Question

If A is a square matrix of order 2 such that $A^2=0$, then

• A. A=$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$, where $\alpha ,\beta ,\gamma$ are numbers such that ${\alpha }^2+\beta \gamma =0$
• B. A=$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$ with $\alpha =\pm \beta$
• C. A=$\left[\begin{array}{cc}\alpha & -\alpha \\ -\beta & -\beta \end{array}\right]$ with ${\alpha }^2+{\beta }^2=1$
• D. none of the above

Answer 1

The correct option is A

A=$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$, where $\alpha ,\beta ,\gamma$ are numbers such that ${\alpha }^2+\beta \gamma =0$

$LetA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\Rightarrow A^2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]A^2=\left[\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right](Since,A^2=0)ifa=0,then,b=c=d=0andA=0ifa\ne 0,a^2+bc=0\Rightarrow a^2=-bc\ne 0b(a+d)=c(a+d)=0\Rightarrow a=-d\Rightarrow a^2=d^{2}HenceA=\left[\begin{array}{cc}a& b\\ c& -a\end{array}\right],where{a}^2+bc=0orA=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right],where{\alpha }^2+\beta \gamma =0$