datavorous/entrance-exam-dataset huggingface dataset:
Let A be a 3×3 invertible matrix. If adj24A=adj3adj2A|, then A2 is equal to

Options:
A 26
B 212
C 512
D 66

The correct answer is:
26

Solution:
We know that adjA=An-1

KA=KnA

where n is order of matrix

Given

adj24A=adj3adj2A

⇒24A2=3adj2A2

⇒24A=3adj2A

⇒243A=33adj2A ⇒243A=332A2A

⇒243A=332323AA

⇒A=243232333 =23·332633 =8

⇒A2=64

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let $A$ be a $3 \times 3$ matrix and $B$ be its adjoint matrix. If $|B|=64,$ then $|A|$ is equal to

Options:
A ±2
B $\pm 4$
C $\pm 8$
D ±12

The correct answer is:
$\pm 8$

Solution:
We have,
$|B|=64$
$\Rightarrow |\operatorname{adj} A \mid=64$
$\Rightarrow|A|^{2}=64$
$|A|=\pm 8$

Topic: Mathematics, Matrices, JEE Main
Let A be a 3×3 matrix and detA=2. If n=detadjadj....adjA⏟2024−times, then the remainder when n is divided by 9 is equal to __________.

Options:

The correct answer is:
7

Solution:
Given,

A=2

And n=detadjadjadj...A⏟2024 times

Now, using the formula adjadjadj...A⏟r times=An−1r we get,

adjadjadj...A⏟2024 times=An−12024

⇒n=A22024

⇒n=222024 as A=2

Now, solving

22024=41012=3+11012=3k+1

⇒222024=23k+1

⇒222024=2·23k

⇒222024=2·9-1k

⇒222024=9m-2

⇒222024=9t-2+9

⇒222024=9m+7

Hence, the remainder is 7

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is ______

Options:

The correct answer is:
27

Solution:
$\begin{aligned} \text { Let } A & =\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right] \\ A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] & =3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\end{aligned}$
$\Rightarrow a_1+a_2+a_3=3$ ...(1)
$\Rightarrow b_1+b_2+b_3=3$ ...(2)
$\Rightarrow c_1+\mathrm{ca}_2+\mathrm{c}_3=3$ ...(3)
Now,
$\begin{aligned}
& |A|=\left(a_1 b_2 c_3+a_2 b_3 c_1+a_3 b_1 c_2\right) \\
& -\left(a_3 b_2 c_1+a_2 b_1 c_3+a_1 b_3 c_2\right)
\end{aligned}$
$\therefore$ From above in formation, clearly $|\mathrm{A}|_{\max }=27$, when $a_1=3, b_2=3, c_3=3$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let A be a $3 \times 3$ matrix such that
$$
\mathrm{A}\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 3 \\
0 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
Then $\mathrm{A}^{-1}$ is:

Options:
A
$\left[\begin{array}{lll}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{array}\right]$

B
$\left[\begin{array}{lll}3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0\end{array}\right]$

C
$\left[\begin{array}{lll}0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1\end{array}\right]$

D
$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3\end{array}\right]$

The correct answer is:
$\left[\begin{array}{lll}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{array}\right]$

Solution:
$$
\text { Given } A\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 3 \\
0 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
Applying $\mathrm{C}_1 \leftrightarrow \mathrm{C}_3$
$$
\mathrm{A}\left[\begin{array}{lll}
3 & 2 & 1 \\
3 & 2 & 0 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]
$$
Again Applying $\mathrm{C}_2 \leftrightarrow \mathrm{C}_3$
$$
\text { A }\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
pre-multiplying both sides by $\mathrm{A}^{-1}$
$$
\begin{aligned}
&\mathrm{A}^{-1} \mathrm{~A}\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]=\mathrm{A}^{-1}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
&\mathrm{I}\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]=\mathrm{A}^{-1} \mathrm{I}=\mathrm{A}^{-1} \\
&\left(\because \mathrm{A}^{-1} \mathrm{~A}=\mathrm{I} \text { and } \mathrm{I}=\text { Identity matrix }\right)
\end{aligned}
$$
$$
\text { Hence, } A^{-1}=\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (11 Apr Online)
Let A, be a 3 ×3 matrix, such that A2-5A+7I=O.
Statement - I : A-1=17 5I-A.
Statement - II : The polynomial A3-2A2-3A+I,can be reduced to 5A-4I. Then :

