datavorous/entrance-exam-dataset huggingface dataset:
If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{c}$ is a vector perpendicular to $\mathbf{b}$, then $\left\{\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|^2}\right\}(\mathbf{b} \times \mathbf{c})+\left\{\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right\} \mathbf{b}+\left\{\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{c}|^2}\right\} \mid=$

Options:
A $\sqrt{14}$
B 14
C 13
D $\sqrt{17}$

The correct answer is:
$\sqrt{14}$

Solution:
Any vector $\mathbf{r}$ can be written in linear combination of two non-parallel vector $\mathbf{b}$ and $\mathbf{c}$, as
$$
\mathbf{r}=\left(\frac{\mathbf{r} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right) \mathbf{b}+\left(\frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|^2}\right) \mathbf{c}+\left(\frac{\mathbf{r} \cdot(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|^2}\right)(\mathbf{b} \times \mathbf{c})
$$
So, $\mathbf{r}=\mathbf{a} \Rightarrow|\mathbf{r}|=|\mathbf{a}|=\sqrt{1+4+9}=\sqrt{14}$.

Topic: Mathematics, Vector Algebra, JEE Main
If $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar vectors, then $\lambda$ is equal to

Options:
A
0

B
$-1$

C
2

D
1

The correct answer is:
0

Solution:
Since $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar therefore $[\vec{a} \vec{b} \vec{c}]=0$
$$
\begin{aligned}
& \text { i.e., }\left|\begin{array}{ccc}
1 & 2 & \lambda \\
-2 & 3 & 1 \\
3 & -1 & 2 \lambda-1
\end{array}\right|=0 \\
& \Rightarrow 1(6 \lambda-2)-2(-4 \lambda-1)+\lambda(-7)=0 \\
& \Rightarrow(6 \lambda-2)+8 \lambda+2+2+2 \lambda-9 \lambda=0 \\
& \Rightarrow 7 \lambda=0 \Rightarrow \lambda=0
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (07 May Online)
If $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=r \hat{i}+\hat{j}+(2 r-1 \hat{k}$ are three vectors such that $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$, then $r$ is equal to

Options:
A
1

B
$-1$

C
0

D
2

The correct answer is:
0

Solution:
Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$, and
$$
\vec{c}=r \hat{i}+\hat{j}+(2 r-1) \hat{k}
$$
Since, $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$ therefore, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.
$$
\begin{aligned}
& \therefore\left|\begin{array}{ccc}
1 & -2 & 3 \\
2 & 3 & -1 \\
r & 1 & 2 r-1
\end{array}\right|=0 \\
& \Rightarrow 1(6 r-3+1)+2(4 r-2+r)+ \\
& \Rightarrow 6 r-2+10 r-4+6-9 r=0 \\
& \Rightarrow r=0
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (19 May Online)
If $\mathbf{a}=\mathbf{i}-2 \mathbf{j}$ and $\mathbf{b}=2 \mathbf{i}+\lambda \mathbf{j}$ are parallel, then $\lambda$ is

Options:
A 4
B 2
C -2
D -4

The correct answer is:
-4

Solution:
$\frac{1}{2}=\frac{-2}{\lambda} \Rightarrow \lambda=-4$

Topic: Mathematics, Vector Algebra, JEE Main
If a→=i^+2k^, b→=i^+j^+k^, c→=7i^-3j^+4k^, r→×b→+b→×c→=0→ and r→·a→=0 then r→.c→ is equal to:

Options:
A 34
B 12
C 36
D 30

The correct answer is:
34

Solution:
Given,

r→×b→+b→×c→=0→

⇒r→×b→-c→×b→=0→

⇒r→-c→×b→=0

⇒r→-c→||b→

Therefore, r→-c→=λb→

⇒r→=c→+λb→

Also,

r→·a→=0 (given)

