datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right]$, then the matrix $\mathrm{X}$ for which $2 \mathrm{X}+3 \mathrm{~A}=$ 0 holds true is

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

The correct answer is:
$\left[\begin{array}{ccc}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]$

Solution:
$\because 2 X+3 A=0$
$\Rightarrow x=\frac{-3}{2} A$
$\Rightarrow x=\frac{-3}{2}\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right]$
$\Rightarrow X=\left[\begin{array}{ccc}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 2)
If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and $\mathbf{A}^{2}-\mathrm{kA}-\mathrm{I}_{2}=\mathrm{O}$, where $\mathrm{I}_{2}$ is the $2 \times 2$ identity matrix, then what is the value of $\mathrm{k}$ ?

Options:
A 4
B $-4$
C 8
D $-8$

The correct answer is:
4

Solution:
$\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1+4 & 2+6 \\ 2+6 & 4+9\end{array}\right]=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]$
$\mathrm{A}^{2}-\mathrm{kA}-\mathrm{I}_{2}=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]-\left[\begin{array}{cc}\mathrm{k} & 2 \mathrm{k} \\ 2 \mathrm{k} & 3 \mathrm{k}\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}5-k-1 & 8-2 k-0 \\ 8-2 k-0 & 13-3 k-1\end{array}\right]$
$=\left[\begin{array}{cc}4-k & 8-2 k \\ 8-2 k & 12-3 k\end{array}\right]$
Given, $A^{2}-k A-I_{2}=0$
$\therefore 4-k=0 \Rightarrow k=4$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
If $\mathrm{A}=\left(\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right)$ then what is determinant of AB?

Options:
A 0
B 1
C 10
D 20

The correct answer is:
0

Solution:
$\begin{array}{l}\text { Let } \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right] \\ \mathrm{AB}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right] \Rightarrow \mathrm{AB}=\left[\begin{array}{ll}3 & 0 \\ 5 & 0\end{array}\right] \\ \text { So, }|\mathrm{AB}|=0 & (\because \text { one column is zero })\end{array}$

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}-1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5\end{array}\right]$ then which of the following is/are correct?
$1.$ $A$ and $B$ commute.
$2. A B$ is a null matrix. 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:
2 only

Solution:
$\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3\end{array}\right] ; \mathrm{B}=\left[\begin{array}{ccc}-1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5\end{array}\right]$
$\mathrm{AB}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\mathrm{BA}=\left[\begin{array}{ccc}-8 & 7 & -10 \\ 48 & -42 & 60 \\ 40 & -35 & 50\end{array}\right]$
as $\mathrm{AB} \neq \mathrm{BA}$
So $\mathrm{A}$ and $\mathrm{B}$ are not commute.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 2)
If $A=\left[\begin{array}{ccc}1 & -2 & -3 \\ 2 & 1 & -2 \\ 3 & 2 & 1\end{array}\right]$, then which one of the following is
correct?

Options:
A $A$ is symmetric matrix
B $A$ is anti- symmetric matrix
C $A$ is singular matrix
D $A$ is non-singular matrix

The correct answer is:
$A$ is non-singular matrix

Solution:
Here we see that its diagonal elements are not zero, so it is not anti-symmetric matrix. Now, $|\mathrm{A}|=1(1+4)+2(2+6)-3(4-3)$
$=5+16-3=18 \neq 0$
Hence, it is non-singular matrix.

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 2)
If $\mathrm{A}=\{1,2,3,4\}$ and $\mathrm{R}=\{(1,1),(1,3),(2,2),(3,1),$,
$(3,4),(4,3),(4,4)\}$ is a relation on $A \times A$, then which one of the following is correct?

Options:
A $R$ is reflexive
B $R$ is symmetric and transitive
C $R$ istransitive but not reflexive
D $R$ is neither reflexive nor transitive

The correct answer is:
$R$ is neither reflexive nor transitive

Solution:
Since, $3 \in \mathbf{A}$
but $(3,3) \notin \mathrm{R}$
So, it is notreflexive. and $(3,4) \in \mathrm{R}$ and $(4,3) \in \mathrm{R}$
but $(3,3) \notin \mathrm{R}$
So, it is also not transitive. Hence, $R$ is neither reflexive nor transitive.

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If $A=\{1,2,5,6\}$ and $B=\{1,2,3\}$, then what is $(A \times B) \cap(B \times A)$ equal to?

