datavorous/entrance-exam-dataset huggingface dataset:
If $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}$ are
three coplanar vectors and $|\overrightarrow{\mathrm{c}}|=\sqrt{6}$, then which one of the following is correct ?

Options:
A $\quad \mathrm{m}=2$ and $\mathrm{n}=\pm 1$
B $\mathrm{m}=\pm 2$ and $\mathrm{n}=-1$
C $\quad \mathrm{m}=2$ and $\mathrm{n}=-1$
D $\quad \mathrm{m}=\pm 2$ and $\mathrm{n}=1$

The correct answer is:
$\quad \mathrm{m}=\pm 2$ and $\mathrm{n}=1$

Solution:
$\quad \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
and $\vec{c}=\hat{i}+m \hat{j}+n \hat{k} ;|\vec{c}|=\sqrt{6}$
Given, $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are coplanar.
So, $\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 2 \\ 1 & \mathrm{~m} & \mathrm{n}\end{array}\right|=0$
$\mathrm{c}_{1} \rightarrow \mathrm{c}_{1}+\mathrm{c}_{2} \quad \mathrm{c}_{3} \rightarrow \mathrm{c}_{3}+\mathrm{c}_{2}$
$\Rightarrow\left|\begin{array}{ccc}0 & 1 & 0 \\ 5 & 3 & 5 \\ 1+\mathrm{m} & \mathrm{m} & \mathrm{m}+\mathrm{n}\end{array}\right|=0$
$\Rightarrow 0+1(5 m+5 n)-(5+5 m)=0$
$\Rightarrow 5 \mathrm{~m}+5 \mathrm{n}-5-5 \mathrm{~m}=0 \Rightarrow 5 \mathrm{n}=5 \Rightarrow \mathrm{n}=1$
$|\vec{c}|=6 \Rightarrow \sqrt{1+m^{2}+n^{2}}=\sqrt{6}$
$\Rightarrow 1+\mathrm{m}^{2}+\mathrm{n}^{2}=6 \Rightarrow 2+\mathrm{m}^{2}=6 \Rightarrow \mathrm{m}^{2}=4 \Rightarrow \mathrm{m}=\pm 2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 1)
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-\hat{k}$, then what is
$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})$ equal to?

Options:
A $2 \hat{i}+3 \hat{j}-\hat{k}$
B $2 \hat{i}-3 \hat{j}-\hat{k}$
C $3 \hat{i}+\hat{j}+\hat{k}$
D $\overrightarrow{0}$

The correct answer is:
$\overrightarrow{0}$

Solution:
Since, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\Rightarrow \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})+\overrightarrow{\mathbf{b}} \times(\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}})+\overrightarrow{\mathbf{c}} \times(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}})$
$=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})-(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})-(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})-(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 1)
If a $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{b}}+\hat{\mathrm{k}}$, and $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}$ are coplanar vectors,
then what is the value of $a+b+c-a b c ?$

Options:
A 0
B 1
C 2
D -2

The correct answer is:
2

Solution:
Vectors a $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}$ are
coplanar vectors. $\Rightarrow\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$
$\Rightarrow$ a $(b c-1)-1(c-1)+1(1-b)=0$
$\Rightarrow \quad a b c-a-c+1+1-b=0$
$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}-\mathrm{abc}=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 2)
If $\overrightarrow{\mathbf{a}}=\hat{\mathrm{i}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{c}}=y \hat{\mathbf{i}}+\mathrm{x} \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$
then $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$ depends on

Options:
A $x$ only
B $y$ only
C Both $x$ and $y$
D Neither x nor $y$

The correct answer is:
Neither x nor $y$

Solution:
Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$
and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$
$\begin{array}{l}\text { Now, }(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & 1 & (1-x) \\ y & x & (1+x-y)\end{array}\right| \\ \begin{aligned}=& i\left(1+x-y-x+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right) \\+& k\left(x^{2}-y\right) \end{aligned} \\ =\hat{i}\left(1-y+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right)+\hat{k}\left(x^{2}-y\right) \\ \text { Now, } \vec{a} .(\vec{b} \times \vec{c})=1\left(1-y+x^{2}\right)+0\left(x+x^{2}-x y-y\right) \\ -1\left(x^{2}-y\right) \\ =1-y+x^{2}-x^{2}+y & \end{array}$
$=1$ which shows that $\vec{a} .(\vec{b} \times \vec{c})$ does not depend
on $x$ and $y$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
If $\mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}}$ where $\mathrm{i}=\sqrt{-1}$ then what is the value of $\mathrm{A}$ ?

