Let $A$ be a square matrix of real numbers such that $A ^ { 4 } = A$. Which of the following is true for every such $A$?
(A) $\operatorname { det } ( A ) \neq - 1$
(B) $A$ must be invertible.
(C) $A$ can not be invertible.
(D) $A ^ { 2 } + A + I = 0$ where $I$ denotes the identity matrix.

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.

If $A = \left( \begin{array} { l l } 1 & 1 \\ 0 & i \end{array} \right)$ and $A ^ { 2018 } = \left( \begin{array} { l l } a & b \\ c & d \end{array} \right)$, then $a + d$ equals:
(A) $1 + i$
(B) 0
(C) 2
(D) 2018.

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 .

Let $A$ and $B$ be two $3 \times 3$ matrices such that $( A + B ) ^ { 2 } = A ^ { 2 } + B ^ { 2 }$. Which of the following must be true?
(A) $A$ and $B$ are zero matrices.
(B) $A B$ is the zero matrix.
(C) $( A - B ) ^ { 2 } = A ^ { 2 } - B ^ { 2 }$
(D) $( A - B ) ^ { 2 } = A ^ { 2 } + B ^ { 2 }$

30. If $\left| \begin{array} { c c c } 6 \mathrm { i } & - 3 \mathrm { i } & 1 \\ 4 & 3 \mathrm { i } & - 1 \\ 20 & 3 & \mathrm { i } \end{array} \right| = \mathrm { x } + \mathrm { iy }$, then:
(A) $x = 3 , y = 1$
(B) $x = 1 , y = 3$
(C) $x = 0 , y = 3$
(D) $x = 0 , y = 0$

6. If
$$A = \left[ \begin{array} { l l } \alpha & 0 \\ 1 & 1 \end{array} \right] \text { and } B = \left[ \begin{array} { l l } 1 & 0 \\ 5 & 1 \end{array} \right] ,$$
then value of $a$ for which $A ^ { 2 } = B$, is:
(a) 1
(b) - 1
(c) 4
(d) no real values

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 .

18. $\mathrm { A } = \left[ \begin{array} { l l l } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right] , \mathrm { B } = \left[ \begin{array} { c c c } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right] , \mathrm { U } = \left[ \begin{array} { c } \mathrm { f } \\ \mathrm { g } \\ \mathrm { h } \end{array} \right] , \mathrm { V } = \left[ \begin{array} { c } \mathrm { a } ^ { 2 } \\ 0 \\ 0 \end{array} \right]$. If there is vector matrix X , such that $\mathrm { AX } = \mathrm { U }$ has infinitely many solutions, then prove that $\mathrm { BX } = \mathrm { V }$ cannot have a unique solution. If afd $\neq 0$ then prove that $\mathrm { BX } = \mathrm { V }$ has no solution.
Sol. $\mathrm { AX } = \mathrm { U }$ has infinite solutions $\Rightarrow | \mathrm { A } | = 0$ $\left| \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm { ab } = 1$ or $\mathrm { c } = \mathrm { d }$ and $\left| \mathrm { A } _ { 1 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 0 & \mathrm { f } \\ 1 & \mathrm { c } & \mathrm { g } \\ 1 & \mathrm {~d} & \mathrm {~h} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } ; \left| \mathrm { A } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { f } & 1 \\ 1 & \mathrm {~g} & \mathrm {~b} \\ 1 & \mathrm {~h} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h }$ $\left| \mathrm { A } _ { 3 } \right| = \left| \begin{array} { l l l } \mathrm { f } & 0 & 1 \\ \mathrm {~g} & \mathrm { c } & \mathrm { b } \\ \mathrm { h } & \mathrm { d } & \mathrm { b } \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } , \mathrm { c } = \mathrm { d } \Rightarrow \mathrm { c } = \mathrm { d }$ and $\mathrm { g } = \mathrm { h }$ $\mathrm { BX } = \mathrm { V }$ $| \mathrm { B } | = \left| \begin{array} { l l l } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right| = 0 \quad$ (since $\mathrm { C } _ { 2 }$ and $\mathrm { C } _ { 3 }$ are equal) $\quad \Rightarrow \mathrm { BX } = \mathrm { V }$ has no unique solution. and $\left| \mathrm { B } _ { 1 } \right| = \left| \begin{array} { l l l } \mathrm { a } ^ { 2 } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ 0 & \mathrm {~g} & \mathrm {~h} \end{array} \right| = 0 ($ since $\mathrm { c } = \mathrm { d } , \mathrm { g } = \mathrm { h } )$ $\left| \mathrm { B } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { a } ^ { 2 } & 1 \\ 0 & 0 & \mathrm { c } \\ \mathrm { f } & 0 & \mathrm {~h} \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { cf } = \mathrm { a } ^ { 2 } \mathrm { df } \quad ($ since $\mathrm { c } = \mathrm { d } )$
$$\left| \mathrm { B } _ { 3 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 1 & \mathrm { a } ^ { 2 } \\ 0 & \mathrm {~d} & 0 \\ \mathrm { f } & \mathrm {~g} & 0 \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { df }$$
since if $\operatorname { adf } \neq 0$ then $\left| \mathrm { B } _ { 2 } \right| = \left| \mathrm { B } _ { 3 } \right| \neq 0$. Hence no solution exist.

