MCQ
Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is
  • A
    $50$
  • B
    $85$
  • $100$
  • D
    $137$

Answer

Correct option: C.
$100$
c
$A=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]$

$=\left[\begin{array}{cc}1 & -a+a b \\ 0 & b^{2}\end{array}\right]$

$\therefore T _{ n }=\left\{ A \in S ; A ^{ n ( n +1)}= I \right\}$

$\therefore$ $b$ must be equal to $1$

$\therefore$ In this case $A ^{2}$ will become identity matrix and a can take any value from $1$ to $100$

$\therefore$ Total number of common element will be $100$ .

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