⇒c→+λb→·a→=0

⇒c→·a→+λb→·a→=0

⇒ λ=-c→·a→b→·a→

Now,

r→·c→=c→+λb→·c→

=c→-c→·a→b→·a→b→·c→

=c→2-c→·a→b→·a→b→·c→

=74-153×8

=74-40=34

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (29 Jan Shift 2)
If $\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=4$, then $(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}) \cdot(2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ is equal to

Options:
A 12
B 2
C 0
D $-12$

The correct answer is:
$-12$

Solution:
$(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}) \cdot(2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=\overrightarrow{\mathbf{a}} \cdot\{\hat{\mathbf{j}} \times(2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})\}$
$$
\begin{array}{l}
=\overrightarrow{\mathbf{a}} \cdot\{-3(\hat{\mathbf{j}} \times \hat{\mathbf{k}})\} \\
=-3(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}) \\
=-12 \quad(\because \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=4 \text { given })
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a} \cdot \mathbf{i}=4$, then $(\mathbf{a} \times \mathbf{j}) \cdot(2 \mathbf{j}-3 \mathbf{k})=$

Options:
A 12
B 2
C 0
D -12

The correct answer is:
-12

Solution:
$\begin{aligned}
& (\mathbf{a} \times \mathbf{j}) \cdot(2 \mathbf{j}-3 \mathbf{k})=\mathbf{a}\{\mathbf{j} \times(2 \mathbf{j}-3 \mathbf{k})\} \\
& =\mathbf{a} \cdot\{-3(\mathbf{j} \times \mathbf{k})\}=-3(\mathbf{a} \cdot \mathbf{i})=-12
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If aii=1n, where n is an even integer, is an arithmetic progression with common difference 1, and ∑i=1nai=192,∑i=1n2a2i=120, then n is equal to

Options:
A 18
B 36
C 96
D 48

The correct answer is:
96

Solution:
Given, ∑i=1nai=192

a1+a2 ⋯ an=192

⇒ n2a1+an=192

a1+an=384n ⋯1

Also given ∑i=1n2a2i=120

⇒ a2+a4+a6 ⋯an ⏟n2 terms=120

⇒ n2×12a2+an=120

a2+an=480n

a1+1+an=480n ⋯2

Now equation 2-equation 1

480n-384n=a1+an+1-a,+an

1n480-384=1

480-384=n

n=96

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If $\quad \overrightarrow{\mathbf{a}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \quad$ and $\overrightarrow{\mathbf{c}}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, then the angle between $2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$ is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{3}$
C $\frac{\pi}{2}$
D $\frac{3 \pi}{2}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
Now,
$\begin{aligned}
2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}} & =2(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
& =\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} & =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& =\hat{\mathbf{i}}+\hat{\mathbf{k}}
\end{aligned}$
Let $\theta$ be the angle between $2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$.
$\begin{array}{ll}\therefore & \cos \theta=\frac{(\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{k}})}{\sqrt{1^2+1^2} \sqrt{1^2+1^2}} \\ \Rightarrow & \cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\end{array}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a→=i^+j^+2k^,b→=i^+2j^+2k^ and c→=1, then the maximum value of a→×b→ b→×c→ c→×a→ is equal to

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
a→×b→ b→×c→ c→×a→=a→ b→ c→2
=a→×b→.c→2
a→×b→.c→ is maximum, if a→×b→ is parallel to c→
Hence, the maximum value =a→×b→2c→2=5

Topic: Mathematics, Vector Algebra, JEE Main
If a→=αi^+βj^+3k^, b→=-βi^-αj^-k^ and c→=i^-2j^-k^ such that a→·b→=1 and b→·c→=-3, then 13((a→×b→)·c→) is equal to _______.

