Q156 / 270 · NDA 2015 (Phase 2) · Mathematics

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

  • A
    $\left[\begin{array}{ccc}-\frac{3}{2} & 0 & -3 \\ -3 & -\frac{9}{2} & -6\end{array}\right]$
  • B
    $\left[\begin{array}{ccc}\frac{3}{2} & 0 & -3 \\ 3 & -\frac{9}{2} & -6\end{array}\right]$
  • C
    $\left[\begin{array}{ccc}\frac{3}{2} & 0 & 3 \\ 3 & \frac{9}{2} & 6\end{array}\right]$
  • Correct answer
    D
    $\left[\begin{array}{rrr}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]$

Solution

$\begin{aligned} & \because 2 \mathrm{X}+3 \mathrm{~A}=0 \\ & \Rightarrow \mathrm{x}=\frac{-3}{2} \mathrm{~A} \\ & \Rightarrow \mathrm{x}=\frac{-3}{2}\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right] \\ & \Rightarrow \mathrm{X}=\left[\begin{array}{ccc}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]\end{aligned}$

Hence, the correct answer is option: 'D'.

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