We say that a matrix $S \in \mathbf{M}_n$ is a symmetry matrix if $S^2 = I_n$. If a matrix $A$ is a product of two symmetry matrices, is the same true for every matrix similar to $A$?

Let $B$ and $C$ be two matrices of $\mathbf{GL}_n$. Let $A \in \mathbf{M}_{2n}$ be the matrix defined by blocks as follows: $$A = \left(\begin{array}{cc} B & 0_n \\ 0_n & C \end{array}\right)$$ Let $S_1$ be the block matrix $$S_1 = \left(\begin{array}{cc} 0_n & P \\ Q & 0_n \end{array}\right),$$ where $P, Q$ are two elements of $\mathbf{GL}_n$. Determine the conditions relating $B, C, P, Q$ for the matrices $S_1$ and $S_2 = S_1 A$ to be symmetry matrices.

138- If determinant $D = \begin{vmatrix} 1 & 1 & 1 \\ bc & ac & ab \\ ac & ab & bc \end{vmatrix}$, then what is the value of $\begin{vmatrix} a+b & b & ab \\ b+c & c & bc \\ a+c & a & ac \end{vmatrix}$?
(1) $-D$ (2) $D$ (3) $(a+b+c)D$ (4) $abcD$

138. If one unit is added to all entries of the second column of matrix $A = \begin{bmatrix} 2 & 3 & 4 \\ 5 & a & 7 \\ 3 & b & 6 \end{bmatrix}$, what number is added to the value of the original determinant of the matrix?
(1) $-3$ (2) $-2$ (3) $3$ (4) $6$
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137- If $A = \begin{bmatrix} 2 & 1 & 5 \\ -3 & 0 & 4 \\ 1 & 0 & 2 \end{bmatrix}$, what are the entries of the first row of $A^3$?

• [(1)] $\begin{bmatrix} 30 & 6 & 94 \end{bmatrix}$
• [(2)] $\begin{bmatrix} 30 & 6 & 78 \end{bmatrix}$
• [(3)] $\begin{bmatrix} 24 & 8 & 46 \end{bmatrix}$
• [(4)] $\begin{bmatrix} 30 & 6 & 46 \end{bmatrix}$

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138- From the matrix relation $\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} X \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 8 \end{bmatrix}$, what is matrix $X$?
\[ (1)\quad \begin{bmatrix} 7 & 9 \\ 4 & 4 \end{bmatrix} \qquad (2)\quad \begin{bmatrix} 7 & 9 \\ 2 & -2 \end{bmatrix} \] \[ (3)\quad \begin{bmatrix} -9 & 7 \\ -4 & -4 \end{bmatrix} \qquad (4)\quad \begin{bmatrix} -9 & -7 \\ 4 & 4 \end{bmatrix} \]

139- What are the solutions of the equation $0 = \begin{vmatrix} -4 & 1 & 1 \\ 1 & 2-x & 1 \\ 3 & 2 & 3-x \end{vmatrix}$?
\[ (1)\quad 1,\ -4 \qquad (2)\quad 1,\ 4 \qquad (3)\quad 1,\ 5 \qquad (4)\quad 2,\ 5 \]

143. Suppose $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 7 & 8 & 4 \\ 3 & 2 & 5 \\ 6 & 9 & 3 \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}$. The sum of the entries of the third row of matrix $A$ is:
(1) $3$ (2) $5$ (3) $12$ (4) $13$

144. Suppose $A = \begin{bmatrix} 1 & -1 \\ 2 & 1 \\ 3 & 1 \end{bmatrix}$. If $BA^T A = 52I$, what is the maximum value of the entries of matrix $B$?
(1) $14$ (2) $18$ (3) $24$ (4) $28$

18. The number of distinct roots of $\left| \begin{array} { l } \sin x \cos x \cos x \\ \cos x \sin x \cos x \\ \cos x \cos x \sin x \end{array} \right| = 0$ in the interval $- \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 4 }$ is:
(A) 0
(B) 2
(C) 1
(D) 3

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 $\mathrm { x } = \pi , \frac { \mathrm { dy } } { \mathrm { dx } } = 0 + \frac { 2 \pi ( - 1 ) ( - 1 ) } { 1 + 0 } = 2 \pi$.

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 $\mathrm { x } = \pi , \frac { \mathrm { dy } } { \mathrm { dx } } = 0 + \frac { 2 \pi ( - 1 ) ( - 1 ) } { 1 + 0 } = 2 \pi$.

18. $\mathrm { A } = \left[ \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right] , \mathrm { B } = \left[ \begin{array} { c c c } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right] , \mathrm { U } = \left[ \begin{array} { c } \mathrm { f } \\ \mathrm { g } \\ \mathrm { h } \end{array} \right] , \mathrm { V } = \left[ \begin{array} { c } \mathrm { a } ^ { 2 } \\ 0 \\ 0 \end{array} \right]$. 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.

