Question

If $A=\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]$, then $A^{\prime }A$ is equal to

• A. $I$
• B. $-iA$
• C. $-I$
• D. $iA$

Answer 1

Answer: (C)

$-I$

$A=\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]$
$A^{\prime }=\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]$, since $A$ is a symmetric matrix
$\therefore A^{\prime }A=\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}{i}^2& 0\\ 0& i^2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=-\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]=-I$
Since $i^2=-1$