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  4. Let A-1=[1&2017&2 1&2017&4 1&2018&8]. Then , |2A|-|2A-1| is equal to

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Q. Let $A^{-1}=\begin{bmatrix}1&2017&2\\ 1&2017&4\\ 1&2018&8\end{bmatrix}$. Then , $|2A|-|2A^{-1}|$ is equal to

KVPYKVPY 2017

Solution:

Given,
$A^{-1}=\begin{bmatrix}1&2017&2\\ 1&2017&4\\ 1&2018&8\end{bmatrix}$
$\left| A^{-1}\right|-\begin{bmatrix}1&2017&2\\ 1&2017&4\\ 1&2018&8\end{bmatrix}$
Apply $R_{1}\rightarrow R_{1}-R_{2}$
$\left| A^{-1}\right|=\begin{bmatrix}0&0&-2\\ 1&2017&4\\ 1&2018&8\end{bmatrix}$
Expand along $R_{1}$
$\left| A^{-1}\right|=-2\left(2018-2017\right)=-2$
$AA^{-1}=1$
$\left| A\right|\left| A^{-1}\right|=1$
$\left| A\right|=\frac{1}{\left| A^{-1}\right|^{2}}=-\frac{1}{2}$
Now, $\left|2A\right|-\left|2A^{-1}\right|=8\left| A\right|-8\left| A^{-1}\right|$
$=8\left[\frac{-1}{2}+2\right]=12$