datavorous/entrance-exam-dataset huggingface dataset:
If $(4,0)$ and $(-4,0)$ are the foci of an ellipse and the semi-
minor axis is 3 , then the ellipse passes through which one of the following points?

Options:
A $(2,0)$
B $(0,5)$
C $(0,0)$
D $(5,0)$

The correct answer is:
$(5,0)$

Solution:
Foci of an ellipse are $(4,0)$ and $(-4,0)$. (Given) $2 a e=8 \Rightarrow a e=4$
and semiminor axis is $3 \quad \therefore b=3$
We know that, $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$\Rightarrow\left(\frac{4}{a}\right)^{2}=\left(1-\frac{9}{a^{2}}\right)\left(\because e=\frac{4}{a}, b=3\right)$
$\Rightarrow \frac{16}{a^{2}}=\frac{a^{2}-9}{a^{2}}$
$\Rightarrow 16=a^{2}-9 \Rightarrow a^{2}=25 \Rightarrow a=5$
Now, standard equation of an ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Thus, the equation of an ellipse is
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
which is satisfied by $(5,0)$. Hence the ellipse passes through $(5,0)$.

Topic: Mathematics, Ellipse, NDA, NDA 2010 (Phase 1)
If $4 \sin ^{2} \theta=1$, where $0 < \theta < 2 \pi$, how many values does $\theta$ take?

Options:
A 1
B 2
C 4
D None of the above

The correct answer is:
2

Solution:
$\tan \left(-585^{\circ}\right)=-\tan 585^{\circ}$
$=-\tan \left[540^{\circ}+45^{\circ}\right]=-\tan \left[3 \pi+\frac{\pi}{4}\right]$
$\quad=-\left[\frac{\tan 3 \pi+\tan \frac{\pi}{4}}{1-\tan 3 \pi \tan \frac{\pi}{4}}\right]=-\left[\frac{0+1}{1-0 \times 1}\right]=-1$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2012 (Phase 2)
If $4 \sin ^{2} \theta=1$, where $0 < \theta < 2 \pi$, how many values does $\theta$ take?

Options:
A 1
B 2
C 4
D None of the above

The correct answer is:
4

Solution:
$4 \sin ^{2} \theta=1$
$\Rightarrow \sin ^{2} \theta=\frac{1}{4} \Rightarrow \sin \theta=\pm \frac{1}{2} \Rightarrow \theta=\pm \frac{\pi}{6}$
Hence $\theta$ take 4 values. (one value for each quadrant)

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
If $4 x-5 y+3 ?$ $0 x-9 y=107$ are two lines of
regression, then what are the values of $\overline{\mathrm{x}}$ and $\overline{\mathrm{v}}$
respectively? $\quad$

Options:
A 12 and 18
B 18 and 12
C 13 and 17
D 17 and 13

The correct answer is:
13 and 17

Solution:
$4 \mathrm{x}-5 \mathrm{y}+33=0$
$20 \mathrm{x}-9 \mathrm{y}=107$
$20 x-25 y+165=0$
$(1) \times 5 \Rightarrow \quad 20 x-9 y-107=0$
$(2) \Rightarrow \frac{y-(+)(+)}{-16 \mathrm{y}+272=0} \Rightarrow 16 \mathrm{y}=272 \Rightarrow \mathrm{y}=17$
$(1) \Rightarrow 4 x-5(17)+33=0$
$\Rightarrow 4 x-85+33=0$
$\Rightarrow 4 x=52 \Rightarrow x=13$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
If $4^{\mathrm{x}}-6.2^{\mathrm{x}}+8=0$, then the values of $\mathrm{x}$ are

