Question

Let \(X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}\),
Y = αI + βX + γX² 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

Correct Answer: 100

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 ,  \(( α - β - γ )² = 100\)