X=O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has

Options:
A exactly two solutions
B infinitely many solutions
C a unique solution
D no solution

The correct answer is:
infinitely many solutions

Solution:
Let AT=A and BT=-B

C=A2B2-B2A2

CT=A2 B2T-B2 A2T

=B2TA2T-A2TB2T

=B2 A2-A2 B2

CT=-C

C is skew symmetric.

So det C=0

so system have infinite solutions.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let A and B be any two 3 × 3 matrices. If A is symmetric and B is skew symmetric, then the matrix AB-BA is

Options:
A skew symmetric
B I or - I, where I is an identity matrix
C symmetric
D neither symmetric nor skew symmetric

The correct answer is:
symmetric

Solution:
A square matrix P is said to be a symmetric matrix, if PT=P and it is said to be a skew symmetric matrix, if PT=-P.

Given AT=A, BT=-B

Now, AB-BAT=ABT-BAT
=BTAT-ATBT

=(-B)A-A(-B)

=AB-BA

⇒AB-BA is a symmetric matrix.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (19 Apr Online)
Let A and B be any two 3×3 symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?

Options:
A A4-B4 is a symmetric matrix
B AB-BA is a symmetric matrix
C B5-A5 is a skew-symmetric matrix
D AB+BA is a skew-symmetric matrix

The correct answer is:
B5-A5 is a skew-symmetric matrix

Solution:
Given that AT=A,BT=-B

Now solving all option one by one we get,

Solving A option

C=A4-B4

CT=A4-B4=A4T-B4T=A4-B4=C

⇒C=CT is a symmetric matrix

Solving B option

C=AB-BA

CT=AB-BAT=ABT-BAT

=BTAT-ATBT=-BA+AB=C

⇒C=CT is a symmetric matrix

Solving C option

C=B5-A5

CT=B5-A5T=B5T-A5T=-B5-A5

⇒C≠CT or C≠-CT

Solving D option

C=AB+BA

CT=AB+BAT=ABT+BAT

=-BA-AB=-C

C=-CT is skew symmetric matrix

∴ Option (C) is not true.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let a and b be any two numbers satisfying 1a2+1b2=14. Then, the foot of perpendicular from the origin on the variable line xa+yb=1 lies on :

Options:
A A circle of radius =2
B A hyperbola with each semi-axis = 2 .
C A hyperbola with each semi-axis =2
D A circle of radius = 2

The correct answer is:
A circle of radius =2

Solution:
The given line is bx+ay=ab.
Foot of perpendicular from origin
xb=ya=--aba2+b2
x=ab2a2+b2, y=a2ba2+b2
x2+y2=a2b2a2+b2a2+b22
x2+y2=a2b2a2+b2
given 1a2+1b2=14
a2+b2a2b2=14
⇒ x2+y2=4, which is equation of circle.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (09 Apr Online)
Let A and B be independent events such that P(A)=p,P(B)=2p. The largest value of p, for which P (exactly one of A,B occurs)=59, is:

Options:
A 49
B 15
C 512
D 29

The correct answer is:
512

Solution:
Probability (exactly one of A,B occurs) =59

PA∩B¯+PA¯∩B=59

PA-PA∩B+PB-PA∩B=59

Given, A and B are independent events.

So, PA∩B=PAPB

⇒P(A)+P(B)-2P(A)P(B)=59

⇒p+2p-4p2=59

⇒36p2-27p+5=0

⇒(12p-5)(3p-1)=0

p=13 or 512

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (26 Aug Shift 1)