Options:
A Both the statements are true
B Both the statements are false
C Statement - I is true, but Statement - II is false
D Statement - I is false, but Statement - II is true

The correct answer is:
Both the statements are true

Solution:
A2=5A-7I.....i
AAA-1-5AA-1-7IA-1=O
AI=5I-7A-1
A-1=17(5I-A)
Hence, the statement I is true.
Now A3-2A2-3A+I
=A5A-7I-2A2-3A+I {fromi}
=5A2-7A-2A2-3A+I
=3A2-10A+I

=35A-7I-10A+I {fromi}

=5A-20I

=5A-4I

Hence, Statement II is true.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2016 (10 Apr Online)
Let A be a 3×3 matrix such that adjadjadj.A=124. Then A-1adj A is equal to

Options:
A 23
B 6
C 12
D 1

The correct answer is:
23

Solution:
Given equation is

adjadjadj.A=124

⇒A(n-1)3=124

∵ n=3

⇒A(3-1)3=124

⇒A8=124

⇒A2=12

⇒A=23

Now, A-1adj A=A-1.adj A=1A×A3-1

=A=23

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let A be a 3×3 matrix such that adjA=2-11-1021-2-1 and B=adjadjA. If |A|=λ and B-1⊤=μ, then the ordered pair λ, μ is equal to

Options:
A 3, 181
B 9, 19
C 3, 81
D 9, 181

The correct answer is:
3, 181

Solution:
adj A=A2=9

⇒A=±3=λ⇒λ=3

⇒B=adj A2=81

⇒B-1T=B-1=B-1=1B=181=μ

∴λ, μ=3, 181.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let A be a 3×3 matrix with detA=4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R2→2R2+5R3 on 2 A, then detB is equal to :

Options:
A 64
B 16
C 128
D 80

The correct answer is:
64

Solution:
Let A=a11a12a13a21a22a23a31a32a33⇒2A=2a112a122a132a212a222a232a312a322a33

R2→2R2+5R3

B=2a112a122a134a21+10a314a22+10a324a23+10a332a312a322a33

B=2a112a122a134a21+10a314a22+10a324a23+10a332a312a322a33

B=2a112a122a134a214a224a232a312a322a33+2a112a122a1310a3110a3210a332a312a322a33

B=2×4×2 a11a12a13a21a22a23a31a32a33+2×10×2 a11a12a13a31a32a33a31a32a33

B=16 a11a12a13a21a22a23a31a32a33 +0

B=16A

Value of the second determinant is zero. Since Row 2 and Row 3 are identical.

|B|=16×4=64∵|A|=4

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Let A be a 3×3 real matrix. If det(2Adj(2Adj(Adj(2A))))=241, then the value of detA2 equals ______.

Options:

The correct answer is:
4

Solution:
∴|KA|=Kn|A|, If A is n×n matrix

|AdjA|=|A|n-1 If A is n×n matrix

|2Adj(2Adj(Adj(2A)))|=241

⇒23|Adj(2Adj(Adj(2 A)))|=241
⇒|Adj(2Adj(Adj(2 A)))|=238
⇒(|2Adj(Adj(2 A))|)2=238
⇒|2Adj(Adj(2 A))|=219
⇒23|Adj(Adj(2 A))|=219
⇒|Adj(Adj(2 A))|=216
⇒(|Adj(2 A)|)2=216
⇒|Adj(2 A)|=28
⇒|(2 A)|2=28
⇒|(2 A)|=24
⇒23| A|=24
⇒|A|=2

So, A2=A2=4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let A be a 3×3 real matrix such that A110=110;A101=-101 and A001=112. If X=x1 x2 x3T and I is an identity matrix of order 3, then the system A-2IX=411 has