Options:
A $\{(1,1),(2,1),(6,1),(3,2)\}$
B $\{(1,1),(1,2),(2,1),(2,2)\}$
C $\{(1,1),(2,2)\}$
D $\{(1,1),(1,2),(2,5),(2,6)\}$

The correct answer is:
$\{(1,1),(1,2),(2,1),(2,2)\}$

Solution:
LetA $=\{1,2,5,6\}$ and $\mathrm{B}=\{1,2,3\}$ $\mathrm{A} \times \mathrm{B}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)$
$(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)\}$
and $\mathrm{B} \times \mathrm{A}=\{(1,1),(1,2),(1,5),(1,6),(2,1),(2,2)$,
$(2,5),(2,6),(3,1),(3,2),(3,5),(3,6)\}$
$\Rightarrow \quad(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})$
$=\{(1,1),(1,2),(2,1),(2,2)\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
If $A=\{1,2\}, B=\{2,3\}$ and $C=\{3,4\}$, then what is the cardinality of $(A \times B) \cap(A \times C)$

Options:
A 8
B 6
C 2
D 1

The correct answer is:
2

Solution:
$\mathrm{A} \times \mathrm{B}=\{(1,2),(1,3),(2,2),(2,3)\}$
$\mathrm{A} \times \mathrm{C}=\{(1,3),(1,4),(2,3),(2,4)\}$
Number of elementsin $(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})=2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
If $A=\{1,3,5,7\}$, then what is the cardinality of the power $\operatorname{set} P(A)$?

Options:
A 8
B 15
C 16
D 17

The correct answer is:
16

Solution:
Number of element in set A is Cordinality of the power set $\mathrm{P}(\mathrm{A})=2^{4}=16$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
If $a^{-1}+b^{-1}+c^{-1}=0$ such that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$,
then what is $\lambda$ equal to?

Options:
A $-a b c$
B $a b c$
C 0
D 1

The correct answer is:
$a b c$

Solution:
Let $a^{-1}+b^{-1}+c^{-1}=0$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow \quad \frac{b c+a c+a b}{a b c}=0$
$\Rightarrow a b+b c+c a=0$ ...(1)
Consider $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$
$\Rightarrow(1+a)[(1+b)(1+c)-1]-(1+c-1)+(1-1-b)=\lambda$
$\Rightarrow(1+a)(c+b+b c)-c-b=\lambda$
$\Rightarrow \mathrm{bc}+\mathrm{ac}+\mathrm{ab}+\mathrm{abc}=\lambda$
$\Rightarrow$ abc $=\lambda \quad$ (using (1))

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 2)
If $\left(a_{1} / x\right)+\left(b_{1} / y\right)=c_{1},\left(a_{2} / x\right)+\left(b_{2} / y\right)=c_{2}$
$\Delta_{1}=\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|, \quad \Delta_{3}=\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|$,
then $(x, y)$ is equal to which one of the following?

Options:
A $\left(\Delta_{2} / \Delta_{1}, \Delta_{3} / \Delta_{1}\right)$
B $\left(\Delta_{3} / \Delta_{1}, \Delta_{2} / \Delta_{1}\right)$
C $\left(-\Delta_{1} / \Delta_{2},-\Delta_{1} / \Delta_{3}\right)$
D $\left(-\Delta_{1} / \Delta_{2},-\Delta_{1} / \Delta_{3}\right)$

The correct answer is:
$\left(-\Delta_{1} / \Delta_{2},-\Delta_{1} / \Delta_{3}\right)$

Solution:
Let $\frac{1}{x}=u, \frac{1}{y}=v$
$\therefore a_{1} u+b_{1} v=c_{1}$ and $a_{2} u+b_{2} v=c_{2}$
Using the method of cross multiplication $\frac{u}{b_{1} c_{2}-b_{2} c_{1}}=\frac{v}{c_{1} a_{2}-c_{2} a_{1}}=\frac{-1}{a_{1} b_{2}-a_{2} b_{1}}$
$\Rightarrow \frac{\frac{1}{x}}{\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|}=\frac{\frac{1}{y}}{\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|}$ $=\frac{-1}{\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|}$
$\begin{aligned} & \frac{\frac{1}{x}}{\Delta_{2}}=\frac{\frac{1}{y}}{\Delta_{3}}=-\frac{1}{\Delta_{1}} \\ \therefore & \frac{1}{x}=-\frac{\Delta_{2}}{\Delta_{1}} \\ \text { and } \frac{1}{y}=-\frac{\Delta_{3}}{\Delta_{1}} \\ \Rightarrow x &=-\frac{\Delta_{1}}{\Delta_{2}} \text { and } y=-\frac{\Delta_{1}}{\Delta_{3}} \end{aligned}$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 2)
If $\overrightarrow{\mathrm{a}}=(2,1,-1), \overrightarrow{\mathrm{b}}=(1,-1,0), \overrightarrow{\mathrm{c}}=(5,-1,1)$, then what is the
unit vector parallel to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$ in the opposite direction?

