If A- - 4i-6 10i- 14i 6-4i - and k - 1 2i- where i- - -1 then kA is equal to

The correct option is A [ 2+3i 5; 7 2 3i ] k=12i=i2i^2= i2 kA= [ (4i 6)( i2) amp; 10i( i2); 14i( i2) amp; (6+4i)( i2) ]=[ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Cube Root of a Complex Number

If A= [ 4i-...

Question

If $A = [\begin{matrix} 4 i - 6 & 10 i 14 i & 6 + 4 i \end{matrix}]$ and $k = \frac{1}{2 i}$ , where $i = \sqrt{- 1}$ then $k A$ is equal to

[\begin{matrix} 2 + 3 i & 5 7 & 2 - 3 i \end{matrix}]

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[\begin{matrix} 2 - 3 i & 5 7 & 2 + 3 i \end{matrix}]

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[\begin{matrix} 2 - 3 i & 7 5 & 2 + 3 i \end{matrix}]

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[\begin{matrix} 2 + 3 i & 5 7 & 2 + 3 i \end{matrix}]

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Solution

The correct option is A $[\begin{matrix} 2 + 3 i & 5 7 & 2 - 3 i \end{matrix}]$
$k = \frac{1}{2 i} = \frac{i}{2 i^{2}} = \frac{- i}{2}$

$k A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (4 i - 6) (\frac{- i}{2}) & 10 i (\frac{- i}{2}) 14 i (\frac{- i}{2}) & (6 + 4 i) (\frac{- i}{2}) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = [\begin{matrix} 2 + 3 i & 5 7 & 2 - 3 i \end{matrix}]$

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If A=[4i−610i14i6+4i] and k=12i, where i=√−1 then kA is equal to

The correct option is A [2+3i572−3i]k=12i=i2i2=−i2kA=⎡⎢ ⎢ ⎢ ⎢⎣(4i−6)(−i2)10i(−i2)14i(−i2)(6+4i)(−i2)⎤⎥ ⎥ ⎥ ⎥⎦=[2+3i572−3i]

If $A = [\begin{matrix} 4 i - 6 & 10 i 14 i & 6 + 4 i \end{matrix}]$ and $k = \frac{1}{2 i}$ , where $i = \sqrt{- 1}$ then $k A$ is equal to