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

The correct answer is:
infinitely many solutions

Solution:
Let the matrix be A=a1b1c1a2b2c2a3b3c3

Now A001=c1c2c3=112

⇒ c1=1,c2=1,c3=2

A101=c1+a1c2+a2c3+a3=-101

⇒ a1=-2,a2=-1,a3=-1

A110=a1+b1a2+b2a3+b3=110

⇒ b1=3,b2=2,b3=1

So matrix A=-231-121-112

⇒A-2I=-431-101-110

i.e. A-2 I=0

Now X=x1x2x3

So, -431-101-110x1x2x3=411

-4x1+3x2+x3=4 ...1

-x1+x3=1 ...2

-x1+x2=1 ...3

Solving the above system of equations we get infinite solutions.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let A be a fixed point (0,6) and B be a moving point (2t,0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is

Options:
A 3x2+2y-6=0
B 2x2-3y+9=0
C 3x2-2y-6=0
D 2x2+3y-9=0

The correct answer is:
2x2+3y-9=0

Solution:
Mid point M=t,3

mAB=-3t

Perpendicular bisector of AB is

y-3=t3x-t

So, C=0, 3-t23

Let P be h, k

h=t2; k=3-t26

⇒k=3-4h26

⇒2x2+3y-9=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let A be a matrix of order 3×3 and detA=2. Then detdet(A)adj5adjA3 is equal to _____.

Options:
A 256×106
B 1024×106
C 512×106
D 256×1011

The correct answer is:
512×106

Solution:
\(|(\operatorname{det}(\mathrm{A})) \operatorname{adj}(5 \operatorname{adj}(\mathrm{A}))|\)
\(=\left|2 \operatorname{adj}\left(5 \operatorname{adj}\left(\mathrm{A}^3\right)\right)\right|\)
\(=2^3 \mid \operatorname{adj}\left(5 \operatorname{adj}\left(\mathrm{A}^3\right) \|\right.\)
\(=2^3 \cdot\left|5 \operatorname{adj}\left(\mathrm{A}^3\right)\right|^2\)
\(=2^3\left(5^3 \cdot\left|\operatorname{adj}\left(\mathrm{A}^3\right)\right|\right)^2\)
\(=2^3 \cdot 5^6 \cdot\left|\operatorname{adj} \mathrm{A}^3\right|^2\)
\(=2^3 \cdot 5^6\left(\left(|\mathrm{~A}|^3\right)^2\right)^2\)
\(=2^3 \cdot 5^6 \cdot 2^{12}=2^{15} \times 5^6\)
\(=2^9 \times 10^6\)
\(=512 \times 10^6\)

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let A be a matrix of order 3×3 such that A=3. Let B=3A-1 and C=adjA2, then the value of A2B3C4 is

Options:
A 316212
B 3212
C 31028
D 312214

The correct answer is:
316212

Solution:
B=3A-1=33A=9
C=adjA2=A28=98
A2B3C4=A2×B3×C4
=32⋅36⋅38212=316212

Topic: Mathematics, Determinants, JEE Main
Let A be a matrix of order 3 × 3 such that det(A) = 2, B = 2A–1 and C=(adjA)163 , then the value of det(A3B2C3) is

Options:

The correct answer is:
2

Solution:
A=2, B=2A-1=23|A|=22
C=adjA163=116⋅22=122
∴det.A3B2C3=23⋅222⋅1223=2

Topic: Mathematics, Determinants, JEE Main
Let A be a matrix of order 3 such that A2=3A-2I where, I is an identity matrix of order 3. If A5=Î±a+Î²I, then Î±Î² is equal to

Options:
A 2025
B -2025
C -930
D -640

The correct answer is:
-930

Solution:
A3=A2A=3A-2IA=3A2-2A=33A-2I-2A=7A-6I
A4=A3A=7A-6IA=7A2-6A=73A-2I-6A=15A-14I
A5=A4A=15A-14IA=15A2-14A=153A-2I-14A=31A-30I
⇒α=31, β=-30

Topic: Mathematics, Matrices, JEE Main
Let $A$ be a matrix such that $A$. $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ is a scalar matrix and $|3 A|=108$. Then $A^2$ equals

