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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Matrices

NDA 2011 · Mathematics · Matrices

If the matrix A=[ array ll α & β β & α array ] is such that A^2=1 ,...

If the matrix A=\left[αββα\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right] is such that A2=1A^2=1, then which one of the following is correct?

Options

  1. Aα=0,β=1\alpha=0, \beta=1 or α=1,β=0\alpha=1, \beta=0
  2. Bα=0,β≠1\alpha=0, \beta \neq 1 or α≠0,β=1\alpha \neq 0, \beta=1
  3. Cα=1,β≠0\alpha=1, \beta \neq 0 or α≠1,β=0\alpha \neq 1, \beta=0
  4. Dα≠0,β≠0\alpha \neq 0, \beta \neq 0

Correct answer

A. α=0,β=1\alpha=0, \beta=1 or α=1,β=0\alpha=1, \beta=0

Step-by-step solution

A=\left[−αβ−βα\begin{array}{ll}-\alpha & \beta \\ -\beta & \alpha\end{array}\right]\therefore \quad A^2=\left[αββα\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right] \times\left[αββα\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]\Rightarrow \quad A^2=\left[α2+β22αβ2αββ2+α2\begin{array}{cc}\alpha^2+\beta^2 & 2 \alpha \beta \\ 2 \alpha \beta & \beta^2+\alpha^2\end{array}\right] Given that A2=1\mathrm{A}^2=1⇒[α2+β22αβ2αββ2+α2]=[1001]⇒α2+β2=1 and αβ=0⇒α=0,β=1 or α=1,β=0.\begin{aligned} & \Rightarrow\left[\begin{array}{cc} \alpha^2+\beta^2 & 2 \alpha \beta \\ 2 \alpha \beta & \beta^2+\alpha^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow \quad \alpha^2+\beta^2=1 \text{ and } \alpha \beta=0 \\ & \Rightarrow \quad \alpha=0, \beta=1 \text{ or } \alpha=1, \beta=0. \end{aligned}$
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