Options:
A -8
B 0
C 4
D 8

The correct answer is:
0

Solution:
$\mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}} \times \frac{1+2 \mathrm{i}}{1+2 \mathrm{i}}(\mathrm{ByRationalizing})$
$=\frac{4+2 \mathrm{i}+8 \mathrm{i}-4}{(1)^{2}-(2 \mathrm{i})^{2}}=\frac{0+10 \mathrm{i}}{5}=2 \mathrm{i}=0+2 \mathrm{i}$
$\Rightarrow \mathrm{A}=0, \mathrm{~B}=2$

Topic: Mathematics, Complex Number, NDA, NDA 2012 (Phase 1)
If A is a $2 \times 3$ matrix and $\mathrm{AB}$ is a $2 \times 5$ matrix, then B must be a

Options:
A $3 \times 5$ matrix
B $5 \times 3$ matrix
C $3 \times 2$ matrix
D $5 \times 2$ matrix

The correct answer is:
$3 \times 5$ matrix

Solution:
A is $2 \times 3$ matrix
$\mathrm{AB}$ is $2 \times 5$ matrix
Let ' $\mathrm{B}$ ' bem $\times \mathrm{n}$ matrix
$[\mathrm{A}]_{2 \times 3}[\mathrm{~B}]_{\mathrm{m} \times \mathrm{n}}=[\mathrm{AB}]_{2 \times 5}$
number of columns of $\mathrm{A}=$ number of rows of $\mathrm{B}$. $\therefore m=3$
we can observe that $\mathrm{n}=5$ from the product. So, $\mathrm{B}$ is $3 \times 5$ matrix.

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 1)
If $\mathrm{A}$ is a $3 \times 3$ matrix such that $|\mathrm{A}|=4$, then what is $\mathrm{A}(\operatorname{adj} \mathrm{A})$ equal to?

Options:
A $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
B $\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$
C $\left[\begin{array}{ccc}16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16\end{array}\right]$
D Cannot be determined, as data is insufficient

The correct answer is:
$\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$

Solution:
We know that $\mathrm{A}^{-1}=\frac{(\mathrm{adjA})}{|\mathrm{A}|}$
or, $\mathrm{AA}^{-1}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
or, $\mathrm{I}_{\mathrm{n}}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
$\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}=|\mathrm{A}|\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
If A is a matrix of order $\mathrm{p} \times \mathrm{q}$ and $\mathrm{B}$ is a matrix of order $\mathrm{s} \times \mathrm{t}$, under which one of the following conditions does AB exist?

Options:
A $\mathrm{p}=\mathrm{t}$
B $\mathrm{p}=\mathrm{s}$
C $q=t$
D $\mathrm{q}=\mathrm{s}$

The correct answer is:
$\mathrm{q}=\mathrm{s}$

Solution:
If A is a matrix of order $\mathrm{p} \times \mathrm{q}$ and $\mathrm{B}$ is a matrix of order s $\times \mathrm{t}$, then $\mathrm{AB}$ will exist when the number of column in $\mathrm{A}$ is equal to the number of rows in $\mathrm{B}$ $\Rightarrow \mathrm{q}=\mathrm{s}$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
If A is a non-null row matrix with 5 columns and B is a nonnull column matrix with 5 rows, how manyrows are there in $\mathrm{A} \times \vec{B}$ ?

Options:
A 1
B 5
C 10
D 25

The correct answer is:
1

Solution:
If there is matrix of $m$ rows and $n$ columns and another with $n$ rows and k columns, their product will be a matrix of $\mathrm{m}$ rows and $\mathrm{k}$ column. A is a non-null matrix with one row and 5 columns and $\mathrm{B}$ is a non-null matrix with 5 rows and one column. Therefore number of row in $\mathrm{A} \times \mathrm{B}$ is 1.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
If $\vec{a}$ is a non-zero vector of modulus a and $\lambda$ is a non-zero
scalar and $\lambda \vec{a}$ is a unit vector then

Options:
A $\quad \lambda=\pm 1$
B $\mathrm{a}=|\lambda|$
C $\mathrm{a}=\frac{1}{|\lambda|}$
D $\mathrm{a}=\frac{1}{\lambda}$ only

The correct answer is:
$\mathrm{a}=\frac{1}{|\lambda|}$

Solution:
As given, $\lambda \overrightarrow{\mathrm{a}}$ is a unit vector.
$\Rightarrow|\lambda \overrightarrow{\mathrm{a}}|=1$
$\Rightarrow|\lambda| \overrightarrow{\mathrm{a}} \mid=1$
$\Rightarrow \mathrm{a}=\frac{1}{|\lambda|} \quad[\because|\overrightarrow{\mathrm{a}}|=\mathrm{a}]$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
If $\vec{a}$ is a position vector of a point $(1,-3)$ and $A$ is another point $(-1,5)$, then what are the coordinates of the point $B$ such that $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ ?

