Question

If ${{A}_{3\times 3}}$ is a matrix such that $\left| A \right|=a$ , $B=\left( adjA \right)$ such that $\left| B \right|=b$ . Find the value of $\dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25}$ where $\dfrac{1}{2}S=\dfrac{a}{b}+\dfrac{{{a}^{2}}}{{{b}^{3}}}+\dfrac{{{a}^{3}}}{{{b}^{5}}}+...............\text{ }up\text{ }to\text{ }\infty $ , and $a=3$ .

Answer 1

Hint: For solving this question we will use the $\left| adjA \right|={{\left| A \right|}^{2}}$ formula from the matrix and determinants concept. After that, we will put the value of $a$ and $b$ in the expression of $S$ and then analyse it and calculate its value and then easily we will find the value of $\dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25}$.

Complete step by step answer:
It is given that ${{A}_{3\times 3}}$ is a matrix such that $\left| A \right|=a=3$ , $B=\left( adjA \right)$ such that $\left| B \right|=b$ and $\dfrac{1}{2}S=\dfrac{a}{b}+\dfrac{{{a}^{2}}}{{{b}^{3}}}+\dfrac{{{a}^{3}}}{{{b}^{5}}}+...............\text{ }up\text{ }to\text{ }\infty $ . And we have to find the value of $\dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25}$ .
Now, we know that it $A$ is a $3\times 3$ square matrix then, $\left| adjA \right|={{\left| A \right|}^{2}}$ . So, as it is given that $\left| A \right|=a=3$ and $B=\left( adjA \right)$ . Then,
$\begin{align}
  & B=\left( adjA \right) \\
 & \Rightarrow \left| B \right|=b=\left| adjA \right|={{\left| A \right|}^{2}} \\
 & \Rightarrow b={{a}^{2}} \\
 & \Rightarrow b=9.......................\left( 1 \right) \\
\end{align}$
Now, as we have the value of both $a$ and $b$ . And $\dfrac{1}{2}S=\dfrac{a}{b}+\dfrac{{{a}^{2}}}{{{b}^{3}}}+\dfrac{{{a}^{3}}}{{{b}^{5}}}+...............\text{ }up\text{ }to\text{ }\infty $ . Then,
$\begin{align}
  & \dfrac{1}{2}S=\dfrac{a}{b}+\dfrac{{{a}^{2}}}{{{b}^{3}}}+\dfrac{{{a}^{3}}}{{{b}^{5}}}+...............\text{ }up\text{ }to\text{ }\infty \\
 & \Rightarrow \dfrac{1}{2}S=\dfrac{3}{9}+\dfrac{{{3}^{2}}}{{{9}^{3}}}+\dfrac{{{3}^{3}}}{{{9}^{5}}}+...............\text{ }up\text{ }to\text{ }\infty \\
 & \Rightarrow \dfrac{S}{2}=\dfrac{1}{3}+\dfrac{{{3}^{2}}}{{{3}^{6}}}+\dfrac{{{3}^{3}}}{{{3}^{10}}}+...............\text{ }up\text{ }to\text{ }\infty \\
 & \Rightarrow \dfrac{S}{2}=\dfrac{1}{3}+\dfrac{1}{{{3}^{4}}}+\dfrac{1}{{{3}^{7}}}+...............\text{ }up\text{ }to\text{ }\infty .......................\left( 2 \right) \\
\end{align}$
Now, as the expression of $\dfrac{S}{2}$ is an infinite G.P. with a common ratio of $\dfrac{1}{{{3}^{3}}}$ so, first we will multiply the $\dfrac{S}{2}$ by $\dfrac{1}{{{3}^{3}}}$ . Then,
$\begin{align}
  & \dfrac{S}{2}=\dfrac{1}{3}+\dfrac{1}{{{3}^{4}}}+\dfrac{1}{{{3}^{7}}}+...............\text{ }up\text{ }to\text{ }\infty \\
 & \Rightarrow \dfrac{1}{{{3}^{3}}}\times \dfrac{S}{2}=\dfrac{1}{{{3}^{4}}}+\dfrac{1}{{{3}^{7}}}+\dfrac{1}{{{3}^{10}}}...............\text{ }up\text{ }to\text{ }\infty ...........................\left( 3 \right) \\
\end{align}$
Now, subtract the equation (3) from the equation (2). Then,
$\begin{align}
  & \dfrac{S}{2}-\dfrac{1}{{{3}^{3}}}\times \dfrac{S}{2}=\left( \dfrac{1}{3}+\dfrac{1}{{{3}^{4}}}+\dfrac{1}{{{3}^{7}}}+...............\text{ }up\text{ }to\text{ }\infty \right)-\left( \dfrac{1}{{{3}^{4}}}+\dfrac{1}{{{3}^{7}}}+\dfrac{1}{{{3}^{10}}}+...............\text{ }up\text{ }to\text{ }\infty \right) \\
 & \Rightarrow \dfrac{S}{2}\left( 1-\dfrac{1}{27} \right)=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{S}{2}\times \dfrac{26}{27}=\dfrac{1}{3} \\
 & \Rightarrow S=\dfrac{9}{13}.........................\left( 4 \right) \\
\end{align}$
Now, from equation (1) put $b=9$ , from equation (4) $S=\dfrac{9}{13}$ and the given data put $a=3$ in the $\dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25}$ and calculate its value. Then,
$\begin{align}
  & \dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25} \\
 & \Rightarrow \dfrac{\left( 3\times 81+9\times 3+1 \right)}{25}\times \dfrac{9}{13} \\
 & \Rightarrow \dfrac{\left( 243+27+1 \right)}{25}\times \dfrac{9}{13} \\
 & \Rightarrow \dfrac{2439}{325}\approx 7.5046 \\
\end{align}$
Now, from the above result, we can say that $\dfrac{\left( a{{b}^{2}}+{{a}^{2}}b+1 \right)S}{25}=\dfrac{2439}{325}\approx 7.5046$ .

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to calculate the answer quickly. Moreover, the student should calculate the value of $b$ correctly and don’t confuse it with the value of $\left| A \right|$ or apply any lengthy method to find it. After that for the value of $S$ G.P. formation should be done correctly and we can apply the formula of summation of infinite G.P having common ratio less than one also to calculate $S$ .