Question

Let $A=\left[\begin{array}{ccc}2& 0& 7\\ 0& 1& 0\\ 1& -2& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-x& 14x& 7x\\ 0& 1& 0\\ x& -4x& -2x\end{array}\right]$ be two matrices such that $AB=(AB{)}^{-1}$ and $AB\ne I,$ where $I$ is an identity matrix of order $3\times 3.$ Then the value of $tr(AB+(AB{)}^2+(AB{)}^3+\cdots +(AB{)}^{100})$ is
$($ Here, $tr\left(A\right)$ denotes the trace of matrix $A,$ i.e., sum of diagonal elements of $A.)$

• A. $150$
• B. $300$
• C. $200$
• D. $100$

Answer 1

The correct option is D $100$

$AB=\left[\begin{array}{ccc}2& 0& 7\\ 0& 1& 0\\ 1& -2& 1\end{array}\right]\left[\begin{array}{ccc}-x& 14x& 7x\\ 0& 1& 0\\ x& -4x& -2x\end{array}\right]$

$\Rightarrow AB=\left[\begin{array}{ccc}5x& 0& 0\\ 0& 1& 0\\ 0& 10x-2& 5x\end{array}\right]$

$(AB{)}^2=\left[\begin{array}{ccc}25x^2& 0& 0\\ 0& 1& 0\\ 0& 10x-2+50x^2-10x& 25x^2\end{array}\right]$
$(AB{)}^2=I$
$\therefore 25x^2=1$ and $50x^2-2=0$
$\Rightarrow x=\pm \frac{1}{5}$

At $x=\frac{1}{5},AB=I$ but $AB\ne I$
$\therefore x=-\frac{1}{5}$
and $AB=\left[\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& -4& -1\end{array}\right]$
$(AB{)}^2=I\Rightarrow (AB{)}^3=AB$
$(AB{)}^4=(AB{)}^2=I$
$\therefore (AB{)}^{2n}=I$ and $(AB{)}^{2n-1}=AB,n\in N$
$\therefore AB+(AB{)}^2+(AB{)}^3+\cdots +(AB{)}^{100}$
$=AB+I+AB+I+\cdots +I$
$=50AB+50I$
Hence, $tr(AB+(AB{)}^2+(AB{)}^3+\cdots +(AB{)}^{100})$
$=tr(50AB+50I)$
$=tr\left(50AB\right)+tr\left(50I\right)$
$=50(-1+1-1)+50(1+1+1)$
$=100$