If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right],$ then show that $|2 A|=4|A|$.

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The given matrix is $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$

$\therefore 2 A=2\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right]$

$\begin{aligned} L H S:|2 A| &=\left|\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right| \\ &=2 \times 4-4 \times 8 \\ &=8-32=-24 \end{aligned}$

Now, $|A|=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]=1 \times 2-2 \times 4=2 \times 8=-6$

$R H S: 4|A|=4 \times(-6)=-24$

$\therefore L. H. S.=\therefore \mathrm{R.} \mathrm{H.}  \mathrm{S}$

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