Options:
A
$\left[\begin{array}{cc}4 & -32 \\ 0 & 36\end{array}\right]$

B
$\left[\begin{array}{cc}4 & 0 \\ -32 & 36\end{array}\right]$

C
$\left[\begin{array}{cc}36 & 0 \\ -32 & 4\end{array}\right]$

D
$\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$

Solution:
A. $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ is a scalar matrix and $|3 \mathrm{~A}|=108$ suppose the scalar matrix is $\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$
$$
\begin{aligned}
&\therefore A \cdot\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right] \\
&\Rightarrow A=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]^{-1} \\
&{\left[\therefore A B=C \Rightarrow A B B^{-1}=C B^{-1} \Rightarrow A=C B^{-1}\right]} \\
&\Rightarrow A=\frac{1}{3}\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{cc}
3 & -2 \\
0 & 1
\end{array}\right] \\
&\Rightarrow A=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{cc}
1 & -\frac{2}{3} \\
0 & \frac{1}{3}
\end{array}\right]
\end{aligned}
$$
$\Rightarrow A=\left[\begin{array}{cc}k & -\frac{2}{3} k \\ 0 & \frac{k}{3}\end{array}\right]$
$\because|3 \mathrm{~A}|=108$
$\Rightarrow 108=\left|\begin{array}{cc}3 k & -2 k \\ 0 & k\end{array}\right|$
$\Rightarrow 3 k^2=108 \Rightarrow k^2=36 \Rightarrow k=\pm 6$
For $k=6$
$A=\left[\begin{array}{cc}6 & -4 \\ 0 & 2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$
For $k=-6$
$\Rightarrow A=\left[\begin{array}{cc}-6 & 4 \\ 0 & -2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Let A be a matrix such that A·1203 is a scalar matrix and 3A=108. Then, A2 equals :

Options:
A 40-3236
B 36-3204
C 360-324
D 4-32036

The correct answer is:
36-3204

Solution:
Let A=abcd

Now, A.1203=abcd1203=a2a+3bc2c+3d is a scalar.

∴ c =0, 2a+3b=0, a=2c+3d⇒a=3d

Given, 3A=108

⇒A=10832=12⇒ad-bc=ad=12

⇒3dd=12⇒d2=4 ∵a=3d

∴ a2=3d2=9d2==9×4=36 and

∵ 2a+3b=0⇒b=-2a3=-2×63⇒b=-4

Now, A2=ab0dab0d=a2ab+bd0d2=36-3204

Topic: Mathematics, Matrices, JEE Main, JEE Main 2018 (15 Apr)
Let $\mathrm{A}$ be a matrix such that $\mathrm{AB}$ is a scalar matrix where $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ and $\operatorname{det}(3 A)=27$. Then $3 A^{-1}+A^2=$

Options:
A $\left[\begin{array}{cc}4 & -6 \\ 0 & 2\end{array}\right]$
B $\left[\begin{array}{cc}9 & -4 \\ 0 & 3\end{array}\right]$
C $\left[\begin{array}{cc}10 & -6 \\ 0 & 2\end{array}\right]$
D $\left[\begin{array}{cc}10 & -6 \\ 0 & 4\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}10 & -6 \\ 0 & 4\end{array}\right]$

Solution:
$\begin{aligned} & A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ & A B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\left[\begin{array}{ll}a & 2 a+3 b \\ c & 2 c+3 d\end{array}\right]=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] \\ & \therefore \quad c=0, a=k \\ & \Rightarrow 2 a+3 b=0 \\ & 2 c+3 d=k \\ & \therefore 3 d=k=a \\ & \therefore 6 d+3 b=0 \Rightarrow b+2 d=0 \\ & \text { Also }|3 A|=27 \Rightarrow 9|A|=27 \Rightarrow|A|=3 \\ & a b-b c=3 \\ & \therefore \quad a d=3 \\ & 3 d \cdot d=3 \Rightarrow d=1 \\ & \therefore \quad b=-2, a=3 \\ & \therefore \quad A=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right] \\ & A^2=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & -8 \\ 0 & 1\end{array}\right] \\ & A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A=\frac{1}{3}\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \\ & \therefore \quad 3 A^{-1}+A^2=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}9 & -8 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}10 & -6 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
Let A be a n×n matrix such that A=2. If the determinant of the matrix Adj2.Adj2 A-1 is 284, then n is equal to _____ .