23. If $A = \left[ \begin{array} { c c c } 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array} \right]$, $6 A ^ { - 1 } = A ^ { 2 } + c A + d I$, then $( c , d )$ is:
(a) $( - 6,11 )$
(b) $( - 11,6 )$
(c) $( 11,6 )$
(d) $( 6,11 )$

Let $(x,y,z)$ be points with integer coordinates satisfying the system of homogeneous equations: $$\begin{array}{r} 3x-y-z=0\\ -3x+z=0\\ -3x+2y+z=0 \end{array}$$ Then the number of such points for which $x^{2}+y^{2}+z^{2}\leq100$ is

Let k be a positive real number and let
$$A = \left[ \begin{array} { c c c } 2 k - 1 & 2 \sqrt { k } & 2 \sqrt { k } \\ 2 \sqrt { k } & 1 & - 2 k \\ - 2 \sqrt { k } & 2 k & - 1 \end{array} \right] \text { and } B = \left[ \begin{array} { c c c } 0 & 2 k - 1 & \sqrt { k } \\ 1 - 2 k & 0 & 2 \sqrt { k } \\ - \sqrt { k } & - 2 \sqrt { k } & 0 \end{array} \right]$$
If $\operatorname { det } ( \operatorname { adj } \mathrm { A } ) + \operatorname { det } ( \operatorname { adj } \mathrm { B } ) = 10 ^ { 6 }$, then $[ \mathrm { k } ]$ is equal to [Note : adj M denotes the adjoint of a square matrix M and $[ \mathrm { k } ]$ denotes the largest integer less than or equal to k].

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 $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = \left[ p _ { i j } \right]$ be a $n \times n$ matrix with $p _ { i j } = \omega ^ { i + j }$. Then $P ^ { 2 } \neq 0$, when $n =$
(A) 57
(B) 55
(C) 58
(D) 56

Let $P = \left[ \begin{array} { c c c } 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{array} \right]$ and $I$ be the identity matrix of order 3. If $Q = \left[ q _ { i j } \right]$ is a matrix such that $P ^ { 50 } - Q = I$, then $\frac { q _ { 31 } + q _ { 32 } } { q _ { 21 } }$ equals
(A) 52
(B) 103
(C) 201
(D) 205

Let $P = \left[\begin{array}{ccc} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q = [q_{ij}]$ is a matrix such that $PQ = kI$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order 3. If $q_{23} = -\frac{k}{8}$ and $\det(Q) = \frac{k^2}{2}$, then
(A) $\alpha = 0, k = 8$
(B) $4\alpha - k + 8 = 0$
(C) $\det(P\operatorname{adj}(Q)) = 2^9$
(D) $\det(Q\operatorname{adj}(P)) = 2^{13}$

The total number of distinct $x \in \mathbb{R}$ for which $\left|\begin{array}{ccc} x & x^2 & 1+x^3 \\ 2x & 4x^2 & 1+8x^3 \\ 3x & 9x^2 & 1+27x^3 \end{array}\right| = 10$ is

Let $S$ be the set of all column matrices $\left[ \begin{array} { l } b _ { 1 } \\ b _ { 2 } \\ b _ { 3 } \end{array} \right]$ such that $b _ { 1 } , b _ { 2 } , b _ { 3 } \in \mathbb { R }$ and the system of equations (in real variables)
$$\begin{aligned} - x + 2 y + 5 z & = b _ { 1 } \\ 2 x - 4 y + 3 z & = b _ { 2 } \\ x - 2 y + 2 z & = b _ { 3 } \end{aligned}$$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $\left[ \begin{array} { l } b _ { 1 } \\ b _ { 2 } \\ b _ { 3 } \end{array} \right] \in S$ ?
(A) $x + 2 y + 3 z = b _ { 1 } , 4 y + 5 z = b _ { 2 }$ and $x + 2 y + 6 z = b _ { 3 }$
(B) $x + y + 3 z = b _ { 1 } , 5 x + 2 y + 6 z = b _ { 2 }$ and $- 2 x - y - 3 z = b _ { 3 }$
(C) $- x + 2 y - 5 z = b _ { 1 } , 2 x - 4 y + 10 z = b _ { 2 }$ and $x - 2 y + 5 z = b _ { 3 }$
(D) $x + 2 y + 5 z = b _ { 1 } , 2 x + 3 z = b _ { 2 }$ and $x + 4 y - 5 z = b _ { 3 }$

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{ - 1,0,1 \}$. Then, the maximum possible value of the determinant of $P$ is $\_\_\_\_$ .