{vmatrix} 2 & 1 \\ 3 & 1 {vmatrix} + {vmatrix} 1 & 1/3 \\ 3 & 1 {vmatrix} + {vmatrix} 1/2 & 1/9 \\ 3 & 1 {vmatrix} + {vmatrix} 1/4 & 1/27 \\ 3 & 1 {vmatrix} + ∞ = [AP EAPCET 23 May 2025 S1]

Correct Answer is : −12-1/2−21

Correct Answer is : −12-1/2−21

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Matrices and Determinants

Section: Mathematics

Question : 10 of 48

Marks: +1, -0

\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & \frac{1}{3} \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} \frac{1}{2} & \frac{1}{9} \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} \frac{1}{4} & \frac{1}{27} \\ 3 & 1 \end{vmatrix} + \dots \infty =

Solution:

\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & \frac{1}{3} \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} \frac{1}{2} & \frac{1}{9} \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} \frac{1}{4} & \frac{1}{27} \\ 3 & 1 \end{vmatrix} + \dots + \infty

= (2-3) + \left(1 - 3 \times \frac{1}{3}\right) + \left(\frac{1}{2} - 3 \times \frac{1}{9}\right) + \left(\frac{1}{4} - 3 \times \frac{1}{27}\right) + \left(\frac{1}{8} - 3 \times \frac{1}{81}\right) + \dots + \infty

= -1 + 0 + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{2^2} - \frac{1}{3^2}\right) + \left(\frac{1}{2^3} - \frac{1}{3^3}\right) + \dots + \infty

= -1 + \left[ \left( \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \infty \right) - \left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \dots + \infty \right) \right]

= -1 + \frac{\frac{1}{2}}{1 - \frac{1}{2}} - \frac{\frac{1}{3}}{1 - \frac{1}{3}}

(Using sum of infinite GP formula)

= -1 + \frac{\frac{1}{2}}{\frac{1}{2}} - \frac{\frac{1}{3}}{\frac{2}{3}} = -1 + 1 - \frac{1}{2} = -\frac{1}{2}

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