Options:

The correct answer is:
5

Solution:
Given,

A=2

Now simplifying,

Adj2.Adj2 A-1

=2·Adj2 A-1n-1

=2nAdj2 A-1n-1

=2n2A-1n-1n-1

=2nn-12nA-1n-1n-1

∴ A-1=1A=12

=2nn-12n-1n-1n-1

=2nn-1+n-13=284

Now comparing both side we get,

nn-1+n-13=84

⇒n-1n+n2-2n+1=84

⇒n-1n2-n+1=4×21

Now if n-1=4⇒n=5 now checking n2-3n+1=25-5+1=21

Hence, n=5

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let $A$ be a non-singular matrix of order 3 . If $\operatorname{det}(3 \operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A)))=3^{-13} \cdot 2^{-10}$ and $\operatorname{det}(3 \operatorname{adj}(2 \mathrm{~A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$, then $|3 \mathrm{~m}+2 \mathrm{n}|$ is equal to $\qquad$

Options:

The correct answer is:
14

Solution:
$\begin{aligned} & |3 \operatorname{adj}(2 \operatorname{adj}(|A| A))|=\mid 3 \operatorname{adj}\left(2|A|^2 \operatorname{adj}(A) \mid\right. \\ & =\left.\left|3.2^2\right| A\right|^4 \operatorname{adj}\left(\left.\operatorname{adj}(A)\left|=2^6 3^3\right| A\right|^{12}|A|^4\right. \\ & =2^6 3^3|A|^{16}=2^{-10} 3^{-13} \\ & \Rightarrow|A|^{16}=2^{-16} 3^{-16} \Rightarrow|A|=2^{-1} 3^{-1}\end{aligned}$
$\begin{aligned} & \text { Now }|3 \operatorname{adj}(2 \mathrm{~A})|=\left|3.2^2 \operatorname{adj}(\mathrm{A})\right| \\ & =2^6 3^3|\mathrm{~A}|^2=2^{-\mathrm{m}} 3^{-\mathrm{n}} \\ & \Rightarrow 2^6 3^3 2^{-2} 3^{-2}=2^{-\mathrm{m}} 3^{-\mathrm{n}} \\ & \Rightarrow 2^{-\mathrm{m}} 3^{-\mathrm{n}}=2^4 3^1 \\ & \Rightarrow \mathrm{m}=-4, \mathrm{n}=-1 \\ & \Rightarrow|3 \mathrm{~m}+2 \mathrm{n}|=|-12-2|=14\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let A be a non-singular matrix of order 3 such that Aadj3A=5AAT, then A-13 is equal to

Options:

The correct answer is:
1.8

Solution:
Aadj3A=5AAT
Taking determinant on both sides, we get,
Aadj3A=5AAT
⇒A3A2=53A2
⇒36A3=53A2⇒A=0, 125729 A≠0
Hence, A-13=1A13=95=1.8

Topic: Mathematics, Determinants, JEE Main
Let A be a non-singular square matrix such that A2=A satisfying Ι-0.8A-1=Ι-αA where Ι is a unit matrix of the same order as that of A, then the value of -4α is equal to

Options:

The correct answer is:
16

Solution:
Ι-0.8A-1=Ι-αA
Ι-0.8AΙ-0.8A-1=Ι-0.8AΙ-αA
Ι=Ι-α+45A+4α5A2
Aα+45=4α5A⇒α5=-45⇒α=-4

Topic: Mathematics, Matrices, JEE Main
Let A be a non-singular symmetric matrix of order 3. If AT=A2-I, then A-I-1 is equal to

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

The correct answer is:
A

Solution:
∵AT=A (as A is symmetric)
∴A=A2-I⇒A2-A=I
⇒I=AA-I=A-IA
⇒A-I-1=A

Topic: Mathematics, Matrices, JEE Main
Let a→ be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^ and i^-j^,j^-k^. If θ is the angle between the vector a→ and the vector b→=2i^-2j^+k^ and a→·b→=6, then the ordered pair θ,|a→×b→| is equal to