Options:
A $\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{3}$
B $\frac{\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}$
C $\frac{2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}$
D None of the above

The correct answer is:
$\frac{2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}$

Solution:
Let $\overrightarrow{\mathrm{a}}=(2,1,-1), \overrightarrow{\mathrm{b}}=(1,-1,0), \overrightarrow{\mathrm{c}}=(5,-1,1)$
$\therefore \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=(-2,1,-2)$
Let $\overrightarrow{\mathrm{n}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$ be the unit vector which is $\|$ to $(-2$
$1,-2$ ) in the opposite direction.
$\therefore \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1$ and $\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \\ -2 & 1 & -2\end{array}\right|=0$
$\Rightarrow x=-2 y, y=y, z=-2 y$
$x^{2}+y^{2}+z^{2}=1 \Rightarrow 4 y^{2}+y^{2}+4 y^{2}=1 \Rightarrow y=\pm \frac{1}{3}$
Hence, the Required vector
$\hat{n}=\frac{2}{3} \hat{i}-\frac{\hat{j}}{3}+\frac{2}{3} \hat{k}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
If $A=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]$ then that is $A+3 A^{-1}$ equal to?

Options:
A $3 I$
B $5 I$
C $7 I$
D None of these where $I$ is the identity matrix order $2 .$

The correct answer is:
$7 I$

Solution:
$\mathrm{A}=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]$
Now, $\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj}(\mathrm{A})$
$=\frac{1}{(10-7)}\left[\begin{array}{cc}5 & -7 \\ -1 & 2\end{array}\right]$
$=\frac{1}{3}\left[\begin{array}{cc}5 & -7 \\ -1 & 2\end{array}\right]$
$\begin{aligned} \Rightarrow \quad & 3 \mathrm{~A}^{-1}=\left[\begin{array}{cc}5 & -7 \\ -1 & 2\end{array}\right] \\ & \text { Now, } \mathrm{A}+3 \mathrm{~A}^{-1}=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]+\left[\begin{array}{cc}5 & -7 \\ -1 & 2\end{array}\right] \end{aligned}$
$=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]=7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=7 I$ where $I$ is Identity Matrix.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$, then what is $\mathrm{A}^{\mathrm{n}}$ equal to?

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

The correct answer is:
$\left[\begin{array}{ll}2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1} \\ 2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1}\end{array}\right]$

Solution:
Given matrix is:
$\mathrm{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}4+4 & 4+4 \\ 4+4 & 4+4\end{array}\right]=\left[\begin{array}{ll}2^{3} & 2^{3} \\ 2^{3} & 2^{3}\end{array}\right]$
$\begin{aligned} \mathrm{A}^{3} &=\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}16+16 & 16+16 \\ 16+16 & 16+16\end{array}\right] \\ &=\left[\begin{array}{ll}32 & 32 \\ 32 & 32\end{array}\right]=\left[\begin{array}{cc}2^{5} & 2^{5} \\ 2^{5} & 2^{5}\end{array}\right] \end{aligned}$
Going this way we get
$\mathrm{A}^{4}=\left[\begin{array}{cc}2^{7} & 2^{7} \\ 2^{7} & 2^{7}\end{array}\right] \Rightarrow \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1} \\ 2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1}\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
If $\mathrm{A}=\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)$, then which one of the following is correct?

Options:
A $A^{2}=-2 A$
B $A^{2}=-4 A$
C $A^{2}=-3 A$
D $\mathrm{A}^{2}=4 \mathrm{~A}$

The correct answer is:
$A^{2}=-4 A$

Solution:
$\mathrm{A}=\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)$
$\mathrm{A}^{2}=\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)$
$=\left(\begin{array}{cc}4+4 & -4-4 \\ -4-4 & 4+4\end{array}\right)=\left(\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right)$
$=-4\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)=-4 \mathrm{~A}$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 2)
If $\mathrm{A}=\left[\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right]$ and $\operatorname{det}\left(\mathrm{A}^{3}\right)=125$, then $\alpha$ is equal to

Options:
A $\pm 1$
B $\pm 2$
C $\pm 3$
D $\pm 5$

The correct answer is:
$\pm 3$

Solution:
$\quad A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right], \operatorname{det}\left(A^{3}\right)=125$
$\left|A^{3}\right|=125 \Rightarrow|A|=5$
$\therefore \alpha 2-4=5 \Rightarrow \alpha^{2}=9 \Rightarrow \alpha=\pm 3$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 1)
If $\mathrm{A}(2,3), \mathrm{B}(1,4), \mathrm{C}(0-2)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ are the vertices of a
parallelogram, then what is the value of $(\mathrm{x}, \mathrm{y})$ ?

