Question

If $S=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]$ and $A=\left[\begin{array}{ccc}b+c& c+a& b-c\\ c-b& c+b& a-b\\ b-c& a-c& a+b\end{array}\right](a,b,c\ne 0)$, then SAS${}^{-1}$ is

• A. symmetric matrix
• B. diagonal matrix
• C. invertible matrix
• D. singular matrix

Answer 1

Answer: (A), (B), (C)

The correct options are
A symmetric matrix
B invertible matrix
C diagonal matrix

$S=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right],\left|S\right|=-1(-1)+1\left(1\right)=2\therefore S^{-1}=\frac{AdjS}{\left|S\right|}=\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$

Now $SA=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}b+c& c+a& b-c\\ c-b& c+a& a-b\\ b-c& a-c& a+b\end{array}\right]=\left[\begin{array}{ccc}0& 2a& 2a\\ 2b& 0& 2b\\ 2c& 2c& 0\end{array}\right]SA{S}^{-1}=\left[\begin{array}{ccc}0& 2a& 2a\\ 2b& 0& 2b\\ 2c& 2c& 0\end{array}\right]\cdot \frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]=\left[\begin{array}{ccc}0& a& a\\ b& 0& b\\ 0& 0& c\end{array}\right]\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$

$=\left[\begin{array}{ccc}2a& 0& 0\\ 0& 2b& 0\\ 0& 0& 2c\end{array}\right]\therefore SA{S}^{-1}$ is a diagonal matrix

${\left(SA{S}^{-1}\right)}^T=\left[\begin{array}{ccc}2a& 0& 0\\ 0& 2b& 0\\ 0& 0& 2c\end{array}\right]=SA{S}^{-1}\therefore$ Symmetric

$\left|SA{S}^{-1}\right|\ne 0\therefore SA{S}^{-1}$ is non-singular and invertible