Options:
A π3,36
B π4,36
C π3,6
D π4,6

The correct answer is:
π4,6

Solution:
Given,

a→ be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^,i^+k^,

So finding the normal vector to the above plane, we get,

n→1=i^j^k^110101=i^-j^-k^

And normal vector to planes i^-j^,j^-k^ will be,

n→2=i^j^k^1-1001-1=i^+j^+k^

Now a→=λn→1×n→2

So, direction ratio of a→=i^j^k^1-1-1111=-2j^+2k^

Hence, direction ratio of a→=0,-2,2=0,-1,1

Now given, b→=2i^-2j^+k^

And a→=λ(-j^+k^)

So, a→·b→=6=λ(2+1)

⇒λ=2

∴a→=-2j^+2k^

Now we know that, a→·b→=|a¯||b|cosθ

⇒6=22×3cosθ

⇒cosθ=12

⇒θ=π4

Now finding, a→×b→=a→b→sinθ=22×3×12=6

Hence, ordered pair θ,a→×b→=π4,6

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let A be a point on the line r→=1-3μi^+μ-1j^+2+5μk^ and B3, 2, 6 be a point in the space. Then the value of μ for which the vector AB→ is parallel to the plane x-4y+3z=1 is

Options:
A 12
B 14
C -14
D 18

The correct answer is:
14

Solution:
Let A is 1-3μ, μ-1, 2+5μ

We know that the vector joining two points x1, y1, z1 and x2, y2, z2 is x2-x1i^+y2-y1j^+z2-z1k^

⇒AB→=3μ+2i^+3-μj^+4-5μk^ which is parallel to plane x-4y+3z=1

⇒13μ+2-43-μ+34-5μ=0

⇒3μ+2-12+4μ+12-15μ=0

⇒ -8μ+2=0

⇒μ=14.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let A be a point on the x-axis. Common tangents are drawn from A to the curves x2+y2=8 and y2=16x. If one of these tangents touches the two curves at Q and R, then QR2 is equal to

Options:
A 64
B 76
C 81
D 72

The correct answer is:
72

Solution:
Let a tangent on parabola y2=16x be,

y=mx+4 m

It is also common to x2+y2=8

If 4m2=81+m2

⇒16m2=8(1+m2)

⇒m4+m2-2=0⇒(m2+2)(m2-1)=0

⇒m=±1

Taking one of the tangents y=x+4

⇒x-y+4=0

Tangent to x2+y2=8 is xx1+yy1-8=0

∴x11=y11=-84

⇒Q(-2,2)

Tangent to y2=16x is yy1=8(x+x1)

⇒8x-yy1+8x1=0

∴81=-y1-1=8x14

⇒y1=8, x1=4

∴R(4,8)

Using the distance formula between two points,

QR=(4+2)2+(8-2)2=62

∴(QR)2=622=72

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let a be a positive real number such that ∫0aex-xdx=10e-9 where, [x] is the greatest integer less than or equal to x. Then, a is equal to:

Options:
A 10-loge1+e
B 10+loge2
C 10+loge3
D 10+loge1+e

The correct answer is:
10+loge2

Solution:
Given:

a>0

Let n≤a<n+1, n∈W

Then,

a=a+a

Here, a=n
Now,

∫0aex-xdx=10e-9

⇒∫0nexdx+∫naex-xdx=10e-9

⇒n∫01exdx+∫naex-ndx=10e-9

⇒ne-1+ea-n-1=10e-9

⇒ne-n+ea-n-1=10e-9

⇒ne-n-1+eae-n=10e-9

On comparing, we get

n=10

and,

-n-1+eae-n=-9

⇒-10-1+ea-10=-9

⇒ea-10=2

⇒a-10logee=loge2

⇒a=10+loge2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let a be a real number such that the function f(x)=ax2+6x-15,x∈R is increasing in -∞,34 and decreasing in 34,∞. Then the function g(x)=ax2-6x+15,x∈R has a

Options:
A local maximum at x=-34
B local minimum at x=-34
C local maximum at x=34
D local minimum at x=34

The correct answer is:
local maximum at x=-34

Solution:
We have,

fx=ax2+6x-15

Graph of fx is shown below as it is increasing in -∞,34 and decreasing in 34,∞.

Then, abscissa of vertex is 34.