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 $| U | = 3$.

31. The sum of the elements of $\mathrm { U } ^ { - 1 }$ is
(A) - 1
(B) 0
(C) 1
(D) 3
Sol. (B) Moreover adj $\mathrm { U } = \left[ \begin{array} { c c c } - 1 & - 2 & 0 \\ - 7 & - 5 & - 3 \\ 9 & 6 & 3 \end{array} \right]$. Hence $\mathrm { U } ^ { - 1 } = \frac { \operatorname { adj } \mathrm { U } } { 3 }$ and sum of the elements of $\mathrm { U } ^ { - 1 } = 0$.

32. The value of $\left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] U \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right]$ is
(A) 5
(B) $5 / 2$
(C) 4
(D) $3 / 2$
Sol. (A) The value of $\left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] \mathrm { U } \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right]$
$$\begin{aligned} & = \left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] \left[ \begin{array} { c c c } 1 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 1 & - 4 & - 3 \end{array} \right] \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right] \\ & = \left[ \begin{array} { l l l } - 1 & 4 & 4 \end{array} \right] \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right] = - 3 + 8 = 5 . \end{aligned}$$

Section - D

1. If roots of the equation $x ^ { 2 } - 10 c x - 11 d = 0$ are $a , b$ and those of $x ^ { 2 } - 10 a x - 11 b = 0$ are $c , d$, then the value of $\mathrm { a } + \mathrm { b } + \mathrm { c } + \mathrm { d }$ is (a, b, c and d are distinct numbers)

$$\begin{array} { l l } \text { As } a + b = 10 c \text { and } c + d = 10 a & + d 11 d , c d = - 11 b \\ & a b = - 11 d ( a + c ) \\ \Rightarrow & a c = 121 \text { and } ( b + d ) = 9 ( a + c ) \\ & a ^ { 2 } - 10 a c - 11 d = 0 \\ & c ^ { 2 } - 10 a c - 11 b = 0 \\ \Rightarrow & a ^ { 2 } + c ^ { 2 } - 20 a c - 11 ( b + d ) = 0 \\ \Rightarrow & ( a + c ) ^ { 2 } - 22 ( 121 ) - 11 \times 9 ( a + c ) = 0 \\ \Rightarrow & ( a + c ) = 121 \text { or } - 22 \text { (rejected) } \\ \therefore & a + b + c + d = 1210 \end{array}$$

1. The value of $5050 \frac { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x } { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 101 } d x }$ is

Sol. $= \frac { 5050 \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x } { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 101 } d x } = 5050 \frac { \mathrm { I } _ { 100 } } { \mathrm { I } _ { 101 } }$
$$\begin{aligned} I _ { 101 } & = \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) \left( 1 - x ^ { 50 } \right) ^ { 100 } d x \\ & = I _ { 100 } - \int _ { 0 } ^ { 1 } x \cdot x ^ { 49 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x \\ & = I _ { 100 } - \left[ \frac { - x \left( 1 - x ^ { 50 } \right) ^ { 101 } } { 101 } \right] _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \frac { \left( 1 - x ^ { 50 } \right) ^ { 101 } } { 5050 } \\ I _ { 101 } & = I _ { 100 } - \frac { I _ { 101 } } { 5050 } \\ \Rightarrow & 5050 \frac { I _ { 100 } } { I _ { 101 } } = 5051 \end{aligned}$$

1. If $\mathrm { a } _ { \mathrm { n } } = \frac { 3 } { 4 } - \left( \frac { 3 } { 4 } \right) ^ { 2 } + \left( \frac { 3 } { 4 } \right) ^ { 3 } + \cdots ( - 1 ) ^ { \mathrm { n } - 1 } \left( \frac { 3 } { 4 } \right) ^ { \mathrm { n } }$ and $\mathrm { b } _ { \mathrm { n } } = 1 - \mathrm { a } _ { \mathrm { n } }$, then find the minimum natural number $\mathrm { n } _ { 0 }$ such that $\mathrm { b } _ { \mathrm { n } } > \mathrm { a } _ { \mathrm { n } } \forall \mathrm { n } > \mathrm { n } _ { 0 }$

