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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Matrices

NDA 2015 · Mathematics · Matrices

If A=[ array ccc 1 & 0 & -2 2 & -3 & 4 array ] , then the matrix X...

If A=\left[10−22−34\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right], then the matrix XX for which 2X+3A=02 X+3 A=0 holds true is

Options

  1. A\left[−320−3−3−92−6\begin{array}{ccc}-\frac{3}{2} & 0 & -3 \\ -3 & -\frac{9}{2} & -6\end{array}\right]
  2. B\left[320−33−92−6\begin{array}{ccc}\frac{3}{2} & 0 & -3 \\ 3 & -\frac{9}{2} & -6\end{array}\right]
  3. C\left[32033926\begin{array}{ccc}\frac{3}{2} & 0 & 3 \\ 3 & \frac{9}{2} & 6\end{array}\right]
  4. D\left[−3203−392−6\begin{array}{rrr}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]

Correct answer

D. \left[−3203−392−6\begin{array}{rrr}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]

Step-by-step solution

∵2X+3A=0⇒x=−32A⇒x=−32[10−22−34]⇒X=[−3203−392−6]\begin{aligned} & \because 2 \mathrm{X}+3 \mathrm{ A}=0 \\ & \Rightarrow \mathrm{x}=\frac{-3}{2} \mathrm{ A} \\ & \Rightarrow \mathrm{x}=\frac{-3}{2}\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right] \\ & \Rightarrow \mathrm{X}=\left[\begin{array}{ccc}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]\end{aligned}
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