Options:
A 1,2
B 1,1
C 1,0
D 2,2

The correct answer is:
1,2

Solution:
Given equation $4^{x}-6.2^{x}+8=0$
$\Rightarrow\left(2^{x}\right)^{2}-6.2^{x}+8=0$
Put $2^{\mathrm{x}}=\mathrm{y}$, we have
$y^{2}-6 y+8=0$
$(y-4)(y-2)=0$
$y=2$,
$\mathrm{So}_{2} \quad 2^{\mathrm{x}}=2 \Rightarrow \mathrm{x}=$
and $2^{x}=2^{2} \Rightarrow x=2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
If $\left[\begin{array}{cc}5 & 0 \\ 0 & 7\end{array}\right]^{-1}\left[\begin{array}{c}\mathrm{x} \\ -\mathrm{y}\end{array}\right]=\left[\begin{array}{c}-1 \\ 2\end{array}\right]$, then which one of the following is correct?

Options:
A $x=5, y=14$
B $x=-5, y=14$
C $x=-5, y=-14$
D $\quad x=5, y=-14$

The correct answer is:
$x=-5, y=-14$

Solution:
The given equation is : $\left[\begin{array}{ll}5 & 0 \\ 0 & 7\end{array}\right]^{-1}\left[\begin{array}{c}\mathrm{x} \\ -\mathrm{y}\end{array}\right]=\left[\begin{array}{c}-1 \\ 2\end{array}\right]$
Multiplying both sides by $\left[\begin{array}{ll}5 & 0 \\ 0 & 7\end{array}\right]$, we get
$\left[\begin{array}{ll}5 & 0 \\ 0 & 7\end{array}\right]\left[\begin{array}{ll}5 & 0 \\ 0 & 7\end{array}\right]^{-1}\left[\begin{array}{c}\mathrm{x} \\ -\mathrm{y}\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 0 & 7\end{array}\right]\left[\begin{array}{c}-1 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{l}-\mathrm{x} \\ -\mathrm{y}\end{array}\right]=\left[\begin{array}{l}-5 \\ 14\end{array}\right]$
$\Rightarrow\left[\begin{array}{c}\mathrm{x} \\ -\mathrm{y}\end{array}\right]=\left[\begin{array}{c}-5 \\ 14\end{array}\right] \quad \Rightarrow \mathrm{x}=-5$ and $\mathrm{y}=-14$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 1)
If $(-5,4)$ divides the line segment between the coordinate axes in the ratio $1: 2$, then what is its equation?

Options:
A $8 x+5 y+20=0$
B $5 x+8 y-7=0$
C $8 x-5 y+60=0$
D $5 x-8 y+57=0$

The correct answer is:
$8 x-5 y+60=0$

Solution:
Let $A(a, 0)$ and $B(0, b)$ be two points on $x$ -axis and $y$ axis respectively
Given $(-5,4)$ divides line $A B$ in the ratio $1: 2$. By section formula we have
$$
\begin{aligned}
&-5=\frac{1 \times 0+2 \times a}{3} \\
\Rightarrow & a=\frac{-15}{2} \text { and } 4=\frac{1 \times b+2 \times 0}{3} \\
\Rightarrow & b=12
\end{aligned}
$$
Thus, $A=\left(\frac{-15}{2}, 0\right)$ and $B=(0,12)$
Hence, equation of line joining $\left(\frac{-15}{2}, 0\right)$ and $(0,12)$ is
$$
\begin{array}{l}
(y-0)=\frac{12-0}{0+\frac{15}{2}}\left(x+\frac{15}{2}\right) \\
\Rightarrow \quad y=\frac{4}{5}(2 x+15)
\end{array}
$$
$\Rightarrow 5 y=(8 x+60) \Rightarrow 8 x-5 y+60=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 1)
If 5 and 7 are the roots of the equation
$\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$, then what is the third root?

Options:
A $-12$
B 9
C 13
D 14

The correct answer is:
$-12$

Solution:
Given, $\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$
$\Rightarrow x\left(x^{2}-56\right)-4(7 x-35)+5(56-5 x)=0$
$\Rightarrow x^{3}-56 x-28 x+140+280-25 x=0$
$\Rightarrow x^{3}-109 x+420=0$
$\Rightarrow(x-5)(x-7)(x+12)=0$
$\Rightarrow x=-12$
Hence, the third root is $-12$.