Options:
A $(2,0)$
B $(0,2)$
C $(-2,0)$
D $(0,-2)$

The correct answer is:
$(0,2)$

Solution:
Let the coordinates of $\mathrm{B}$ be $(\mathrm{x}, \mathrm{y})$.
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}$
P.V. of $\mathrm{A}$ is $(-1,5) \mathrm{so}, \overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+5 \hat{\mathrm{j}}, \overrightarrow{\mathrm{OB}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}$
$\therefore \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$
$\Rightarrow(\mathrm{x}+1) \hat{\mathrm{i}}+(\mathrm{y}-5) \overline{\mathrm{j}}=\overline{\mathrm{i}}-3 \overline{\mathrm{j}}$
$\Rightarrow \mathrm{x}+1=1$ and $\mathrm{y}-5=-3$
$\Rightarrow \mathrm{x}=0$ and $\mathrm{y}=2$
$\therefore \quad$ Coordinates of $\mathrm{B}$ are $(0,2)$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
If $\mathrm{A}$ is a relation on a set $\mathrm{R}$, then which one of the following is correct?

Options:
A $\quad R \subseteq A$
B $\mathrm{A} \subseteq \mathrm{R}$
C $A \subseteq(R \times R)$
D $\quad \mathrm{R} \subseteq(\mathrm{A} \times \mathrm{A})$

The correct answer is:
$A \subseteq(R \times R)$

Solution:
Since A is a relation on a set $R$
$\therefore \mathrm{A} \subseteq(\mathrm{R} \times \mathrm{R})$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
If $A$ is a square matrix of order 3 with $|A| \neq 0$, then which one of the following is correct?

Options:
A $|a d j A|=|A|$
B $|a d j A|=|A|^{2}$
C $|a d j A|=|A|^{3}$
D $|a d j A|^{2}=|A|$

The correct answer is:
$|a d j A|=|A|^{2}$

Solution:
$|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1} \quad$ \{n is order of square matrix $\}$
If A is square matrix of order 3 , then $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{2}$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 2)
If $\mathrm{A}$ is a square matrix of order $\mathrm{n}>1$, then which one of the $\begin{array}{ll}\text { following is correct? } \end{array}$

Options:
A $\operatorname{det}(-\mathrm{A})=\operatorname{det} \mathrm{A}$
B $\operatorname{det}(-\mathrm{A})=(-1)^{\mathrm{n}} \operatorname{det} \mathrm{A}$
C $\operatorname{det}(-\mathbf{A})=-\operatorname{det} \mathbf{A}$
D $\operatorname{det}(-\mathrm{A})=\mathrm{n} \operatorname{det} \mathrm{A}$

The correct answer is:
$\operatorname{det}(-\mathrm{A})=(-1)^{\mathrm{n}} \operatorname{det} \mathrm{A}$

Solution:
$\mathrm{A}$ is square matrix of order $\mathrm{n}>1$. $\operatorname{det}(-\mathrm{A})=(-1)^{\mathrm{n}} \operatorname{det} \mathrm{A}$

Topic: Mathematics, Determinants, NDA, NDA 2019 (Phase 1)
If A is a square matrix such that $\mathrm{A}^{2}=\mathrm{I}$ where $\mathrm{I}$ is the identity matrix, then what is $\mathrm{A}^{-1}$ equal to?

Options:
A $\mathrm{A}+1$
B Null matrix
C $\mathrm{A}$
D Transpose of A

The correct answer is:
$\mathrm{A}$

Solution:
Let $\mathrm{A}^{2}=\mathrm{I}$
$\Rightarrow \mathrm{A}^{2} \mathrm{~A}^{-1}=\mathrm{IA}^{-1}$
$\Rightarrow \mathrm{A}=\mathrm{A}^{-1}$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 1)
If A is a square matrix such that $\mathrm{A}-\mathrm{A}^{\mathrm{T}}=0$, then which one of the following is correct?