Options:

The correct answer is:
2

Solution:
a→·b→=1⇒-αβ-αβ-3=1

⇒-2αβ=4⇒αβ=-2 …1

b→·c→=-3⇒-β+2α+1=-3

β-2α=4 …2

Solving 1 and 2, (α,β)=(-1,2)

13[a→b→c→]=13αβ3-β-α-11-2-1

=13-123-21-11-2-1

R1→R1+R3

=13002-21-11-2-1=1324-1=2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If $\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}} \quad$ and $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}$, then $\lambda+\mu$ is equal to:

Options:
A 0
B 1
C 1
D 3

The correct answer is:
0

Solution:
We have
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|$
$=\hat{\mathbf{i}}(-1)-\hat{\mathbf{j}}(-1)+\hat{\mathbf{k}}(1-1)$
$=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$
$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|=\hat{\mathbf{k}}(-1)=-\hat{\mathbf{k}}$
Now, $\lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+\hat{\mathbf{k}})$
$=(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$
$\because \quad \lambda \overrightarrow{\mathbf{a}}+\mu \overrightarrow{\mathbf{b}}=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}$
$\Rightarrow(\lambda+\mu) \hat{\mathbf{i}}+(\lambda+\mu) \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}=-\hat{\mathbf{k}}$
Equating the coefficient of $\hat{\mathbf{i}}$
$\therefore \quad \lambda+\mu=0$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+\mathbf{j}, \mathbf{c}=\mathbf{i}$ and $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\lambda \mathbf{a}+\mu \mathbf{b}$, then $\lambda+\mu=$

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
a. $\mathbf{c}=1$ and $\mathbf{b} \cdot \mathbf{c}=1$
Given that $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=(\mathbf{c} . \mathbf{a}) \mathbf{b}-(\mathbf{c} . \mathbf{b}) \mathbf{a}=\mu \mathbf{b}+\lambda \mathbf{a}$
where $\mu=\mathbf{c} \cdot \mathbf{a}=1, \lambda=-(\mathbf{c} . \mathbf{b})=-1$
$\Rightarrow \mu+\lambda=1-1=0$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=\mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{c}=\mathbf{i}-\mathbf{j}-\mathbf{k}$, then $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ is

Options:
A $\mathbf{i}-\mathbf{j}+\mathbf{k}$
B $2 \mathbf{i}-2 \mathbf{j}$
C $3 \mathbf{i}-\mathbf{j}+\mathbf{k}$
D $2 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$

The correct answer is:
$2 \mathbf{i}-2 \mathbf{j}$

Solution:
$\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} . \mathbf{c}) \mathbf{b}-(\mathbf{a} . \mathbf{b}) \mathbf{c}$
$\begin{aligned} & \ \mathbf{a} \cdot \mathbf{c}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}-\mathbf{k})=1-1+1=1 \\ & \mathbf{a} \cdot \mathbf{b}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}+\mathbf{k})=1-1-1=-1 \\ & \therefore \begin{array}{l}\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(1) \mathbf{b}-(-1) \mathbf{c} \\ \quad=\mathbf{b}+\mathbf{c}=\mathbf{i}-\mathbf{j}+\mathbf{k}+\mathbf{i}-\mathbf{j}-\mathbf{k}=2 \mathbf{i}-2 \mathbf{j}\end{array}\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $a_{i j}=\frac{1}{2}(3 i-2 j)$ and $A=\left[a_{i j}\right]_{2 \times 2}$, then $A$ is equal to

Options:
A $\left[\begin{array}{cc}1 / 2 & 2 \\ -1 / 2 & 1\end{array}\right]$
B $\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$
C $\left[\begin{array}{cc}2 & 2 \\ 1 / 2 & -1 / 2\end{array}\right]$
D None of these

The correct answer is:
$\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$

Solution:
$\Rightarrow a_{11}=1 / 2, \quad a_{12}=-1 / 2$ and $a_{21}=2, a_{22}=1$
$\therefore \quad A=\left[a_{i j}\right]_{2 \times 2}=$ $\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$
$\therefore \quad A=\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main
If $A$ is $(2,5), B$ is $(4,-11)$ and $C$ lies on $9 x+7 y+4=0$, then the locus of the centroid of the $\triangle A B C$ is a straight line parallel to the straight line is

Options:
A $7 x-9 y+4=0$
B $9 x-7 y-4=0$
C $9 x+7 y+4=0$
D $7+9 y+4=0$

The correct answer is:
$9 x+7 y+4=0$

Solution:
According to question,
$x_1=\frac{2+4+x}{3} \Rightarrow x=3 x_1-6$
$\begin{aligned}
& y_1=\frac{5-11+y}{3} \Rightarrow y=3 y_1+6 \\
& \therefore 9\left(3 x_1-6\right)+7\left(3 y_1+6\right)+4=0
\end{aligned}$
Hence locus is $27 x+21 y-8=0$, which is parallel to $9 x+7 y+4=0$.

