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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Matrices

NDA 2021 · Mathematics · Matrices

Let A=[ array cc 0 & 2 -2 & 0 array ] and (m I+n A)^2=A where m, n...

Let A=\left[02−20\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right] and(m I+n A)^2=Awherewherem, narepositiverealnumbersandare positive real numbers andI is the identity matrix. What is (m+n)(m+n) equal to?

Options

  1. A0
  2. B12\frac{1}{2}
  3. C1
  4. D32\frac{3}{2}

Correct answer

D. 32\frac{3}{2}

Step-by-step solution

A=[02−20]MI⇒m[1001]=[m00m]nA⇒n[02−20]=[02n−2n0]Here, MI+nA⇒[m00m]+[02n−2n0]=[m2n−2nm]Given (MI+nA)2A=[m2n−2nm][m2n−2nm]=[02−20]⇒4mn=2 and m2−4n2=0⇒m+n=1+12=32\begin{aligned} & A=\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right] \\ & \mathrm{MI} \Rightarrow m\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right] \\ & n \mathrm{ A} \Rightarrow n\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 2 n \\ -2 n & 0\end{array}\right] \\ & \text{Here, } \mathrm{MI}+n \mathrm{ A} \Rightarrow\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right]+\left[\begin{array}{cc}0 & 2 n \\ -2 n & 0\end{array}\right]=\left[\begin{array}{cc}m & 2 n \\ -2 n & m\end{array}\right] \\ & \text{Given }(\mathrm{MI}+n \mathrm{ A})^2 \\ & \mathrm{ A}=\left[\begin{array}{cc}m & 2 n \\ -2 n & m\end{array}\right]\left[\begin{array}{cc}m & 2 n \\ -2 n & m\end{array}\right]=\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right] \\ & \Rightarrow 4 m n=2 \text{ and } m^2-4 n^2=0 \\ & \Rightarrow m+n=1+\frac{1}{2}=\frac{3}{2}\end{aligned}
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