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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Matrices

NDA 2017 · Mathematics · Matrices

If A=[ array cc 4 i-6 & 10 i 14 i & 6+4 i array ] and k= 1 2 i^...

If A=\left[4i−610i14i6+4i\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right] andk=\frac{1}{2 i^{\prime}}wherewherei=\sqrt{-1}$, then kA is equal to

Options

  1. A\left[2+3i572−3i\begin{array}{cc}2+3 i & 5 \\ 7 & 2-3 i\end{array}\right]
  2. B\left[2−3i572+3i\begin{array}{cc}2-3 \mathrm{i} & 5 \\ 7 & 2+3 \mathrm{i}\end{array}\right]
  3. C\left[2−3i752+3i\begin{array}{cc}2-3 i & 7 \\ 5 & 2+3 i\end{array}\right]
  4. D\left[2+3i572+3i\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2+3 \mathrm{i}\end{array}\right]

Correct answer

A. \left[2+3i572−3i\begin{array}{cc}2+3 i & 5 \\ 7 & 2-3 i\end{array}\right]

Step-by-step solution

A=\left[4i−610i14i6+4i\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right]andandK=\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[4i−610i14i6+4i\begin{array}{cc}4 \mathrm{i}-6 & 10 \mathrm{i} \\ 14 \mathrm{i} & 6+4 \mathrm{i}\end{array}\right]=\left[(4i−6)(−i2)10i(−i2)14i(−i2)(6+4i)(−i2)\begin{array}{cc}(4 i-6)\left(\frac{-i}{2}\right) & 10 i\left(\frac{-i}{2}\right) \\ 14 i\left(\frac{-i}{2}\right) & (6+4 i)\left(\frac{-i}{2}\right)\end{array}\right]=\left[−2i2+3i−5i2−7i2−3i−2i2\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[2+3i572−3i\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]
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