A ``basic row operation'' on a matrix means adding a multiple of one row to another row. Consider the matrices $$A = \left(\begin{array}{rrr} x & 5 & x \\ 1 & 3 & -2 \\ -2 & -2 & 2 \end{array}\right) \quad \text{and} \quad B = \left(\begin{array}{rrr} 0 & 0 & 21 \\ 1 & -1 & -14 \\ 0 & \frac{4}{3} & 4 \end{array}\right)$$ It is given that $B$ can be obtained from $A$ by applying finitely many basic row operations. Then, the value of $x$ is:
(A) 3
(B) $-3$
(C) $-1$
(D) 2.

Let $n \geq 3$. Let $A = \left( \left( a _ { i j } \right) \right) _ { 1 \leq i , j \leq n }$ be an $n \times n$ matrix such that $a _ { i j } \in \{ 1 , - 1 \}$ for all $1 \leq i , j \leq n$. Suppose that $$\begin{aligned} & a _ { k 1 } = 1 \text { for all } 1 \leq k \leq n \text { and } \\ & \sum _ { k = 1 } ^ { n } a _ { k i } a _ { k j } = 0 \text { for all } i \neq j \end{aligned}$$ Show that $n$ is a multiple of 4.

Let $M$ be a $3 \times 3$ matrix with all entries being 0 or 1 . Then, all possible values for $\operatorname { det } ( M )$ are
(A) $0 , \pm 1$
(B) $0 , \pm 1 , \pm 2$
(C) $0 , \pm 1 , \pm 3$
(D) $0 , \pm 1 , \pm 2 , \pm 3$.

Consider all $2 \times 2$ matrices whose entries are distinct and taken from the set $\{ 1,2,3,4 \}$. The sum of determinants of all such matrices is
(A) 24 .
(B) 10 .
(C) 12 .
(D) 0 .

Define a polynomial $f ( x )$ by $$f ( x ) = \left| \begin{array} { l l l } 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{array} \right|$$ for all $x \in \mathbb { R }$, where the right hand side above is a determinant. Then the roots of $f ( x )$ are of the form
(A) $\alpha , \beta \pm i \gamma$ where $\alpha , \beta , \gamma \in \mathbb { R } , \gamma \neq 0$ and $i$ is a square root of $- 1$.
(B) $\alpha , \alpha , \beta$ where $\alpha , \beta \in \mathbb { R }$ are distinct.
(C) $\alpha , \beta , \gamma$ where $\alpha , \beta , \gamma \in \mathbb { R }$ are all distinct.
(D) $\alpha , \alpha , \alpha$ for some $\alpha \in \mathbb { R }$.

If matrix $$A = \left[ \begin{array} { l l l } a & b & c \\ b & c & a \\ c & a & b \end{array} \right]$$ where $\mathrm { a } , \mathrm { b } , \mathrm { c }$ are real positive numbers, $\mathrm { abc } = 1$ and $\mathrm { A } ^ { \mathrm { T } } \mathrm { A } = \mathrm { I }$, then find the value of $\mathrm { a } ^ { 3 } + \mathrm { b } ^ { 3 } + \mathrm { c } ^ { 3 }$.

4. If M is a $3 \times 3$ matrix, where $\mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { I }$ and $\operatorname { det } ( \mathrm { M } ) = 1$, then prove that $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0$.
Sol. $\quad ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { M } ^ { \mathrm { T } } ( \mathrm { I } - \mathrm { M } )$
$$\Rightarrow \left| ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } \right| = | \mathrm { M } - \mathrm { I } | = \left| \mathrm { M } ^ { \mathrm { T } } \right| | \mathrm { I } - \mathrm { M } | = | \mathrm { I } - \mathrm { M } | \Rightarrow | \mathrm { M } - \mathrm { I } | = 0 .$$
Alternate: $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } \right) = \operatorname { det } \left( \mathrm { MM } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \right)$
$$= \operatorname { det } \left( \mathrm { I } - \mathrm { M } ^ { \mathrm { T } } \right) = - \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } \right) = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \Rightarrow \operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0 \text {. }$$

1. If $\mathrm { y } ( \mathrm { x } ) = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ then find $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $\mathrm { x } = \pi$.