-B2A=34

⇒-(6)2a=34

⇒a=-4

∴g(x)=-4x2-6x+15

Local maxima exists at the vertex of gx, whose abscissa is

x=-B2A

⇒x=--6-8=-34

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let A be a set containing \( 10 \) distinct elements. Then the total number of distinct function from A to A, is

Options:
A \( 10 ! \)
B \( 10^{10} \)
C \( 2^{10} \)
D \( 2^{10}-1 \)

The correct answer is:
\( 10^{10} \)

Solution:
If a set A has n elements and another set B contains r elements then total number of functions from A to B is rn as for each of the n elements of input,there would be r choices of output.

Total number of distinct functions from A to A are rn i.e. 1010 .

Topic: Mathematics, Permutation Combination, JEE Main
Let $\mathbf{A}$ be a square matrix all of whose entries are integers. Then which one of the following is true?

Options:
A
If $\operatorname{det} A=\pm 1$, then $A^{-1}$ exists but all its entries are not necessarily integers

B
If $\operatorname{det} A \neq \pm 1$, then $A^{-1}$ exists and all its entries are non-integers

C
If $\operatorname{det} A=\pm 1$, then $A^{-1}$ exists and all its entries are integers

D
If $\operatorname{det} A=\pm 1$, then $A^{-1}$ need not exist

The correct answer is:
If $\operatorname{det} A=\pm 1$, then $A^{-1}$ exists and all its entries are integers

Solution:
Each entry of $A$ is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix $A$ is integer.
Now $\operatorname{det} A=\pm 1$ and $A^{-1}=\frac{1}{\operatorname{det}(A)}(\operatorname{adj} A)$
$\Rightarrow$ all entries in $\mathrm{A}^{-1}$ are integers.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2008
Let $A$ be a square matrix of order 2 such that $|A|=2$ and the sum of its diagonal elements is -3 . If the points $(x, y)$ satisfying $\mathrm{A}^2+x \mathrm{~A}+y \mathrm{I}=\mathrm{O}$ lie on a hyperbola, whose $\text { length of semi major axis is } x \text { and semi minor axis is } y$, eccentricity is $\mathrm{e}$ and the length of the latus rectum is $l$, then $81\left(e^ 4+l^2\right)$ is equal to

Options:

The correct answer is:
233

Solution:
Given $|A|=2$
trace $A=-3$
and $\mathrm{A}^2+\mathrm{xA}+\mathrm{yI}=0$
$\Rightarrow \mathrm{x}=3, \mathrm{y}=2$
$\begin{aligned}
& \text {so, } 81 e^{4}=169 \\
& L R=8 / 9 \\
& 81 e^{4}+81 L^{2}=169+64=233 \end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Let $A$ be a square matrix of order 2 such that $|A|=2$ and the sum of its diagonal elements is -3 . If the points $(x, y)$ satisfying $\mathrm{A}^2+x \mathrm{~A}+y \mathrm{I}=\mathrm{O}$ lie on a hyperbola, whose $\text { length of semi major axis is } x \text { and semi minor axis is } y$, eccentricity is $\mathrm{e}$ and the length of the latus rectum is $l$, then $81\left(e^ 4+l^2\right)$ is equal to

Options:

The correct answer is:
233

Solution:
Given $|A|=2$
trace $A=-3$
and $\mathrm{A}^2+\mathrm{xA}+\mathrm{yI}=0$
$\Rightarrow \mathrm{x}=3, \mathrm{y}=2$
$\begin{aligned}
& \text {so, } 81 e^{4}=169 \\
& L R=8 / 9 \\
& 81 e^{4}+81 L^{2}=169+64=233 \end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let A be a square matrix of order 3, AT be the transpose matrix of matrix A and AAT=4I. If d=2AT+AAT+adjA2, then the value of 12d is equal to A<0

Options:

The correct answer is:
96

Solution:
AAT=4I
⇒AT=4A-1=4×adjAA
⇒adjA=AAT4…i
∵AAT=4I⇒AAT=4I⇒A2=64
⇒A=-8 As A<0
From i
adjA=-2AT⇒2AT+adjA=O
d=4I2=2I=8

⇒12d=96

Topic: Mathematics, Determinants, JEE Main