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 1)
If 5x-2y+k=0 is a tangent to the parabola y2=6x, then their point of contact is

Options:
A 65,65
B 65,625
C 625,65
D 625,625

The correct answer is:
625,65

Solution:
y2=6x

⇒2ydydx=6

⇒dydx=36x→slope of tangent

Also,

5x-2y+k=0

⇒y=52x+k2

Comparing with the standard form of line in slope intercept form, we get

m=52→slope

Therefore,

36x=52

⇒32×3·x=52

⇒3x=52

⇒x=65

⇒x=625

And, y=65.

Hence, point of contact is 625,65.

Topic: Mathematics, Application of Derivatives, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
If $\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y$, where $i=\sqrt{-1}$, then what is $x$ equal to?

Options:
A 3
B 2
C 1
D 0

The correct answer is:
0

Solution:
$\left|\begin{array}{lcc}6 \mathrm{i} & -3 \mathrm{i} & 1 \\ 4 & 3 \mathrm{i} & -1 \\ 20 & 3 & \mathrm{i}\end{array}\right|$
$=6 \mathrm{i}\left[3 \mathrm{i}^{2}+3\right]+3 \mathrm{i}[4 \mathrm{i}+20]+1[12-60 \mathrm{i}]$
$=6 \mathrm{i}[-3+3]+12 \mathrm{i}^{2}+60 \mathrm{i}+12-60 \mathrm{i}$
$=-12+12=0=\mathrm{x}+\mathrm{iy}$
$\therefore \mathrm{x}=0$

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 2)
If 7 points out of 12 are in the same straight line, then what is the number of triangles formed ?

Options:
A 84
B 175
C 185
D 201

The correct answer is:
185

Solution:
Number of triangles formed from 12 point $={ }^{12} \mathrm{C}_{3}$ Since 7 parts are collinear, then ${ }^{7} \mathrm{C}_{3}$ triangles will not be formed so. $={ }^{12} \mathrm{C}_{3}-{ }^{7} \mathrm{C}_{3}$
$=\frac{12 !}{3 ! 9 !}-\frac{7 !}{3 ! 4 !}=\frac{12.11 .10}{3.2 .1}-\frac{7.6 .5}{3.2 .1}$
$=220-35=185$

Topic: Mathematics, Permutation Combination, NDA, NDA 2008 (Phase 1)
If $\left|\begin{array}{ccc}8 & -5 & 1 \\ 5 & \mathrm{x} & 1 \\ 6 & 3 & 1\end{array}\right|=2$ then what is the value of $\mathrm{x}$ ?

Options:
A 4
B 5
C 6
D 8

The correct answer is:
8

Solution:
Since $\left|\begin{array}{ccc}8 & -5 & 1 \\ 5 & x & 1 \\ 6 & 3 & 1\end{array}\right|=2$
$\Rightarrow 8(x-3)+5(5-6)+1(15-6 x)=2$
$\Rightarrow 8 x-24-5+15-6 x=2$
$\Rightarrow 2 x=16$
$\Rightarrow x=8$

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 1)
If $\theta=\frac{\pi}{8}$, then what is the value of $(2 \cos \theta+1)^{10}(2 \cos 2 \theta-1)^{10}(2 \cos \theta-1)^{10}(2 \cos 4 \theta-1)^{10 ?}$

Options:
A $\underline{0}$
B $\underline{1}$
C 2
D 4

The correct answer is:
$\underline{1}$

Solution:
$[(2 \cos \theta+1)(2 \cos \theta-1)]^{10}[2 \cos 2 \theta-1]^{10}[2 \cos 4 \theta-1]^{10}$
when, $\theta=\frac{\pi}{8}$ (given)
$=\left(4 \cos ^{2} \frac{\pi}{8}-1\right)^{10}\left(2 \times \frac{1}{\sqrt{2}}-1\right)^{10}(-1)^{10}$
$=\left(4 \times \frac{2+\sqrt{2}}{4}-1\right)^{10}(\sqrt{2}-1)^{10}$
$=[(\sqrt{2}+1)(\sqrt{2}-1)]^{10}=1^{10}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 2)
If $\alpha=\frac{\pi}{8}$, what is the value of $\cos \alpha \cos 2 \alpha \cos 4 \alpha$ ?

