Question

If $A=\left[\begin{array}{cc}2& 1\\ 4& 1\end{array}\right],B=\left[\begin{array}{cc}3& 4\\ 2& 3\end{array}\right]\text{and}C=\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right],$ then the value of $tr\left(A\right)+tr\left(\frac{ABC}{2}\right)+tr\left(\frac{A(BC{)}^2}{4}\right)+tr\left(\frac{A(BC{)}^3}{8}\right)+...\infty$ is

Answer 1

Let, $tr\left(A\right)+tr\left(\frac{ABC}{2}\right)+tr\left(\frac{A(BC{)}^2}{4}\right)+tr\left(\frac{A(BC{)}^3}{8}\right)+...\infty =X$
$BC=\left[\begin{array}{cc}3& 4\\ 2& 3\end{array}\right]\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right]\Rightarrow BC=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I\Rightarrow X=tr\left(A\right)+tr\left(\frac{A}{2}\right)+tr\left(\frac{A}{2^2}\right)+...\infty \Rightarrow X=tr\left(A\right)+\left(\frac{1}{2}\right)tr\left(A\right)+\left(\frac{1}{2^2}\right)tr\left(A\right)+...\infty$
this is sum of infinite G.P.
$\{\because a+ar+a{r}^2+\cdots +\infty =\frac{a}{1-r},|r|<1\}$
$=\frac{tr\left(A\right)}{1-\left(1/2\right)}=2tr\left(A\right)=2(2+1)=6$