If A=pmatrix0&-22&0pmatrix and (I_2+A)(I_2-A)^-1=pmatrixa&-b&apmatrix, then 13(a^2+b^2) is…

Question

If $A=\begin{pmatrix}0&-	an\frac	heta2\	an\frac	heta2&0\end{pmatrix}$ and $(I_2+A)(I_2-A)^{-1}=\begin{pmatrix}a&-b\b&a\end{pmatrix}$, then $13(a^2+b^2)$ is ......... .

Answer: 13

Accepted Answer

$a^2+b^2=\big|(I_2+A)(I_2-A)^{-1}\big|=\dfrac{|I_2+A|}{|I_2-A|}$.

$I_2+A=\begin{pmatrix}1&-	an\frac	heta2\	an\frac	heta2&1\end{pmatrix}$, $|I_2+A|=1+	an^2\frac	heta2$; similarly $|I_2-A|=1+	an^2\frac	heta2$.

So $a^2+b^2=1$, and $13(a^2+b^2)=13$.

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