Options:
A 0
B $\frac{1}{4}$
C 8
D 4

The correct answer is:
0

Solution:
As given : $\alpha=\frac{\pi}{8}$
$\cos \alpha \cos 2 \alpha \cos 4 \alpha=\cos \frac{\pi}{8} \cdot \cos \frac{\pi}{4} \cos \frac{\pi}{2}=0$
$\left(\because \cos \frac{\pi}{2}=0\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
If $8 x-9 y=20$ and $7 x-10 y=9$, then what is $2 x-y$ equal to?

Options:
A 10
B 11
C 12
D 13

The correct answer is:
10

Solution:
$8 x-9 y=20$ or $80 x-90 y=200$...(1)
$7 x-10 y=9$ or $63 x-90 y=81$...(2)
Subtracting $(2)$ from $(1)$, we get $17 \mathrm{x}=119$
$\mathrm{x}=7$
$8 \times 7-9 y=20$
$9 y=36$
$y=4$
$2 x-y=2 \times 7-4=10$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2013 (Phase 2)
If $(10 x 010)_{2}-(11 y 1)_{2}=(10 z 11)_{2}$, then what are the possible values of the binary digits $x, y, z$ respectively?

Options:
A $0,0,1$
B $0,1,0$
C $1,1,0$
D $0,0,0$

The correct answer is:
$0,1,0$

Solution:
$\begin{aligned} & \therefore 133=(10000101)_{2} \\ &(10 x 010)_{2}-(11 y 1)_{2}=(10 z 11)_{2} \\ & \Rightarrow\left(2^{5} \times 1+0+x \times 2^{3}+0 \times 2^{2}+1 \times 2^{1}+0\right) \\ &=\left(2^{3} \times 1+2^{2} \times 1+y \times 2^{1}+1 \times 2^{0}\right) \\ &=2^{4} \times 1+0+2^{2} \times z+2^{1} \times 1+2^{0} \end{aligned}$
$\Rightarrow(34+8 x)-(13+2 y)=19+4 z$
$\Rightarrow 2=-8 x+2 y+4 z$
$\Rightarrow x=0, y=1, z=0$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $1.3+2.3^{2}+3.3^{3}+\ldots .+\mathrm{n} .3^{n}=\frac{(2 \mathrm{n}-1) 3^{\mathrm{a}}+\mathrm{b}}{3}$ then a and b
are respectively

Options:
A $\mathrm{n}, 2$
B $\mathrm{n}, 3$
C $\mathrm{n}+1,2$
D $\mathrm{n}+1,3$

The correct answer is:
$\mathrm{n}+1,3$

Solution:
$1.3+2.3^{2}+3.3^{3}+\ldots .+\mathrm{n} .3^{\mathrm{n}}=\frac{(2 \mathrm{n}-1) 3^{\mathrm{a}}+\mathrm{b}}{4}$
Let us put $3=\mathrm{x}$. L.H.S: $S=x+2 x^{2}+3 x^{3}+\ldots .+n \cdot x^{n}$
$x s=x^{2}+2 x^{3}+3 x^{4}+\ldots+n \cdot x^{n+1}$
$(1)-(2) \Rightarrow S-x S=\left(x+2 x^{2}+3 x^{3}+\ldots .+n \cdot x^{n}\right)-\left(x^{2}+2 x^{3}\right.$
$\left.+3 x^{4}+\ldots+n \cdot x^{n+1}\right)$
$\Rightarrow S(1-x)=x+x^{2}+x^{3}+\ldots .+x^{n}-n x^{n+1}$
$\Rightarrow S(1-x)=\frac{x\left(1-x^{n}\right)}{1-x}-n x^{n+1}$
$\Rightarrow S=\left(\frac{1}{x-1}\right)\left(\frac{-x\left(x^{n}-1\right)+n x^{n+1}(x-1)}{x-1}\right)$
Put $\mathrm{x}=3$
$\Rightarrow S=\frac{1}{2}\left(\frac{-3^{n+1}+3+2 n \cdot 3^{n+1}}{2}\right)=\left(\frac{3^{n+1}(2 n-1)+3}{4}\right)$