Options:
A A must be a null matrix
B A must be a unit matrix
C A must be a scalar matrix
D None of the above

The correct answer is:
None of the above

Solution:
Since, A is a square matrix and $\mathrm{A}-\mathrm{A}^{\mathrm{T}}=0 \Rightarrow \mathrm{A}=\mathrm{A}^{\mathrm{T}}$
A is a symmetric matrix. Considering following two points.
$1.$ No two rows or two columns should be identical
and
$2.$ There should be two l's and one 0 , in every row or column. Such determinant can be found.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
If $\mathrm{A}$ is a square matrix, then the value of $\operatorname{adj} \mathrm{A}^{\mathrm{T}}-(\operatorname{adj} \mathrm{A})^{\mathrm{T}}$ is equal to

Options:
A A
B $2|\mathrm{~A}| \mathrm{I}$, where $\mathrm{I}$ is the identity matrix
C null matrix whose order is same as that of A
D unit matrix whose order is same as that of A

The correct answer is:
null matrix whose order is same as that of A

Solution:
We know, Adj $\mathrm{A}^{\mathrm{T}}=(\operatorname{adj} \mathrm{A})^{\mathrm{T}}$
$\therefore \mathrm{adj} \mathrm{A}^{\mathrm{T}}-(\mathrm{adj} \mathrm{A})^{\mathrm{T}}=\operatorname{adj} \mathrm{A}^{\mathrm{T}}-\mathrm{adj} \mathrm{A}^{\mathrm{T}}=0$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 2)
If $\mathrm{A}$ is a square matrix, then what is $\operatorname{adj}\left(\mathrm{A}^{-1}\right)-(\operatorname{adj} \mathrm{A})^{-1}$ equal to?

Options:
A $2|\mathrm{~A}|$
B Nullmatrix
C Unitmatrix
D None of the above

The correct answer is:
Nullmatrix

Solution:
Let $[A]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$A^{-1}=\frac{1}{|A|}($ adj. $A)$
$\quad=\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -c \\ -b & a\end{array}\right]^{T}=\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$\operatorname{adj}\left(A^{-1}\right)=\frac{1}{(a d-b c)}\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]^{T}=\frac{1}{(a d-b c)}\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ ...(1)
$$
\begin{array}{l}a d j A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right] \\ |\operatorname{adj} A|=(a d-b c)\end{array}
$$
$(\operatorname{adj} A)^{-1}=\frac{1}{|a d j A|}\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]^{T}=\frac{1}{a d-b c}\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ ...(2)
Subtracting eqn. (1) and (2),
$\operatorname{adj}\left(A^{-1}\right)-(\operatorname{adj} A)^{-1}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$=$ null matrix.

Topic: Mathematics, Matrices, NDA, NDA 2016 (Phase 1)
If $A$ is a square matrix, then what is adj $A^{T}-(a d j A)^{\mathrm{T}}$ equal to?

Options:
A $2|A|$
B $2|A| I$
C Null Matrix
D Unit Matrix

The correct answer is:
Null Matrix

Solution:
We know $\left(\operatorname{adj} \mathrm{A}^{\mathrm{T}}\right)=(\operatorname{adj} \mathrm{A})^{\mathrm{T}}$
$\Rightarrow\left(\operatorname{adj} \mathrm{A}^{\mathrm{T}}\right)-(\operatorname{adj} \mathrm{A})^{\mathrm{T}}=$ Null matrix

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 2)
If $A$ is a subset of $B$, then which one of the following is correct?

Options:
A $A^{c} \subset B^{c}$
B $B^{c} \subset A^{c}$
C $A^{c}=B^{t}$
D $A \subseteq A \cap B$

The correct answer is:
$B^{c} \subset A^{c}$

Solution:
Suppose $U=\{a, b, c, d, e, f, g, h\}$ $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}$
$B=\{a, b, c, d, e\}$ Given $A \subseteq B$
$\mathrm{A}^{\mathrm{c}}=\{\mathrm{e}, \mathrm{f}, \mathrm{g}, \mathrm{h}\}, \mathrm{B}^{\mathrm{C}}=\{\mathrm{f}, \mathrm{g}, \mathrm{h}\}$
Hence, $\mathrm{B}^{\mathrm{C}} \subseteq \mathrm{A}^{\circ}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
If $\mathrm{A}$ is an identity matrix of order 3 , then its inverse $\left(\mathrm{A}^{-1}\right)$

