Mathematics - Matrices Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Given, }[abc{]}_{1\times 3}{\left[\begin{array}{lll}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]}_{3\times 3}=\left[\begin{array}{lll}0& 0& 0\end{array}\right]\\ & \Rightarrow \left[\begin{array}{c}a+8b+7c\\ 9a+2b+3c\\ 7a+7b+7c\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]\\ & \Rightarrow a+8b+7c=0\\ & \Rightarrow 9a+2b+3c=0\\ & \Rightarrow a+b+c=0\\ & \end{array}$$ On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get $$7a+c=0$$ Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get $$\begin{array}{ll}& 6a-b=0\\ \therefore & b=6a\text{ and }c=-7a\end{array}$$ If $b=6,a=1$ and $c=-7$ $$\begin{array}{ll}\therefore & a{x}^2+bx+c=0\\ \Rightarrow & x^2+6x-7=0\\ \Rightarrow & (x+7)(x-1)=0\\ \therefore & x=1,-7\\ \Rightarrow & {\sum }_{n=0}^{\infty }{\left(\frac{1}{1}-\frac{1}{7}\right)}^n\Rightarrow {\sum }_{n=0}^{\infty }{\left(\frac{6}{7}\right)}^n\\ \Rightarrow & 1+\frac{6}{7}+{\left(\frac{6}{7}\right)}^n+\dots \infty \\ & =\frac{1}{1-\frac{6}{7}}=\frac{1}{1/7}=7\end{array}$$