Question

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

• A. $6$
• B. $9$
• C. $12$
• D. $15$

Answer 1

Answer: (A)

$6$

$B=\left[\begin{array}{cc}3& 4\\ 2& 3\end{array}\right]$ $;$ $C=\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right]$

$\Rightarrow BC=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ $\Rightarrow BC=I$

$\therefore ABC=A.I=A$

$A{\left(BC\right)}^2=A{I}^2=A$

$\therefore tr\left(A\right)+tr\left(\frac{ABC}{2}\right)+tr.\left(\frac{A{\left(BC\right)}^2}{4}\right)+tr.\left(\frac{A{\left(BC\right)}^3}{8}\right)+........$

$=3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+......+A$

$=3\left(\frac{1}{1-\frac{1}{2}}\right)$

$=\frac{3}{\frac{1}{2}}$

$=6$

Hence, the answer is $6.$