Let M=pmatrix0&-&0pmatrix, where is a non-zero real number, and N=_k=1^49M^2k.

Question

Let $M=\begin{pmatrix}0&-\alpha\\alpha&0\end{pmatrix}$, where $\alpha$ is a non-zero real number, and $N=\displaystyle\sum_{k=1}^{49}M^{2k}$. If $(I-M^2)N=-2I$, then the positive integral value of $\alpha$ is\_\_\_\_

Answer: 1

Accepted Answer

$M^2=\begin{pmatrix}-\alpha^2&0\0&-\alpha^2\end{pmatrix}=-\alpha^2I$.

$N=\displaystyle\sum_{k=1}^{49}(-\alpha^2)^kI = \left[-\alpha^2+\alpha^4-\cdots-\alpha^{98}ight]I = \frac{-\alpha^2(1-(-\alpha^2)^{49})}{1+\alpha^2}I = \frac{-\alpha^2(\alpha^{98}+1)}{1+\alpha^2}I.$

$I-M^2=(1+\alpha^2)I$, so $(I-M^2)N=-\alpha^2(\alpha^{98}+1)I=-2I \Rightarrow \alpha^2(\alpha^{98}+1)=2$.

Testing $\alpha=1$: $1(1+1)=2$ .

So the positive integral value is $\alpha=1$.

Solutions · 1