Topic: Mathematics, Mathematical Induction, NDA, NDA 2017 (Phase 2)
If $1 / 4,1 / x, 1 / 10$ are in HP, then what is the value of $x$?

Options:
A 5
B 6
C 7
D 8

The correct answer is:
7

Solution:
Since $\frac{1}{4}, \frac{1}{x}, \frac{1}{10}$ are in H.P.
$\therefore 4, \mathrm{x}, 10$ are in $\mathrm{AP}$
$\Rightarrow 2 \mathrm{x}=4+10 \Rightarrow \mathrm{x}=\frac{14}{2}=7$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 1)
If $38 \cdot 55 \mathrm{~kJ}$ of heat is absorbed when $6 \cdot 0 \mathrm{~g}$ of $\mathrm{O}_{2}$ react with CIF according to reaction $2 \mathrm{CIF}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}_{(\mathrm{g})}+0 \mathrm{~F}_{2}(\mathrm{~g})$
What is the standard enthalpy of reaction?

Options:
A $72.28 \mathrm{~K}$
B $205.6 \mathrm{k} \mathrm{J}$
C $102.8 \mathrm{k}]$
D $49.80 \mathrm{k}]$

The correct answer is:
$205.6 \mathrm{k} \mathrm{J}$

Solution:
(A)
$\begin{array}{l}
2 \mathrm{ClF}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}_{(\mathrm{g})}+\mathrm{OF}_{2(\mathrm{~g})} \\
\text { For } 6.0 \mathrm{~g} \text { of } \mathrm{O}_{2} \Rightarrow \Delta \mathrm{H}=+38.55 \mathrm{~kJ} \\
\therefore \text { For } 32.0 \mathrm{~g} \text { of } \mathrm{O}_{2} \Rightarrow \Delta \mathrm{H}^{0}=\frac{38.55 \times 32}{6}=205.6 \mathrm{~kJ}
\end{array}$

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2020 (16 Oct Shift 1)
If (11101011) is converted decimal system, then the
resulting number is

Options:
A 235
B 175
C 160
D 126

The correct answer is:
235

Solution:
$(11101011)_{2}$
$=\left(1 \times 2^{7}+1 \times 2^{6}+1 \times 2^{5}+0 \times 2^{4}+1 \times 2^{3}\right.$
$+0 \times 2^{2}+1 \times 2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
If $(a, 0),(0, b)$ and $(1,1)$ are collinear, what is $(a+b-a b)$ equal to?

Options:
A 2
B 1
C 0
D $-1$

The correct answer is:
0

Solution:
Since, $(a, 0),(0, b)$ and $(1,1)$ are collinear.
$\left|\begin{array}{lll}a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1\end{array}\right|=0$
$\Rightarrow a(b-1)+1(0-b)=0$
$\Rightarrow a b-a-b=0$
$\Rightarrow \quad a+b-a b=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
If $A=\left[\begin{array}{cc}\omega & 0 \\ 0 & \omega\end{array}\right]$, where $\omega$ is cube root of unity, then what is $A^{100}$ equal to?

