Mathematics - Matrices Question with Solution | TestHub

Given, $X=\left[\begin{array}{lll}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$, so $X^2=\left[\begin{array}{lll}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Now, finding $Y=\alpha l+\beta X+\gamma X^2$ &$Z={\alpha }^{2}I-\alpha \beta X+\left({\beta }^2-\alpha \gamma \right)X^2$  by putting the value of $X \& X^2$ we get,

 $Y=\left[\begin{array}{lll}\alpha & \beta & \gamma \\ 0& \alpha & \beta \\ 0& 0& \alpha \end{array}\right] \& Z=\left[\begin{array}{ccc}{\alpha }^2& -\alpha \beta & {\beta }^2-\alpha \gamma \\ 0& {\alpha }^2& -\alpha \beta \\ 0& 0& {\alpha }^2\end{array}\right]$

We know that $Y·Y^{-1}=\mathrm{I}$  

$\Rightarrow \left[\begin{array}{lll}\alpha & \beta & \gamma \\ 0& \alpha & \beta \\ 0& 0& \alpha \end{array}\right]\left[\begin{array}{ccc}\frac{1}{5}& \frac{-2}{5}& \frac{1}{5}\\ 0& \frac{1}{5}& \frac{-2}{5}\\ 0& 0& \frac{1}{5}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}\frac{\alpha }{5}& \frac{-2\alpha }{5}+\frac{\beta }{5}& \frac{\alpha }{5}-\frac{2\beta }{5}+\frac{\gamma }{5}\\ 0& \frac{\alpha }{5}& \frac{-2\alpha }{5}+\frac{\beta }{5}\\ 0& 0& \frac{\alpha }{5}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

On comparing L.H.S and R.H.S we get,

$\Rightarrow \frac{\alpha }{5}=1\Rightarrow \alpha =5$

$\Rightarrow -\frac{2}{5}\alpha +\frac{\beta }{5}=0\Rightarrow \beta =10$

$\Rightarrow \frac{\alpha }{5}-\frac{2\beta }{5}+\frac{\gamma }{5}=0\Rightarrow \gamma =15$

So, ${\left(\alpha -\beta +\gamma \right)}^2={\left(5-10+15\right)}^2=100$