Sol. $\quad \mathrm { a } _ { \mathrm { n } } = \frac { 3 } { 4 } - \left( \frac { 3 } { 4 } \right) ^ { 2 } + \left( \frac { 3 } { 4 } \right) ^ { 3 } + \cdots + ( - 1 ) ^ { \mathrm { x } - 1 } \left( \frac { 3 } { 4 } \right) ^ { \mathrm { n } }$
$$\begin{aligned} & = \frac { \frac { 3 } { 4 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) } { 1 + \frac { 3 } { 4 } } = \frac { 3 } { 7 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) \\ & \mathrm { b } _ { \mathrm { n } } > \mathrm { a } _ { \mathrm { n } } \Rightarrow 2 \mathrm { a } _ { \mathrm { n } } < 1 \\ \Rightarrow & \frac { 6 } { 7 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) < 1 \\ \Rightarrow & 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } < \frac { 7 } { 6 } \\ \Rightarrow & - \frac { 1 } { 6 } < \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \Rightarrow \text { minimum natural number } \mathrm { n } _ { 0 } = 6 \end{aligned}$$

1. If $\mathrm { f } ( \mathrm { x } )$ is a twice differentiable function such that $\mathrm { f } ( \mathrm { a } ) = 0 , \mathrm { f } ( \mathrm { b } ) = 2 , \mathrm { f } ( \mathrm { c } ) = - 1 , \mathrm { f } ( \mathrm { d } ) = 2 , \mathrm { f } ( \mathrm { e } ) = 0$, where $\mathrm { a } < \mathrm { b } < \mathrm { c } < \mathrm { d } < \mathrm { e }$, then the minimum number of zeroes of $\mathrm { g } ( \mathrm { x } ) = \left( \mathrm { f } ^ { \prime } ( \mathrm { x } ) \right) ^ { 2 } + \mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) \mathrm { f } ( \mathrm { x } )$ in the interval $[ \mathrm { a } , \mathrm { e } ]$ is

Sol. $g ( x ) = \frac { d } { d x } \left( f ( x ) \cdot f ^ { \prime } ( x ) \right)$ to get the zero of $g ( x )$ we take function
$$\mathrm { h } ( \mathrm { x } ) = \mathrm { f } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } )$$
between any two roots of $\mathrm { h } ( \mathrm { x } )$ there lies at least one root of $\mathrm { h } ^ { \prime } ( \mathrm { x } ) = 0$ $\Rightarrow \mathrm { g } ( \mathrm { x } ) = 0$
$$\begin{aligned} & \mathrm { h } ( \mathrm { x } ) = 0 \\ \Rightarrow \quad & \mathrm { f } ( \mathrm { x } ) = 0 \text { or } \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0 \\ & \mathrm { f } ( \mathrm { x } ) = 0 \text { has } 4 \text { minimum solutions } \\ & \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0 \text { minimum three solution } \\ & \mathrm { h } ( \mathrm { x } ) = 0 \text { minimum } 7 \text { solution } \\ \Rightarrow \quad & \mathrm { h } ^ { \prime } ( \mathrm { x } ) = \mathrm { g } ( \mathrm { x } ) = 0 \text { has minimum } 6 \text { solutions. } \end{aligned}$$