Topic: Mathematics, Straight Lines, JEE Main
If A is a 2Ã2 matrix such that A 1-1=-1 2 and A2 1-1=10, then trace of A is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

Options:
A 1
B 0
C 2
D 5

The correct answer is:
1

Solution:
Let, A=abcd
A1-1=-12âabcd1-1=-12
On solving, we get,
a-b=-1, c-d=2
Also,
A21-1=AA1-1=A-12=10
This gives, -a+2b=1, -c+2d=0
âb=0, a=-1, d=2, c=4

Topic: Mathematics, Matrices, JEE Main
If A is a 3×3 matrix and A=2, then 3adj3AA2 is equal to

Options:
A 312·611
B 312·610
C 310·611
D 311·610

The correct answer is:
311·610

Solution:
We need to find the value of 3adj3AA2

We know that kA=knA where n is the order of the matrix and k is a constant.

⇒3A=332

⇒3adj3AA2=33adj33·2A2

We know that adjkA=kn-1adjA

=3333·22adjA2

Now we know that adjA=An-1

=3333·223adjA2

=3333·26A23-1

=3333·2624

=210·321

Hence, the required answer is 210·321=610·311

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If A is a 3×3 matrix such that 5 adjA=5, then A is equal to

Options:
A ±125
B ±5
C ±15
D ±1

The correct answer is:
±15

Solution:
5 Adj A=5
125 |adj A|=5
Since [A] is 3×3
adj A=125
A2=125
A= ±15.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2015 (11 Apr Online)
If $A$ is a $3 \times 3$ non-singular matrix and if $|\mathrm{A}|=3$, then $\left|(2 \mathrm{~A})^{-1}\right|$ is

Options:
A 24
B 3
C $\frac{1}{3}$
D $\frac{1}{24}$

The correct answer is:
$\frac{1}{24}$

Solution:
Given, $|A|_{3 \times 3} \neq 0$ and $|A|=3$
Then,
$$
\begin{aligned}
&\left|(2 \mathrm{~A})^{-1}\right|=\left|\frac{1}{2 \mathrm{~A}}\right|=\frac{1}{|2 \mathrm{~A}|} \\
&=\frac{1}{(2)^{3}} \cdot \frac{1}{|\mathrm{~A}|} \quad\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right) \\
&=\frac{1}{8} \cdot \frac{1}{3}=\frac{1}{24}
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main
If A is a 3×3 non-singular matrix such that AA'=A'A and B=A-1A', then BB' equals, where X' denotes the transpose of the matrix X.

Options:
A B-1
B B-1'
C I+B
D I

The correct answer is:
I

Solution:
A'=AT
AA'=A'A
B=A-1A'
B'=A-1A''=A''·A-1'=AA-1'
BB'=A-1A'·AA-1'
=A-1·A·A'·A-1' ∵ AA'=A'A

=I·A'·A-1'

=I·A-1A' ∵ CD'=D'C'
=I·I'
=I2 ∵ I=I'
=I

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (06 Apr)
If $A$ is a $n \times n$ matrix, then $\operatorname{adj}(\operatorname{adj} A)=$

Options:
A $|A|^{n-1} A$
B $|A|^{n-2} A$
C $|A|^n n$
D None of these

The correct answer is:
$|A|^{n-2} A$

Solution:
We know by the fundamental concept that $\operatorname{adj}(\operatorname{adj} A)=\left|A\right|^{n-2} A$.