Sol. $\mathrm { y } = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta = \cos \mathrm { x } \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ so that $\frac { d y } { d x } = - \sin x \int _ { \pi ^ { 2 } / 16 } ^ { x ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } d \theta + \frac { 2 x \cos x \cdot \cos x } { 1 + \sin ^ { 2 } x }$ Hence, at $\mathrm { x } = \pi , \frac { \mathrm { dy } } { \mathrm { dx } } = 0 + \frac { 2 \pi ( - 1 ) ( - 1 ) } { 1 + 0 } = 2 \pi$.

6. T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it has corresponding vertices $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } , \mathrm { D } ^ { \prime }$. T is now compressed to S with face ABCD remaining same and $\mathrm { A } ^ { \prime }$, $B ^ { \prime } , C ^ { \prime } , D ^ { \prime }$ shifted to $A ^ { \prime \prime } , B ^ { \prime \prime } , C ^ { \prime \prime } , D ^ { \prime \prime }$ in $S$. The volume of parallelopiped $S$ is reduced to $90 \%$ of T. Prove that locus of $\mathrm { A } ^ { \prime \prime }$ is a plane.
Sol. Let the equation of the plane ABCD be $\mathrm { ax } + \mathrm { by } + \mathrm { cz } + \mathrm { d } = 0$, the point $\mathrm { A } ^ { \prime \prime }$ be $( \alpha , \beta , \gamma )$ and the height of the parallelopiped ABCD be h . $\Rightarrow \frac { | \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } | } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } } } = 0.9 \mathrm {~h} . \Rightarrow \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } = \pm 0.9 \mathrm {~h} \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } }$ ⇒ the locus of $\mathrm { A } ^ { \prime \prime }$ is a plane parallel to the plane ABCD .

4. If M is a $3 \times 3$ matrix, where $\mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { I }$ and $\operatorname { det } ( \mathrm { M } ) = 1$, then prove that $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0$.
Sol. $\quad ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { M } ^ { \mathrm { T } } ( \mathrm { I } - \mathrm { M } )$
$$\Rightarrow \left| ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } \right| = | \mathrm { M } - \mathrm { I } | = \left| \mathrm { M } ^ { \mathrm { T } } \right| | \mathrm { I } - \mathrm { M } | = | \mathrm { I } - \mathrm { M } | \Rightarrow | \mathrm { M } - \mathrm { I } | = 0 .$$
Alternate: $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } \right) = \operatorname { det } \left( \mathrm { MM } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \right)$
$$= \operatorname { det } \left( \mathrm { I } - \mathrm { M } ^ { \mathrm { T } } \right) = - \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } \right) = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \Rightarrow \operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0 \text {. }$$

1. If $\mathrm { y } ( \mathrm { x } ) = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ then find $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $\mathrm { x } = \pi$.

Sol. $\mathrm { y } = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta = \cos \mathrm { x } \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ so that $\frac { d y } { d x } = - \sin x \int _ { \pi ^ { 2 } / 16 } ^ { x ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } d \theta + \frac { 2 x \cos x \cdot \cos x } { 1 + \sin ^ { 2 } x }$ Hence, at $\mathrm { x } = \pi , \frac { \mathrm { dy } } { \mathrm { dx } } = 0 + \frac { 2 \pi ( - 1 ) ( - 1 ) } { 1 + 0 } = 2 \pi$.

30. The value of $| \mathrm { U } |$ is
(A) 3
(B) - 3
(C) $3 / 2$
(D) 2
Sol. (A) Let $U _ { 1 }$ be $\left[ \begin{array} { c } x \\ y \\ z \end{array} \right]$ so that $\left[ \begin{array} { l l l } 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right] \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right] \Rightarrow \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { c } 1 \\ - 2 \\ 1 \end{array} \right]$ Similarly $U _ { 2 } = \left[ \begin{array} { c } 2 \\ - 1 \\ - 4 \end{array} \right] , U _ { 3 } = \left[ \begin{array} { c } 2 \\ - 1 \\ - 3 \end{array} \right]$. Hence $U = \left[ \begin{array} { c c c } 1 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 1 & - 4 & - 3 \end{array} \right]$ and $| U | = 3$.