Options:
A $A$
B $-A$
C Null matrix
D Identity matrix

The correct answer is:
$A$

Solution:
Given, $A=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]=\left[\begin{array}{cc}\omega^{2} & 0 \\ 0 & \omega^{2}\end{array}\right]$
$A^{3}=\left[\begin{array}{cc}\omega^{2} & 0 \\ 0 & \omega^{2}\end{array}\right]\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]=\left[\begin{array}{cc}\omega^{3} & 0 \\ 0 & \omega^{3}\end{array}\right]$
$\begin{aligned} \text { Similarly, } A^{100} &=\left[\begin{array}{cc}\omega^{100} & 0 \\ 0 & \omega^{100}\end{array}\right] \\ &=\left[\begin{array}{cc}\left(\omega^{3}\right)^{33} \cdot \omega^{1} & 0 \\ 0 & \left(\omega^{3}\right)^{33} \cdot \omega^{1}\end{array}\right] \\ &=\left[\begin{array}{cc}\omega & 0 \\ 0 & \omega\end{array}\right]\left(\because \omega^{3}=1\right)=\mathrm{A} \end{aligned}$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 2)
If $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right], B=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right], C=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$,
then which one of the following is not correct?

Options:
A $A^{2}=B^{2}$
B $\quad B^{2}=C^{2}$
C $A B=C$
D $A B=B A$

The correct answer is:
$A B=B A$

Solution:
Let $\mathrm{A}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right], \mathrm{C}=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots . .$ (1)
$\mathrm{B}^{2}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots \ldots . .(2)$
From (1) and (2), we have $\mathrm{A}^{2}=\mathrm{B}^{2}$
Now, $C^{2}=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
$\ldots \ldots . .(3)$
From (2) and (3), we have $\mathrm{B}^{2}=\mathrm{C}^{2}$
Now, $\mathrm{AB}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\mathrm{C}$
Now, we find
$\mathrm{BA}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right] \neq \mathrm{C}$
Hence $\mathrm{AB} \neq \mathrm{BA}$

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 2)
If $\mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$, then the matrix $\mathrm{A}$ is $/ \mathrm{an}$

Options:
A Singular matrix
B Involutory matrix
C Nilpotent matrix
D Idempotent matrix

The correct answer is:
Involutory matrix

Solution:
$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}(1)(1)+(0)(0) & (1)(0)+(0)(1) \\ (0)(1)+(1)(0) & (0)(0)+(1)(1)\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{A}$
$\therefore$ It is involuntary matrix.

Topic: Mathematics, Matrices, NDA, NDA 2019 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right]$, then the value of $\mathbf{A}^{4}$ is

Options:
A $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
B $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
C $\left[\begin{array}{cc}0 & 0 \\ 1 & 1\end{array}\right]$
D $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

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

Solution:
$\mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$\mathrm{A}^{4}=\mathrm{A}^{2} \cdot \mathrm{A}^{2}$
$=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \cdot\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \cdot\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 1)
If $\mathrm{A}=\{0,1\}$ and $\mathrm{B}=\{1,0\}$, then what is $\mathrm{A} \times \mathrm{B}$ equal to?

Options:
A $\{(0,1),(1,0)\}$
B $\{(0,0),(1,1)\}$
C $\{(0,1),(1,0),(1,1)\}$
D $\mathrm{A} \times \mathrm{A}$

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

Solution:
Let $A=\{0,1\}, B=\{1,0\}$
$\mathrm{A} \times \mathrm{B}=\{(0,1),(1,1),(0,0),(1,0)\}$
$\mathrm{A} \times \mathrm{A}=\{(0,0),(0,1),(1,1),(1,0)\}$
$\therefore \mathrm{A} \times \mathrm{B}=\mathrm{A} \times \mathrm{A}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
If $\left|\begin{array}{lll}a & b & 0 \\ 0 & a & b \\ b & 0 & a\end{array}\right|=0$, then which one of the following is correct?

Options:
A $\frac{a}{b}$ is one of the cube roots of unity.
B $\frac{a}{b}$ is one of the cube roots of $-1$.
C $a$ is one of the cube roots of unity.
D $b$ is one of the cube roots of unity.

The correct answer is:
$\frac{a}{b}$ is one of the cube roots of $-1$.