Options:
A is equal to null matrix
B is equal to $\mathrm{A}$
C is equal to $3 \mathrm{~A}$
D does not exist

The correct answer is:
is equal to $\mathrm{A}$

Solution:
Given, A is an identity matrix. $\therefore \mathrm{A}=\mathrm{I}$
We know, $\mathrm{I}^{-1}=\mathrm{I}$
$\therefore \mathrm{A}^{-1}=\mathrm{A}$

Topic: Mathematics, Matrices, NDA, NDA 2019 (Phase 1)
If $A$ is an invertible matrix, then what is det $\left(A^{-1}\right)$ equal to?

Options:
A $\operatorname{det} A$
B $\frac{1}{\operatorname{det} A}$
C 1
D None of the above

The correct answer is:
$\frac{1}{\operatorname{det} A}$

Solution:
$\operatorname{det}\left(\mathrm{A}^{-1}\right)=\frac{1}{\operatorname{det} \mathrm{A}}$

Topic: Mathematics, Determinants, NDA, NDA 2014 (Phase 2)
If A is an orthogonal matrix of order 3 and $\mathrm{B}=\left[\begin{array}{ccc}1 & 2 & 3 \\ -3 & 0 & 2 \\ 2 & 5 & 0\end{array}\right]$, then which of the following is/are correct?
$1.$ $|\mathrm{AB}|=\pm 47$
$2. \mathrm{AB}=\mathrm{BA}$
Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
The determinent of a orthogonal matrix is always $\pm 1$ $|\mathrm{A}|=\pm 1$
$\mathrm{B}=\left[\begin{array}{ccc}1 & 2 & 3 \\ -3 & 0 & 2 \\ 2 & 5 & 0\end{array}\right]$
$|B|=-10-2(-4)+3(-15)$
$=-47$
$|\mathrm{AB}|=|\mathrm{A}||\mathrm{B}|$
$=(\pm 1)(-47)$
$=\pm 47$

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 2)
If A is any $2 \times 2$ matrix such that $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \mathrm{A}=\left[\begin{array}{cc}-1 & 0 \\ 6 & 3\end{array}\right]$
then what is A equal to?

Options:
A $\left[\begin{array}{ll}-5 & 1 \\ -2 & 2\end{array}\right]$
B $\left[\begin{array}{cc}-5 & -2 \\ 1 & 2\end{array}\right]$
C $\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$
D $\left[\begin{array}{cc}5 & 2 \\ -2 & -1\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$

Solution:
Let $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\mathrm{B}$
Then $\mathrm{BA}=\left[\begin{array}{cc}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow A=B^{-1}\left[\begin{array}{ll}-1 & 0 \\ -6 & 3\end{array}\right]$
$|\mathrm{B}|=3$
$\operatorname{adj} \mathrm{B}=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{A}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{rr}-1 & 0 \\ 6 & 3\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}-3-12 & -6 \\ 6 & 3\end{array}\right]$
$=\left[\begin{array}{rr}-5 & -2 \\ 2 & 1\end{array}\right]$
Aliter:
$\operatorname{Let} A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
then $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}\mathrm{a}+2 \mathrm{c} & \mathrm{b}+2 \mathrm{~d} \\ 0+3 \mathrm{c} & 0+3 \mathrm{~d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow 3 \mathrm{c}=6$ or $\mathrm{c}=2$
$3 \mathrm{~d}=3$ or $\mathrm{d}=1$
$\mathrm{a}+2 \times 2=-1$ or $\mathrm{a}=-5$
$\mathrm{~b}+2 \times 1=0, \mathrm{~b}=-2$
So, $\mathrm{A}=\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
If $A$ is any matrix, then the product $A A$ is defined only when $A$ is a matrix of order $m \times n$ where:

Options:
A $m>n$
B $m < n$
C $m=n$
D $m \leq n$

The correct answer is:
$m=n$

Solution:
A A is defined only when $A$ is a matrix of order $m \times n$ where $m=n$ $\mathrm{A} \times \mathrm{A}=(\mathrm{m} \times \mathrm{n})(\mathrm{m} \times \mathrm{n})=(\mathrm{m} \times \mathrm{n})(\mathrm{n} \times \mathrm{n})$ if $\mathrm{m}=\mathrm{n}$
$=\mathrm{m} \times \mathrm{n}=\mathrm{n} \times \mathrm{n}$ or $\mathrm{m} \times \mathrm{m}$
$=\mathrm{A}$ is a square matrix.

