STD 12 - 3 and 4 . determinant and metrices — JEE कक्षा 11 साइन्स — Question

Given,

$\begin{aligned} A^{-1} &=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] \\\left|A^{-1}\right| &=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] \end{aligned}$

Apply $R_1 \rightarrow R_1-R_2$

$\left|A^{-1}\right|=\left[\begin{array}{ccc} 0 & 0 & -2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{array}\right]$

Expand along $R_1$

$\begin{aligned} \left|A^{-1}\right| &=-2(2018-2017)=-2 \\ A A^{-1} &=I \\ |A|\left|A^{-1}\right| &=1 \\ |A| &=\frac{1}{\left|A^{-1}\right|^2}=-\frac{1}{2} \\ \text { Now, }|2 A| &-\left|2 A^{-1}\right|=8|A|-8\left|A^{-1}\right| \\ &=8\left[\frac{-1}{2}+2\right]=12\end{aligned}$