Let $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$, $Y = \alpha l + \beta X + \gamma X^{2}$ and $Z = \alpha^{2} I - \alpha\beta X + (\beta^{2} - \alpha\gamma) X^{2}$, where $\alpha$, $\beta$, $\gamma \in \mathbb{R}$. If $Y^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{5}…

Question

Let $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$, $Y = \alpha l + \beta X + \gamma X^{2}$ and $Z = \alpha^{2} I - \alpha\beta X + (\beta^{2} - \alpha\gamma) X^{2}$, where $\alpha$, $\beta$, $\gamma \in \mathbb{R}$. If $Y^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \ 0 & \frac{1}{5} & -\frac{2}{5} \ 0 & 0 & \frac{1}{5} \end{bmatrix}$, then $(\alpha - \beta + \gamma)^{2}$ is equal to ______.

Aicharya · Accepted Answer

Given, $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$, so $X^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ Now, finding $Y = \alpha l + \beta X + \gamma X^2$ & $Z = \alpha^2 I - \alpha \beta X + (\beta^2 - \alpha \gamma)X^2$ by putting the value of $X$ & $X^2$ we get, $Y = \begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix}$ & $Z = \begin{bmatrix} \alpha^2 & -\alpha \beta & \beta^2 - \alpha \gamma \ 0 & \alpha^2 & -\alpha \beta \ 0 & 0 & \alpha^2 \end{bmatrix}$ We know that $Y \cdot Y^{-1} = I$ $\Rightarrow \begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix} \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \ 0 & \frac{1}{5} & -\frac{2}{5} \ 0 & 0 & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ $\Rightarrow \begin{bmatrix} \frac{\alpha}{5} & -\frac{2\alpha}{5} + \frac{\beta}{5} & \frac{\alpha}{5} - \frac{2\beta}{5} + \frac{\gamma}{5} \ 0 & \frac{\alpha}{5} & -\frac{2\alpha}{5} + \frac{\beta}{5} \ 0 & 0 & \frac{\alpha}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ 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, $(\alpha - \beta + \gamma)^2 = (5 - 10 + 15)^2 = 100$

Let $X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$, $Y = \alpha l + \beta X +…

Solution