Options:
A $(1,-3)$
B $(2,4)$
C (1,1)
D (0,0)

The correct answer is:
$(1,-3)$

Solution:
As given : $\mathrm{A}(2,3), \mathrm{B}(1,4), \mathrm{C}(0,-2)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ are the
vertices of a parallelogram. Diagonals of a parallelogram bisect each other So, mid-point are same for both diagonals $\mathrm{AC}$ and $\mathrm{BD}$.
$\frac{2+0}{2}=\frac{1+x}{2}$ and $\frac{3-2}{2}=\frac{4+y}{2}$
$\Rightarrow \quad x=1$ and $y=-3$
$\Rightarrow \quad \mathrm{D}(\mathrm{x}, \mathrm{y})=(1,-3)$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 1)
If $A=\{2,3\}, B=\{4,5\}, C=\{5,6\}$, then what is the number of elements of $A \times(B \cap C)$ ?

Options:
A 2
B 4
C 6
D 8

The correct answer is:
2

Solution:
Given $\mathrm{A}=\{2,3\}, \mathrm{B}=\{4,5\}, \mathrm{C}=\{5,6\}$
: $B \cap C=\{5\}$
$\Rightarrow \quad A \times(B \cap C)=\{2,3\} \times\{5\}=\{(2,5),(3,5)\}$
Hence, required number of elements in $A \times(B \cap C)=2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
If $a, 2 a+2,3 a+3$ are in GP, then what is the fourth term of the GP?

Options:
A $-13.5$
B $13.5$
C $-27$
D 27

The correct answer is:
$-13.5$

Solution:
Since, a, $2 \mathrm{a}+2$ and $3 \mathrm{a}+3$ are in GP $\therefore(2 a+2)^{2}=a(3 a+3)$
$\Rightarrow 4 \mathrm{a}^{2}+4+8 \mathrm{a}=3 \mathrm{a}^{2}+3 \mathrm{a} \Rightarrow \mathrm{a}^{2}+5 \mathrm{a}+4=0$
$\Rightarrow a(a+4)+1(a+4)=0 \Rightarrow(a+4)(a+1)=0$
$\Rightarrow a+4=0$ or $a+1=0$
$\Rightarrow a=-4$ or $-1$
Let the fourth term be $\mathrm{x}$.
$\frac{a}{2 a+2}=\frac{3 a+3}{x}$
$\Rightarrow x=\frac{(3 a+3)(2 a+2)}{a}$
When $\mathrm{a}=-4$
$\mathrm{x}=-13.5$
and $\mathrm{a}=-1, \mathrm{x}=0$
So, the fourth term is $-13.5$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
If $|\vec{a}|=2$ and $|\vec{b}|=3$, then $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}$ is equal to

Options:
A 72
B 64
C 48
D 36

The correct answer is:
36

Solution:
$\begin{aligned} & \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n} \\ & \therefore|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=6 \sin \theta \\ & \&|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}| \cos \theta=6 \cos \theta \\ &|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=(6 \sin \theta)^{2}+(6 \cos \theta)^{2} \\ &=36\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ &=36 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
If $a^{2}+b^{2}+c^{2}=0$, then what is $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
equal to ?

Options:
A $\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$
B $-a^{2} b^{2} c^{2}$
C $\mathrm{abc}$
D $3 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$

