If $A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\ {14i}&{6 + 4i} \end{array}} \right]$ and $k = \frac{1}{{2i}}$, where $i = \sqrt { - 1}$, then kA is equal to

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If $A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\ {14i}&{6 + 4i} \end{array}} \right]$ and $k = \frac{1}{{2i}}$, where $i = \sqrt { - 1}$, then kA is equal to

TuteeHUB · Accepted Answer

$\left[ {\begin{array}{*{20}{c}} {2 - 3i}&5\ 7&{2 + 3i} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\ 7&{2 - 3i} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {2 - 3i}&7\ 5&{2 + 3i} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\ 7&{2 + 3i} \end{array}} \right]$

1.	If \(A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\\ {14i}&{6 + 4i} \end{array}} \right]\) and \(k = \frac{1}{{2i}}\), where \(i = \sqrt { - 1}\), then kA is equal to
A.	\(\left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\\ 7&{2 - 3i} \end{array}} \right]\)
B.	\(\left[ {\begin{array}{*{20}{c}} {2 - 3i}&5\\ 7&{2 + 3i} \end{array}} \right]\)
C.	\(\left[ {\begin{array}{*{20}{c}} {2 - 3i}&7\\ 5&{2 + 3i} \end{array}} \right]\)
D.	\(\left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\\ 7&{2 + 3i} \end{array}} \right]\)
Answer» B. \(\left[ {\begin{array}{*{20}{c}} {2 - 3i}&5\\ 7&{2 + 3i} \end{array}} \right]\)

If \(A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\\ {14i}&{6 + 4i} \end{array}} \right]\) and \(k = \frac{1}{{2i}}\), where \(i = \sqrt { - 1}\), then kA is equal to

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