Topic: Mathematics, Matrices, JEE Main
If $A$ is a non- singular matrix, then $A(\operatorname{adj} A)=$

Options:
A $A$
B $1$
C $|A| I$
D $|A|^2$

The correct answer is:
$|A| I$

Solution:
$A(\operatorname{adj} A)=A \cdot A^{-1}|A|=|A| I$.

Topic: Mathematics, Matrices, JEE Main
If A is a non-null diagonal matrix of order 3 such that A4=A2, then the possible number of matrices A are

Options:
A 27
B 26
C 8
D 7

The correct answer is:
26

Solution:
Let, A=d1000d2000d3
A2=d12000d22000d32 & A4=d14000d24000d34
∵A4=A2⇒d14=d12⇒d1=0,1,-1
d24=d22⇒d2=0,1,-1
d34=d32⇒d3=0,1,-1
∵ A can't be a null matrix so the total number of possible matrices is =33-1=26

Topic: Mathematics, Matrices, JEE Main
If \( A \) is a non-null diagonal matrix of order \( 3 \) such that \( A^{4}=A^{2} \), then the possible number of matrices \( A \) are

Options:
A \( 27 \)
B \( 26 \)
C \( 8 \)
D \( 7 \)

The correct answer is:
\( 26 \)

Solution:
Let, A=d1000d2000d3
A2=d12000d22000d32 & A4=d14000d24000d34
∵A4=A2⇒d14=d12⇒d1=0,1,-1
d24=d22⇒d2=0,1,-1
d34=d32⇒d3=0,1,-1
∵ A can't be a null matrix so the total number of possible matrices is =33-1=26

Topic: Mathematics, Matrices, JEE Main
If $A$ is a non-singular matrix, then $A(\operatorname{adj} A)=$

Options:
A $A$
B $1$
C $|A| I$
D $|A|^2$

The correct answer is:
$|A| I$

Solution:
$A(\operatorname{adj} A)=A \cdot A^{-1}|A|=|A| I$.

Topic: Mathematics, Matrices, JEE Main
If $A$ is a singular matrix, then $\operatorname{adj} A$ is

Options:
A Singular
B Non-singular
C Symmetric
D Not defined

The correct answer is:
Singular

Solution:
Since $|\operatorname{adj} A|=|A|^{n-2}$, therefore $|A|=0$ $\Rightarrow|\operatorname{adj} A|=0 \Rightarrow \operatorname{adj} A$ is also singular.

Topic: Mathematics, Matrices, JEE Main
If $A$ is a skew symmetric matrix and $n$ is a positive integer, then $A^n$ is

Options:
A A symmetric matrix
B Skew-symmetric matrix
C Hermitian
D Skew-hermitian

The correct answer is:
Skew-symmetric matrix

Solution:
Given A $\rightarrow$ skew symmetric matrix
$\mathrm{n} \rightarrow$ positive integer
$\mathrm{A}^{\mathrm{n}} \rightarrow$ is symmetric if $\mathrm{n}$ is an odd integer and is skew symmetric if $n$ is an even integer

Topic: Mathematics, Matrices, JEE Main
If A is a skew symmetric matrix of order 3, B is a 3Ã1 column matrix and C=BTAB, then which of the following is false?

Options:
A C is singular
B C is non singular
C C is a symmetric matrix
D C is a skew symmetric matrix

The correct answer is:
C is non singular

Solution:
Given, C=BTAB
CT=BTABT
⇒CT=BTATBTT
⇒CT=-BTAB=-C⇒C is skew symmetric
∵ order of C is 1×1⇒C=0
⇒C is singular and symmetric also.