Solution:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & 0 \\ 0 & \mathrm{a} & \mathrm{b} \\ \mathrm{b} & 0 & \mathrm{a}\end{array}\right|$
$=a\left[a^{2}-0\right]-b\left[-b^{2}\right]+0$
$=a^{3}+b^{3}=0$
$\Rightarrow a^{3}=-b^{3}$
$\Rightarrow\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{3}=-1$
Hence, $\frac{\mathrm{a}}{\mathrm{b}}$ is one of the cube roots of $-1$

Topic: Mathematics, Determinants, NDA, NDA 2014 (Phase 2)
If $\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
such that $A^{2}=B$, then what is the value of $\alpha$?

Options:
A $-1$
B 1
C 2
D 4

The correct answer is:
1

Solution:
Let $\mathrm{A}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$
$\Rightarrow \quad \mathrm{A}^{2}=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$
$\Rightarrow \quad \mathrm{A}^{2}=\left[\begin{array}{cc}\alpha^{2} & 0 \\ \alpha+1 & 1\end{array}\right]$
But it is given that
$A^{2}=B$
$\Rightarrow \quad\left[\begin{array}{ll}\alpha^{2} & 0 \\ \alpha+1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$\Rightarrow \quad \alpha+1=2$
$\Rightarrow \quad \alpha=1$

Topic: Mathematics, Matrices, NDA, NDA 2011 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ is a $2 \times 2$ matrix and $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-\mathrm{x}+2$ is a polynomial, then what is $\mathrm{f}(\mathrm{A})$?

Options:
A $\left[\begin{array}{ll}1 & 7 \\ 1 & 7\end{array}\right]$
B $\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$
C $\left[\begin{array}{ll}2 & 6 \\ 0 & 6\end{array}\right]$
D $\left[\begin{array}{ll}2 & 6 \\ 0 & 7\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$

Solution:
Given that, $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\left[\begin{array}{lr}1 & 2+6 \\ 0 & 9\end{array}\right]=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]$
Since, $f(x)=x^{2}-x+2$
Putting A in place of $x$ $f(A)=A^{2}-A+2 I$
$=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1-1+2 & 8-2+0 \\ 0-0+0 & 9-3+2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]$, then what is $\mathrm{B}^{-1} \mathrm{~A}^{-1}$ equal to?

Options:
A $\left[\begin{array}{cc}1 & -3 \\ -1 & 2\end{array}\right]$
B $\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]$
C $\left[\begin{array}{cc}-1 & 3 \\ -1 & -2\end{array}\right]$
D $\left[\begin{array}{cc}-1 & -3 \\ 1 & -2\end{array}\right]$

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

Solution:
$|\mathrm{A}|=-1,|\mathrm{~B}|=1$
$\mathrm{~A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} \mathrm{A}=\frac{1}{-1}\left[\begin{array}{cc}1 & -1 \\ -2 & 1\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{\mid \mathrm{B}} \mid \mathrm{adj} \mathrm{B}=\frac{1}{1}\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]$
$\mathrm{B}^{-1} \mathrm{~A}^{-1}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 1)
If $\mathrm{A}^{-1}=\left[\begin{array}{cr}1 & -2 \\ -2 & 2\end{array}\right]$, what is det (A)?

Options:
A 2
B $-2$
C $\frac{1}{2}$
D $-\frac{1}{2}$

The correct answer is:
$-\frac{1}{2}$

Solution:
We know that det $(\mathrm{A})=\frac{1}{\operatorname{det}\left(\mathrm{A}^{-1}\right)}$
$\mathbf{A}^{-1}=\left[\begin{array}{rr}1 & -2 \\ -2 & 2\end{array}\right]$
$\Rightarrow\left|\mathrm{A}^{-1}\right|=\left|\begin{array}{rr}1 & -2 \\ -2 & 2\end{array}\right|=2-4=-2 \Rightarrow \operatorname{det}(\mathrm{A})=-\frac{1}{2}$

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 2)