Mathematics - Matrices Question with Solution | TestHub

(1) 

$P^2=\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

$\mathit{PEP}=\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 8& 13& 18\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$=\left[\begin{array}{ccc}1& 2& 3\\ 8& 13& 18\\ 2& 3& 4\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$=\left[\begin{array}{ccc}1& 3& 2\\ 8& 18& 13\\ 2& 4& 3\end{array}\right]=F$

(2)

Here, $F=PEP\Rightarrow PF=P^{2}EP\Rightarrow PF=EP$, $\left|E\right|=0, \left|F\right|=0$

$\left|EQ+PF{Q}^{-1}\right|=\left|EQ+EP{Q}^{-1}\right|=\mid E\left(Q+P{Q}^{-1}|=|E\left|\right|Q+P{Q}^{-1}\mid =0\right.$

$\left|EQ\right|+\left|PF{Q}^{-1}\right|=0+0=0$

(3)

Since, $\left|E\right|=0\Rightarrow \left|EF\right|=0$

Thus, $\left|{\left(EF\right)}^3\right|>{\left|EF\right|}^2$ is not possible

(4) $P=\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\Rightarrow \left|P\right|=-1$

$P^{-1}=\frac{ Adj P}{\left|P\right|}=-\left[\begin{array}{ccc}-1& 0& 0\\ 0& 0& -1\\ 0& -1& 0\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$P^{-1}EP=\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 2& 4& 3\\ 8& 18& 13\end{array}\right]$

$=\left[\begin{array}{ccc}1& 3& 2\\ 8& 18& 13\\ 2& 4& 3\end{array}\right]=F$

$F+P^{-1}EP=F+F=2F=\left[\begin{array}{ccc}2& 6& 4\\ 16& 36& 26\\ 4& 8& 6\end{array}\right]$

Sum of Diagonal elements of $F+P^{-1}EP=44$

$E+P^{-1}FP=E+\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 8& 18& 13\\ 2& 4& 3\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 8& 13& 18\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 8& 13& 18\\ 2& 3& 4\end{array}\right]$

$=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 8& 13& 18\end{array}\right]+\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 8& 13& 18\end{array}\right]$

$=\left[\begin{array}{ccc}2& 4& 6\\ 4& 6& 8\\ 16& 26& 36\end{array}\right]$

$\mathit{Tr}\left(E+P^{-1}FP\right)=44$