Topic: Mathematics, Matrices, JEE Main
If $A$ is a skew-symmetric matrix of order $n$, and $C$ is a column matrix of order $n \times 1$, then $C^T A C$ is

Options:
A A Identity matrix of order $n$
B A unit matrix of order one
C A zero matrix of order one
D None of these

The correct answer is:
A zero matrix of order one

Solution:
Since A is a Skew-Symmetric matrix
$\mathrm{A}^{\mathrm{T}}=-\mathrm{A}$
$\mathrm{C}$ is a column matrix of order $\mathrm{n} \times 1$
So order of $C^{\mathrm{T}}=1 \times n$
So order of ${C}^T A C=(1 \times n) \times(n \times n) \times(n \times 1)$ $=1 \times 1$
Since $\mathrm{C}^{\mathrm{T}} \mathrm{AC}$ has order $=1 \times 1$
$\therefore\left(\mathrm{C}^{\mathrm{T}} \mathrm{AC}\right)^{\mathrm{T}}=\mathrm{C}^{\mathrm{T}} \mathrm{AC}$
Let's assume $\mathrm{C}^{\mathrm{T}} \mathrm{AC}=\mathrm{B}$
$\left(C^T A C\right)^T=B^T$
or $C^T A^T\left(C^T\right)^T=B^T$
or $\mathrm{C}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}} \mathrm{C}=\mathrm{B}^{\mathrm{T}}$
or $\mathrm{C}^{\mathrm{T}}(-\mathrm{A}) \mathrm{C}=\mathrm{B}^{\mathrm{T}}\left[\because \mathrm{A}^{\mathrm{T}}=-\mathrm{A}\right]$
or $-\mathrm{C}^{\mathrm{T}} \mathrm{AC}=\mathrm{B}\left[\because \mathrm{B}^{\mathrm{T}}=\mathrm{B}\right]$
or $-B=B$
or $2 B=0$
or $B=0$
(3) a zero matrix of order 1

Topic: Mathematics, Matrices, JEE Main
If $A$ is a square matrix for which $a_{i j}=i^2-j^2$, then $A$ is

Options:
A Zero matrix
B Unit matrix
C Symmetric matrix
D Skew symmetric matrix

The correct answer is:
Skew symmetric matrix

Solution:
$a_{i j}=i^2-j^2$ is a square matrix. For a skew symmetric matrix $a_{i j}=-a_{j i}$
$\Rightarrow a_{i j}=i^2-j^2$ and $a_{j i}=j^2-i^2$
$\Rightarrow a_{i j}+a_{j i}=0 \Rightarrow a_{i j}=-a_{j i}$.
Hence, $a_{i j}$ is a skew symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
If A is a square matrix of order 2Ã2 and B=1234, such that AB=BA, then A can be

Options:
A 1467
B 1476
C 2224
D 3449

The correct answer is:
1467

Solution:
Let, A=abcd then abcd1234=1234abcd
âa+3b=a+2c,2a+4b=b+2d
c+3d=3a+4c,2c+4d=3b+4d
â3b=2c,a+c=d,
Hence, B=ab3b2a+3b2 where a,bâR
Now, we see A is correct

Topic: Mathematics, Matrices, JEE Main
If $A$ is a square matrix such that $A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$, then $\operatorname{det}(\operatorname{adj} A)$ is equal to

Options:
A $4$
B $16$
C $64$
D $256$

The correct answer is:
$16$

Solution:
$\begin{aligned} & \text { Given } A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right] \\ & \text { Now, }|A(\operatorname{adj} A)|=\left|\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right| \\ & \Rightarrow|A||\operatorname{adj} A|=4^3 \\ & \because \quad|\operatorname{adj} A|=|A|^{n-1} \\ & \text { Here, } \quad n=3 \\ & \therefore \quad|\operatorname{adj} A|=|A|^2 \\ & \left.\Rightarrow \quad|A| A A\right|^2=64 \\ & \Rightarrow \quad|A|^3=(4)^3 \\ & \Rightarrow \quad|A|=4 \\ & \end{aligned}$
$\begin{aligned} & \text { Now, } \quad|\operatorname{adj} A|=|A|^2=(4)^2 \\ & \therefore \quad|\operatorname{adj} A|=16 \\ & \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
If $A$ is a square matrix, then $A+A^{\top}$ is

Options:
A Non singular matrix
B Symmetric matrix
C Skew-symmetric matrix
D Unit matrix

The correct answer is:
Symmetric matrix

Solution:
$A+A^T$ is a square matrix.
$\left(A+A^T\right)^T=A^T+\left(A^T\right)^T=A^T+A$
Hence $A$ is a symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
If A is a symmetric and B is a skew symmetric matrix, then which of the following is correct?

