Question

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

Answer 1

Given $A=\left[\begin{array}{cc}1& 2\\ 4& 2\end{array}\right]\therefore 2A=2\left[\begin{array}{cc}1& 2\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}2\times 1& 2\times 2\\ 2\times 4& 2\times 2\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 8& 4\end{array}\right]$
$LHS=|2A|=\left|\begin{array}{cc}2& 4\\ 8& 4\end{array}\right|=2\times 4-8\times 4=-24$
Now. $|A|=\left|\begin{array}{cc}1& 2\\ 4& 2\end{array}\right|=1\times 2-4\times 2=-6$
$\therefore RHS=4|A|=4\times (-6)=-24\therefore LHS=RHS$