Let p be an odd prime number and $\mathrm { T } _ { \mathrm { p } }$ be the following set of $2 \times 2$ matrices: $$\mathrm { T } _ { \mathrm { p } } = \left\{ \mathrm { A } = \left[ \begin{array} { l l } \mathrm { a } & \mathrm {~b} \\ \mathrm { c } & \mathrm { a } \end{array} \right] : \mathrm { a } , \mathrm {~b} , \mathrm { c } \in \{ 0,1,2 , \ldots , \mathrm { p } - 1 \} \right\}$$
The number of $A$ in $T _ { p }$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname { det } ( \mathrm { A } )$ divisible by p is
A) $( p - 1 ) ^ { 2 }$
B) $2 ( p - 1 )$
C) $( p - 1 ) ^ { 2 } + 1$
D) $2 p - 1$

Let p be an odd prime number and $\mathrm { T } _ { \mathrm { p } }$ be the following set of $2 \times 2$ matrices: $$\mathrm { T } _ { \mathrm { p } } = \left\{ \mathrm { A } = \left[ \begin{array} { l l } \mathrm { a } & \mathrm {~b} \\ \mathrm { c } & \mathrm { a } \end{array} \right] : \mathrm { a } , \mathrm {~b} , \mathrm { c } \in \{ 0,1,2 , \ldots , \mathrm { p } - 1 \} \right\}$$
The number of $A$ in $T _ { p }$ such that the trace of $A$ is not divisible by $p$ but $\operatorname { det } ( A )$ is divisible by $p$ is [Note : The trace of a matrix is the sum of its diagonal entries.]
A) $( \mathrm { p } - 1 ) \left( \mathrm { p } ^ { 2 } - \mathrm { p } + 1 \right)$
B) $\mathrm { p } ^ { 3 } - ( \mathrm { p } - 1 ) ^ { 2 }$
C) $( p - 1 ) ^ { 2 }$
D) $( \mathrm { p } - 1 ) \left( \mathrm { p } ^ { 2 } - 2 \right)$

Let p be an odd prime number and $\mathrm { T } _ { \mathrm { p } }$ be the following set of $2 \times 2$ matrices: $$\mathrm { T } _ { \mathrm { p } } = \left\{ \mathrm { A } = \left[ \begin{array} { l l } \mathrm { a } & \mathrm {~b} \\ \mathrm { c } & \mathrm { a } \end{array} \right] : \mathrm { a } , \mathrm {~b} , \mathrm { c } \in \{ 0,1,2 , \ldots , \mathrm { p } - 1 \} \right\}$$
The number of $A$ in $T _ { p }$ such that $\operatorname { det } ( A )$ is not divisible by $p$ is
A) $2 p ^ { 2 }$
B) $p ^ { 3 } - 5 p$
C) $p ^ { 3 } - 3 p$
D) $p ^ { 3 } - p ^ { 2 }$

Let $\omega$ be the complex number $\cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 }$. Then the number of distinct complex numbers $z$ satisfying $\left| \begin{array} { c c c } z + 1 & \omega & \omega ^ { 2 } \\ \omega & z + \omega ^ { 2 } & 1 \\ \omega ^ { 2 } & 1 & z + \omega \end{array} \right| = 0$ is equal to

Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $$\left[ \begin{array} { c c c } 1 & a & b \\ \omega & 1 & c \\ \omega ^ { 2 } & \omega & 1 \end{array} \right]$$ where each of $a , b$, and $c$ is either $\omega$ or $\omega ^ { 2 }$. Then the number of distinct matrices in the set $S$ is
(A) 2
(B) 6
(C) 4
(D) 8