Section - E

1. Match the following:

Normals are drawn at points $\mathrm { P } , \mathrm { Q }$ and R lying on the parabola $\mathrm { y } ^ { 2 } = 4 \mathrm { x }$ which intersect at $( 3,0 )$. Then
(i) Area of $\triangle \mathrm { PQR }$
(A) 2
(ii) Radius of circumcircle of $\triangle \mathrm { PQR }$
(B) $5 / 2$
(iii) Centroid of $\triangle \mathrm { PQR }$
(iv) Circumcentre of $\triangle \mathrm { PQR }$
(C) $( 5 / 2,0 )$
(D) $( 2 / 3,0 )$
Sol. As normal passes through $( 3,0 )$ $\Rightarrow \quad 0 = 3 \mathrm {~m} - 2 \mathrm {~m} - \mathrm { m } ^ { 3 }$ $\Rightarrow \mathrm { m } ^ { 3 } = \mathrm { m } \Rightarrow \mathrm { m } = 0 , \pm 1$ $\therefore \quad$ Centroid $\equiv \left( \frac { \left( \mathrm { m } _ { 1 } ^ { 2 } + \mathrm { m } _ { 2 } ^ { 2 } + \mathrm { m } _ { 3 } ^ { 2 } \right) } { 3 } , - \frac { 2 \left( \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } + \mathrm { m } _ { 3 } \right) } { 3 } \right) = \left( \frac { 2 } { 3 } , 0 \right)$ Circum radius $= \left| \frac { - 2 m _ { 1 } + 2 m _ { 2 } } { 2 } \right| = 2$ units.
$$\begin{aligned} & \mathrm { Q } \equiv \left( \mathrm {~m} _ { 2 } ^ { 2 } , - 2 \mathrm {~m} _ { 2 } \right) \equiv ( 1 , - 2 ) \\ & \mathrm { R } \equiv \left( \mathrm {~m} _ { 3 } ^ { 2 } , - 2 \mathrm {~m} _ { 3 } \right) \equiv ( 1,2 ) \end{aligned}$$
Area of $\triangle \mathrm { PQR } = \frac { 1 } { 2 } \times 4 \times 1 = 2$ sq. units. $\mathrm { R } = \frac { \mathrm { QR } } { 2 \sin \angle \mathrm { QPR } } = \frac { 4 } { 2 \sin \left( 2 \tan ^ { - 1 } 2 \right) }$ $\Rightarrow \frac { 4 } { 2 \times \sin \left( \tan ^ { - 1 } \frac { 4 } { 1 - 4 } \right) } = \frac { 4 } { 2 \times \frac { 4 } { 5 } } = \frac { 5 } { 2 }$ ∴ circumcentre $\equiv \left( \frac { 5 } { 2 } .0 \right)$.

The number of distinct real values of $\lambda$ for which the system of linear equations $$x + y + z = 0$$ $$x + \lambda y + z = 0$$ $$x + y + \lambda z = 0$$ has a non-trivial solution is
(A) 0
(B) 1
(C) 2
(D) 3

Let $\mathscr { A }$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices in $\mathscr { A }$ is
(A) 12
(B) 6
(C) 9
(D) 3

Let $\mathscr { A }$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices $A$ in $\mathscr { A }$ for which the system of linear equations
$$A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$$
has a unique solution, is
(A) less than 4
(B) at least 4 but less than 7
(C) at least 7 but less than 10
(D) at least 10

Let $\mathscr { A }$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices $A$ in $\mathscr { A }$ for which the system of linear equations
$$A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$$
is inconsistent, is
(A) 0
(B) more than 2
(C) 2
(D) 1

The number of $3 \times 3$ matrices A whose entries are either 0 or 1 and for which the system $A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$ has exactly two distinct solutions, is
A) 0
B) $2 ^ { 9 } - 1$
C) 168
D) 2

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 $\omega \neq 1$ 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

For $3 \times 3$ matrices $M$ and $N$, which of the following statement(s) is (are) NOT correct?
(A) $\quad N ^ { T } M N$ is symmetric or skew symmetric, according as $M$ is symmetric or skew symmetric
(B) $M N - N M$ is skew symmetric for all symmetric matrices $M$ and $N$
(C) $M N$ is symmetric for all symmetric matrices $M$ and $N$
(D) $\quad ( \operatorname { adj } M ) ( \operatorname { adj } N ) = \operatorname { adj } ( M N )$ for all invertible matrices $M$ and $N$

Let $M$ and $N$ be two $3 \times 3$ matrices such that $MN = NM$. Further, if $M \neq N^2$ and $M^2 = N^4$, then
(A) determinant of $\left(M^2 + MN^2\right)$ is 0
(B) there is a $3 \times 3$ non-zero matrix $U$ such that $\left(M^2 + MN^2\right)U$ is the zero matrix
(C) determinant of $\left(M^2 + MN^2\right) \geq 1$
(D) for a $3 \times 3$ matrix $U$, if $\left(M^2 + MN^2\right)U$ equals the zero matrix then $U$ is the zero matrix

Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if
(A) the first column of $M$ is the transpose of the second row of $M$
(B) the second row of $M$ is the transpose of the first column of $M$
(C) $M$ is a diagonal matrix with nonzero entries in the main diagonal
(D) the product of entries in the main diagonal of $M$ is not the square of an integer

Let $X$ and $Y$ be two arbitrary, $3 \times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3 \times 3$, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?
(A) $\quad Y ^ { 3 } Z ^ { 4 } - Z ^ { 4 } Y ^ { 3 }$
(B) $X ^ { 44 } + Y ^ { 44 }$
(C) $X ^ { 4 } Z ^ { 3 } - Z ^ { 3 } X ^ { 4 }$
(D) $X ^ { 23 } + Y ^ { 23 }$

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) $4$