Options:
A ABAT is a symmetric matrix
B ABT+BAT is a symmetric matrix
C A+BA-B is a skew symmetric matrix
D A+IB-I is a skew symmetric matrix

The correct answer is:
ABT+BAT is a symmetric matrix

Solution:
AT=A,BT=-B
ABATT=ATTBTAT=-ABAT (Skew symmetric)
ABT+BATT=ABTT+BATT
=BAT+ABT
=BA-AB=BAT+ABT (Symmetric)
A+BA-BT=A-BTA+BT
=AT-BTAT+BT=A+BA-B (Symmetric)
A+IB-IT=B-ITA+IT
=BT-ITAT+IT
=-B-IA+I (neither symmetric nor skew symmetric)

Topic: Mathematics, Matrices, JEE Main
If A is a symmetric matrix and B is skew- symmetric matrix such that A+B=235-1 , then AB is equal to:

Options:
A -4214
B 4-21-4
C 4-2-1-4
D -4-2-14

The correct answer is:
4-2-1-4

Solution:
Given A is symmetric and B is skew symmetric, hence
A=abbc and B=0d-d0
Now A+B=ab+db-dc=235-1
⇒a=2,  b=4,  c=-1 and d=-1
⇒A=244-1 and B=0-110
Hence AB=244-10-110=4-2-1-4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If $A$ is a symmetric matrix, then matrix $M^{\prime} A M$ is

Options:
A Symmetric
B Skew-symmetric
C Hermitian
D Skew-Hermitian

The correct answer is:
Symmetric

Solution:
$\left(M^{\prime} A M\right)^{\prime}=M^{\prime} A^{\prime} M=M^{\prime} A M$
$\{A$ is symmetric. Hence $M^{\prime} A M$ is a symmetric matrix).

Topic: Mathematics, Matrices, JEE Main
If $\vec{a}$ is a vector of magnitude 50 , collinear with the vector $\vec{b}=6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}$ and makes an acute angle with the positive direction of $Z$ - axis, then $\vec{a}$ is equal to

Options:
A $-24 \hat{i}+32 \hat{j}+30 \hat{k}$
B $24 \hat{i}-32 \hat{j}-30 \hat{k}$
C $-12 \hat{i}+16 \hat{j}-15 \hat{k}$
D $12 \hat{i}-16 \hat{j}-15 \hat{k}$

The correct answer is:
$-24 \hat{i}+32 \hat{j}+30 \hat{k}$

Solution:
Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e.,

$\begin{array}{l}

\vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right) \\

\Rightarrow|a|=|m| \sqrt{36+64+\frac{225}{4}} \\

\Rightarrow 50=\frac{25}{2}|m| \\

\Rightarrow|m|=4 \\

\Rightarrow m=\pm 4

\end{array}$

Since, a makes an acute angle with the positive direction of $Z$ - axis, so its $z$ component must be positive and hence, $m$ must be $-4$.

$\therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)=-24 \hat{i}+32 \hat{j}+30 \hat{k}$

Topic: Mathematics, Vector Algebra, JEE Main
If A is an invertible matrix of order 3 and B is another matrix of the same order as of A, such that B=2, ATAB=ABBT. If AB-1adjATB-1=K, then the value of 4K is equal to

Options:

The correct answer is:
0.25

Solution:
∵ATAB=ABBT
Taking determinant on both sides, we get,
ATAB=ABBT
⇒A=2
Now, AB-1adjATB-1
|A|×1|B|×1|adj(ATB)|
A×1B×1ATB2=AB3A2=116

Topic: Mathematics, Determinants, JEE Main