Which of the following values of $\alpha$ satisfy the equation $$\left| \begin{array} { c c c } ( 1 + \alpha ) ^ { 2 } & ( 1 + 2 \alpha ) ^ { 2 } & ( 1 + 3 \alpha ) ^ { 2 } \\ ( 2 + \alpha ) ^ { 2 } & ( 2 + 2 \alpha ) ^ { 2 } & ( 2 + 3 \alpha ) ^ { 2 } \\ ( 3 + \alpha ) ^ { 2 } & ( 3 + 2 \alpha ) ^ { 2 } & ( 3 + 3 \alpha ) ^ { 2 } \end{array} \right| = - 648 \alpha$$?
(A) $-4$
(B) $9$
(C) $-9$
(D) $4$

How many $3 \times 3$ matrices $M$ with entries from $\{ 0,1,2 \}$ are there, for which the sum of the diagonal entries of $M ^ { T } M$ is 5 ?
[A] 126
[B] 198
[C] 162
[D] 135

Let $R = \left\{ \left( \begin{array} { c c c } a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{array} \right) : a , b , c , d \in \{ 0,3,5,7,11,13,17,19 \} \right\}$. Then the number of invertible matrices in $R$ is

Let $S = \left\{ A = \left( \begin{array} { l l l } 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{array} \right) : a , b , c , d , e \in \{ 0,1 \} \right.$ and $\left. | A | \in \{ - 1,1 \} \right\}$, where $| A |$ denotes the determinant of $A$. Then the number of elements in $S$ is $\_\_\_\_$ .

Let $I = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)$ and $P = \left( \begin{array} { l l } 2 & 0 \\ 0 & 3 \end{array} \right)$. Let $Q = \left( \begin{array} { l l } x & y \\ z & 4 \end{array} \right)$ for some non-zero real numbers $x , y$, and $z$, for which there is a $2 \times 2$ matrix $R$ with all entries being non-zero real numbers, such that $Q R = R P$.
Then which of the following statements is (are) TRUE?

(A)	The determinant of $Q - 2 I$ is zero
(B)	The determinant of $Q - 6 I$ is 12
(C)	The determinant of $Q - 3 I$ is 15
(D)	$y z = 2$

If $A = \left( \begin{array} { c } \alpha - 1 \\ 0 \\ 0 \end{array} \right) , B = \left( \begin{array} { c } \alpha + 1 \\ 0 \\ 0 \end{array} \right)$ be two matrices, then $A B ^ { T }$ is a non-zero matrix for $| \alpha |$ not equal to
(1) 2
(2) 0
(3) 1
(4) 3

$$\left|\begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array}\right| = \alpha(a+b)(b+c)(c+a) \neq 0$$
then $\alpha$ is equal to
(1) $a+b+c$
(2) $abc$
(3) 4
(4) 1

Let $P$ and $Q$ be $3 \times 3$ matrices with $P \neq Q$. If $P^{3} = Q^{3}$ and $P^{2}Q = Q^{2}P$, then the determinant of $(P^{2}+Q^{2})$ is equal to
(1) $-2$
(2) 1
(3) 0
(4) $-1$

If $\alpha , \beta \neq 0 , f ( n ) = \alpha ^ { n } + \beta ^ { n }$ and $$\begin{vmatrix} 3 & 1 + f(1) & 1 + f(2) \\ 1 + f(1) & 1 + f(2) & 1 + f(3) \\ 1 + f(2) & 1 + f(3) & 1 + f(4) \end{vmatrix} = K ( 1 - \alpha ) ^ { 2 } ( 1 - \beta ) ^ { 2 } ( \alpha - \beta ) ^ { 2 }$$, then $K$ is equal to
(1) 1
(2) - 1
(3) $\alpha \beta$
(4) $\frac { 1 } { \alpha \beta }$

If $\Delta _ { r } = \left| \begin{array} { c c c } r & 2 r - 1 & 3 r - 2 \\ \frac { n } { 2 } & n - 1 & a \\ \frac { 1 } { 2 } n ( n - 1 ) & ( n - 1 ) ^ { 2 } & \frac { 1 } { 2 } ( n - 1 ) ( 3 n + 4 ) \end{array} \right|$, then the value of $\sum _ { r = 1 } ^ { n - 1 } \Delta _ { r }$
(1) Is independent of both $a$ and $n$
(2) Depends only on $a$
(3) Depends only on $n$
(4) Depends both on $a$ and $n$