Question

Let $X = \begin{bmatrix} 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ \end{bmatrix}$,
Y = α I + β X + γ X 2 and
Z = α²l - αβX + (β² - αϒ)X² ,α,β,ϒ ∈ 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 ( α - β + ϒ )² is equal to ________.

Answer 1

The correct answer is 100
∵ $$X = \begin{bmatrix} 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ \end{bmatrix}
$∴$ $X^2 = \begin{bmatrix} 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ \end{bmatrix}$
$∴ Y = αl + βX + γX² =$ $\begin{bmatrix} \alpha & \beta & \gamma \\\ 0 & \alpha & \beta \\\ 0 & 0 & \alpha \\\ \end{bmatrix}$
∵ γ.γ-1 = l
$∴$ $\begin{bmatrix} \alpha & \beta & \gamma \\\ 0 & \alpha & \beta \\\ 0 & 0 & \alpha \\\ \end{bmatrix} \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \\\ 0 & \frac{1}{5} & -\frac{2}{5} \\\ 0 & 0 & \frac{1}{5} \\\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{bmatrix}$
∴ $\begin{bmatrix} \frac{\alpha}{5} & \beta - \frac{2\alpha}{5} & \alpha - \frac{2\beta + \gamma}{5} \\\ 0 & \frac{\alpha}{5} & \beta - \frac{2\alpha}{5} \\\ 0 & 0 & \frac{\alpha}{5} \\\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{bmatrix}$
$∴ α = 5, β = 10 , γ = 15$
Therefore , $( α - β - γ )² = 100$