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Let $A^{-1}=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] .$ Then $|2 \mathrm{~A}|-\left|2 \mathrm{~A}^{-1}\right|$ is equal to
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Verified Answer
The correct answer is:
12
$\begin{array}{l}
2^{3}|\mathrm{~A}|-2^{3} \frac{1}{|\mathrm{~A}|} \\
\left|\mathrm{A}^{-1}\right|=\left|\begin{array}{lll}
1 & 2017 & 2 \\
1 & 2017 & 4 \\
1 & 2018 & 8
\end{array}\right| \\
\frac{1}{|\mathrm{~A}|}=-2 \Rightarrow|\mathrm{A}|=\frac{-1}{2}
\end{array}$
Put the value answer is $=12$
2^{3}|\mathrm{~A}|-2^{3} \frac{1}{|\mathrm{~A}|} \\
\left|\mathrm{A}^{-1}\right|=\left|\begin{array}{lll}
1 & 2017 & 2 \\
1 & 2017 & 4 \\
1 & 2018 & 8
\end{array}\right| \\
\frac{1}{|\mathrm{~A}|}=-2 \Rightarrow|\mathrm{A}|=\frac{-1}{2}
\end{array}$
Put the value answer is $=12$
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