Question

Let $A=\left[\begin{array}{cc}1& \frac{3}{2}\\ 1& 2\end{array}\right],B=\left[\begin{array}{cc}4& -3\\ -2& 2\end{array}\right]\text{and}{C}_r=\left[\begin{array}{cc}r{.3}^r& 2^r\\ 0& (r-1)3^r\end{array}\right]$ be 3 given matrices. Compute the value of ${\sum }_{r=1}^{50}tr.\left(\right(AB{)}^r{C}_r).($ where $tr.\left(A\right)$ denotes trace of matrix A $)$

• A. $3({49.3}^{50}+1)$
• B. $3({49.3}^{49}+1)$
• C. $3({49.3}^{48}+1)$
• D. None of these

Answer 1

The correct option is A $3({49.3}^{50}+1)$

Given, $A=\left[\begin{array}{cc}1& \frac{3}{2}\\ 1& 2\end{array}\right],B=\left[\begin{array}{cc}4& -3\\ -2& 2\end{array}\right]$
So, $AB=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow AB=I$
Now, $(AB{)}^r{C}_r=I\left[\begin{array}{cc}r{.3}^r& 2^r\\ 0& (r-1)3^r\end{array}\right]$
$\Rightarrow (AB{)}^r{C}_r=\left[\begin{array}{cc}r{.3}^r& 2^r\\ 0& (r-1)3^r\end{array}\right]$
${\sum }_{r=1}^{50}tr.\left(\right(AB{)}^r{C}_r)={\sum }_{r=1}^{50}tr\left(C_r\right)={\sum }_{r=1}^{50}(2r-1)3^r$
It is an arithmetico-geometric series
$S_{50}=1.3+{3.3}^2+{5.3}^3+......+{99.3}^{50}$
$3S_{50}=3^2+{3.3}^3+......+{97.3}^{50}+{99.3}^{51}$
$\Rightarrow -2S=3+{2.3}^2+2.3^3+.....+{2.3}^{50}-{99.3}^{51}$
$\Rightarrow -2S=3-{99.3}^{51}+2(3^2+3^3+.....+3^{50})$
$\Rightarrow -2S=3-{99.3}^{51}+3^{51}-3^2$
$\Rightarrow S=3(1+{49.3}^{50})$