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 1)
If A is matrix of order $3 \times 2$ and $\mathrm{B}$ is matrix of order $2 \times 3$, then what is $|\mathrm{kAB}|$ equal to (where $\mathrm{k}$ is any scalar quantity)?

Options:
A $\mathrm{k} \mid \mathrm{AB}$
B $\mathrm{k}^{2} \mid \mathrm{AB}$
C $\mathrm{k}^{3} \mid \mathrm{AB}$
D $|\mathrm{AB}|$

The correct answer is:
$\mathrm{k}^{3} \mid \mathrm{AB}$

Solution:
As given of $\mathrm{A}$ and $\mathrm{B}$ are $3 \times 2$ and $2 \times 3$ respectively. $\Rightarrow$ order of $\mathrm{AB}$ is $3 \times 3$ $\Rightarrow|\mathrm{kAB}|=\mathrm{k}^{3} \mid \mathrm{AB}$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 1)
If A is semy square matrix of order 3 and det $\mathrm{A}=5$, then what is $\operatorname{det}\left[(2 \mathrm{~A})^{-1}\right]$ equal to?

Options:
A $1 / 10$
B $2 / 5$
C $8 / 5$
D $1 / 40$

The correct answer is:
$1 / 40$

Solution:
$|A|=5 \Rightarrow|2 A|=2^{3} \times 5=40$
$\Rightarrow\left|A^{-1}\right|=\frac{1}{|A|} \Rightarrow\left|(2 A)^{-1}\right|=\frac{1}{|2 A|}=\frac{1}{40}$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 2)
If $\left|\begin{array}{ccc}a & b & c \\ l & m & n \\ p & q & r\end{array}\right|=2$, then what is the value of the determinant $\left|\begin{array}{ccc}6 a & 3 b & 15 c \\ 2 l & m & 5 n \\ 2 p & q & 5 r\end{array}\right| ?$

Options:
A 10
B 20
C 40
D 60

The correct answer is:
60

Solution:
Let $\left|\begin{array}{ccc}a & b & c \\ l & m & n \\ p & q & r\end{array}\right|=2$
Consider $\left|\begin{array}{ccc}6 a & 3 b & 15 c \\ 2 l & m & 5 n \\ 2 p & q & 5 r\end{array}\right|=2 \times 5\left|\begin{array}{ccc}3 a & 3 b & 3 c \\ l & m & n \\ p & q & r\end{array}\right|$
$=10 \times 3\left|\begin{array}{lll}a & b & c \\ l & m & n \\ p & q & r\end{array}\right|$
$=30\left|\begin{array}{lll}a & b & c \\ l & m & n \\ p & q & r\end{array}\right|=30 \times 2=60$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 1)
If a line is perpendicular to the line $5 \mathrm{x}-\mathrm{y}=0$ and forms a triangle of area 5 square units with co-ordinate axes, then its equation is

Options:
A $\quad x+5 y \pm 5 \sqrt{2}=0$
B $x-5 y \pm 5 \sqrt{2}=0$
C $5 x+y \pm 5 \sqrt{2}=0$
D $5 x-y \pm 5 \sqrt{2}=0$

The correct answer is:
$\quad x+5 y \pm 5 \sqrt{2}=0$

Solution:
$\quad 5 \mathrm{x}-\mathrm{y}=0$
$\mathrm{y}=5 \mathrm{x}$...(1)
Slope $=5$
Slope of perpendicular line will be $-\frac{1}{5}$
Let equation of line is $y=\left(-\frac{1}{5}\right) x+c$...(2)
Putting $\mathrm{y}=0$ $x=5 c$
$O B=5 c$
Intersecting point $\mathrm{A}$
Putting $\mathrm{x}=0$
$\mathrm{y}=-\frac{1}{5} \times 0+\mathrm{c}$
$\mathrm{y}=\mathrm{c}$
area $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{c} \times 5 \mathrm{c}$
$5=\frac{1}{2} \times 5 \mathrm{c}^{2}$
$c=\pm \sqrt{2}$
$\mathrm{y}=\left(-\frac{1}{5}\right) \mathrm{x} \pm \sqrt{2}$
$5 y+x \pm 5 \sqrt{2}=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)