The correct answer is:
$-a^{2} b^{2} c^{2}$

Solution:
We know that if $\mathrm{x}+\mathrm{y}$ ' Hereboth in nominator and denominator $a^{4}-b^{4}+b^{4}-c^{4}+c^{4}-a^{4}=0$
and $a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0$
Hence, $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
$=\frac{3\left(a^{4}-b^{4}\right)\left(b^{4}-c^{4}\right)\left(c^{4}-a^{4}\right)}{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}$
$=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right.$
$=\left(-c^{2}\right)\left(-a^{2}\right)\left(-b^{2}\right)$
$=-a^{2} b^{2} c^{2}$
$\left(\because a^{2}+b^{2}+c^{2}=0\right)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 2)
If $|\overrightarrow{\mathrm{a}}|=\sqrt{2},|\overrightarrow{\mathrm{b}}|=\sqrt{3}$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{6}$, then what is $|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$
equal to?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Consider $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=2\left(|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}\right)$
By putting the values of $|\overrightarrow{\mathrm{a}}|,|\overrightarrow{\mathrm{b}}|$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|$, we get
$6+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=2(2+3)$
$\Rightarrow \quad|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 1)
If $\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}=\overrightarrow{0}$ and $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$, then
what is the value of $\lambda ?$

Options:
A 2
B 3
C 4
D 6

The correct answer is:
6

Solution:
$\vec{a}+2 \vec{b}+3 \vec{c}=\overrightarrow{0}$ ...(1)
$\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=\lambda(\vec{b} \times \vec{c})$ ...(2)
$(1) \times \vec{b} \Rightarrow \vec{b} \times \vec{a}+2 \vec{b} \times \vec{b}+3 \vec{b} \times \vec{c}=0$
$\Rightarrow-\vec{a} \times \vec{b}+0+3 \vec{b} \times \vec{c}=0$
$\Rightarrow \vec{a} \times \vec{b}=3 \vec{b} \times \vec{c}$ ...(3)
$(1) \times \vec{c} \Rightarrow \vec{c} \times \vec{a}+2 \vec{c} \times \vec{b}+3 \vec{c} \times \vec{c}=0$
$\Rightarrow \vec{c} \times \vec{a}-2 \vec{b} \times \vec{c}+0=0$
$\Rightarrow \vec{c} \times \vec{a}=2 \vec{b} \times \vec{c}$ ...(4)
$\therefore(2) \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$
$\Rightarrow 3 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+2(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$
$\Rightarrow 6(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$
$\therefore \lambda=6$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 1)
If $|\vec{a}|=2,|\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$, then what is the value of
$\vec{a} \cdot \vec{b} ?$

Options:
A 4
B 6
C 8
D 10

The correct answer is:
6

Solution:
We know that
$|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2} \times|\overrightarrow{\mathbf{b}}|^{2}$
$\therefore \quad 64+|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^{2}=(4 \times 25)$
$\Rightarrow|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^{2}=36$
$\Rightarrow \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=6$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 1)
If $|\vec{a}|=2,|\vec{b}|=5$ and $|\vec{a} \times \vec{b}|$, then what is $\vec{a} \cdot \vec{b}$ equal to ?

Options:
A 6
B 7
C 8
D 9

The correct answer is:
6

Solution:
Given $|\mathrm{a}|=2,|\mathrm{~b}|=5$ and $|\mathrm{a} \times \mathrm{b}|=8$
Also $|\mathrm{a} \times \mathrm{b}|=|\mathrm{a}| \cdot|\mathrm{b}| \cdot|\sin \theta|$
$\Rightarrow|\sin \theta|=\frac{8}{2 \times 5}=\frac{4}{5}$
$\Rightarrow|\cos \theta|=\frac{3}{5} \Rightarrow \cos \theta=\pm \frac{3}{5}$
$\therefore \mathrm{a} \cdot \mathrm{b}=|\mathrm{a}| \cdot|\mathrm{b}| \cos \theta=2 \times 5 \times \frac{3}{5}=6$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are three
vectors such that $\vec{a}+t \vec{b}$ is perpendicular to $\vec{c}$, then what is t equal to ?

Options:
A 8
B 6
C 4
D 2

The correct answer is:
8

Solution:
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2-\mathrm{t}) \hat{\mathrm{i}}+(2+2 \mathrm{t}) \hat{\mathrm{j}}+(3+\mathrm{t}) \hat{\mathrm{k}}$
$(\vec{a}+t \vec{b})$ and $\vec{c}$ is perpendicular. Therefore,
$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0$
$3(2-\mathrm{t})+2+2 \mathrm{t}=0$
$6-3 \mathrm{t}+2 \mathrm{t}+2=0$
$\mathrm{t}=8$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}$ are perpendicular,
then what is the value of $\lambda ?$

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
$\quad \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}$
Given, $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are $\perp^{\mathrm{r}}$.
$\therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0$
$\Rightarrow 6+6-4 \lambda=0$
$\Rightarrow 4 \lambda=12 \Rightarrow \lambda=3 .$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 1)