Question

Let S be set of $2\times 2$ matrices given by $S=\{A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],\text{where a, b, c, d}\in I\}\text{such that}{A}^T=A^{-1},then$

• A. Number of matrices in set S is equal to 8.
• B. Number of matrices in set S such that $|A-I_2|\ne 0$, is equal to 4.(where $I_2$ is an identity matrix of order 2)
• C. Symmetric matrices are more than skew symmetric matrices in set S.
• D. All matrices in set S are singular.

Answer 1

Answer: (A), (C)

Symmetric matrices are more than skew symmetric matrices in set S.

$A^T=A^{-1}$
$A{A}^T=I$
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$a=0,b=\pm 1,d=0,c=\pm 1ora=\pm 1,b=0,d=\pm 1,c=0$
So total 8 matrices are possible.
$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right],\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$
$|A-I_2|=|A-A{A}^T|=|A||I_2-A^T|=|A||(I_2-A^T{)}^T|=|A||I_2-A|$
$|A-I_2|=|A||I_2-A|$
As $|A-I_2|\ne 0\Rightarrow |A|=1$
So total 3 such matrices are possible
except 1 case
$A=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ where $|A-I_2|=0$