Question

$A=\left[\begin{array}{cc}t& t+1\\ t-1& t\end{array}\right]$ is a matrix such that $A{A}^T=I_2$ then trace of the matrix is

• A. $2$
• B. $0$
• C. $4$
• D. $1$

Answer 1

The correct option is B $0$

$A=\left[\begin{array}{cc}t& t+1\\ t-1& t\end{array}\right]$

$A^T=\left[\begin{array}{cc}t& t-1\\ t+1& t\end{array}\right]$

$A{A}^T=\left[\begin{array}{cc}t& t+1\\ t-1& t\end{array}\right]\left[\begin{array}{cc}t& t-1\\ t+1& t\end{array}\right]$

$=\left[\begin{array}{cc}{t}^2+{(t+1)}^2& t^2-t+t^2+t\\ t^2-1+t^2+t& {(t-1)}^2+t^2\end{array}\right]$

$=\left[\begin{array}{cc}{t}^2+{(t+1)}^2& 2t^2\\ 2t^2& {(t-1)}^2+t^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$2t^2=0$

$t=0$

$\therefore$Trace$=t+t=2t=0$