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If $A=\left[\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right]$ and $k=\frac{1}{2 i}$ where $i=\sqrt{-1}$, then
$\mathrm{kA}$ is equal to
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$\mathrm{kA}$ is equal to
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Verified Answer
The correct answer is:
$\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$
$A=\left[\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right]$ and $K=\frac{1}{2 i}$
$\mathrm{K}=\frac{1}{2 \mathrm{i}}=\frac{\mathrm{i}}{2 \mathrm{i}(\mathrm{i})}=\frac{\mathrm{i}}{2 \mathrm{i}^{2}}=\frac{-\mathrm{i}}{2} .$
$\therefore \mathrm{KA}=\frac{-\mathrm{i}}{2}\left[\begin{array}{cc}4 \mathrm{i}-6 & 10 \mathrm{i} \\ 14 \mathrm{i} & 6+4 \mathrm{i}\end{array}\right]$
$=\left[\begin{array}{cc}(4 \mathrm{i}-6)\left(\frac{-\mathrm{i}}{2}\right) & 10 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) \\ 14 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) & \left.(6+4 \mathrm{i})\left(\frac{-\mathrm{i}}{2}\right)\right]\end{array}\right.$
$=\left[\begin{array}{cc}-2 \mathrm{i}^{2}+3 \mathrm{i} & -5 \mathrm{i}^{2} \\ -7 \mathrm{i}^{2} & -3 \mathrm{i}-2 \mathrm{i}^{2}\end{array}\right]=\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$
$\mathrm{K}=\frac{1}{2 \mathrm{i}}=\frac{\mathrm{i}}{2 \mathrm{i}(\mathrm{i})}=\frac{\mathrm{i}}{2 \mathrm{i}^{2}}=\frac{-\mathrm{i}}{2} .$
$\therefore \mathrm{KA}=\frac{-\mathrm{i}}{2}\left[\begin{array}{cc}4 \mathrm{i}-6 & 10 \mathrm{i} \\ 14 \mathrm{i} & 6+4 \mathrm{i}\end{array}\right]$
$=\left[\begin{array}{cc}(4 \mathrm{i}-6)\left(\frac{-\mathrm{i}}{2}\right) & 10 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) \\ 14 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) & \left.(6+4 \mathrm{i})\left(\frac{-\mathrm{i}}{2}\right)\right]\end{array}\right.$
$=\left[\begin{array}{cc}-2 \mathrm{i}^{2}+3 \mathrm{i} & -5 \mathrm{i}^{2} \\ -7 \mathrm{i}^{2} & -3 \mathrm{i}-2 \mathrm{i}^{2}\end{array}\right]=\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$
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