Let X=pmatrix0&1&00&0&10&0&0pmatrix, Y= I+ X+ X^2, Z=^2I- X+(^2-)X^2, ,,R.

Question

Let $X=\begin{pmatrix}0&1&0\0&0&1\0&0&0\end{pmatrix}$, $Y=\alpha I+\beta X+\gamma X^2$, $Z=\alpha^2I-\alpha\beta X+(\beta^2-\alpha\gamma)X^2$, $\alpha,\beta,\gamma\in\mathbb{R}$. If $Y^{-1}=\begin{pmatrix}\frac15&-\frac25&\frac15\0&\frac15&-\frac25\0&0&\frac15\end{pmatrix}$, then $(\alpha-\beta+\gamma)^2$ is equal to\_\_\_

Answer: 100

Accepted Answer

Since $X^3=O$, $Y=\alpha I+\beta X+\gamma X^2$ is upper triangular with all diagonal entries $\alpha$; from $Y^{-1}$'s diagonal being $	frac15$, $\alpha=5$.

Using $YZ=(\alpha I+\beta X+\gamma X^2)(\alpha^2I-\alpha\beta X+(\beta^2-\alpha\gamma)X^2)$, one can verify (by expanding and using $X^3=0$) that $YZ=\alpha^3I$, so $Y^{-1}=\dfrac{Z}{\alpha^3}$.

Comparing entries: the $(1,2)$ entry of $Y^{-1}$ is $-\dfrac{\alpha\beta}{\alpha^3}=-\dfrac25 \Rightarrow \dfrac{\beta}{\alpha^2}=\dfrac25\Rightarrow\beta=\dfrac25(25)=10$.

The $(1,3)$ entry gives $\dfrac{\beta^2-\alpha\gamma}{\alpha^3}=\dfrac15 \Rightarrow \beta^2-\alpha\gamma=25 \Rightarrow 100-5\gamma=25\Rightarrow\gamma=15$.

So $\alpha-\beta+\gamma=5-10+15=10$, and $(\alpha-\beta+\gamma)^2=100$.

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