IfA=[ … ],(θ = π 24) and A^5=[ … ], where i=√-1, then which one of the following is not true?

$${A}^{2}=[\begin{matrix}\mathrm{cos}\theta i\mathrm{sin}\theta \\ i\mathrm{sin}\theta \mathrm{cos}\theta \end{matrix}][\begin{matrix}\mathrm{cos}\theta i\mathrm{sin}\theta \\ i\mathrm{sin}\theta \mathrm{cos}\theta \end{matrix}]$$A2=[cosθisinθisinθcosθ​][cosθisinθisinθcosθ​]

$$=[\begin{matrix}{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta 2i\mathrm{sin}\theta \mathrm{cos}\theta \\ 2i\mathrm{sin}\theta \mathrm{cos}\theta {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \end{matrix}]$$=[cos2θ−sin2θ2isinθcosθ2isinθcosθcos2θ−sin2θ​]

$$=[\begin{matrix}\mathrm{cos}2\theta i\mathrm{sin}2\theta \\ i\mathrm{sin}2\theta \mathrm{cos}2\theta \end{matrix}]$$=[cos2θisin2θisin2θcos2θ​]

$${A}^{3}=[\begin{matrix}\mathrm{cos}2\theta i\mathrm{sin}2\theta \\ i\mathrm{sin}2\theta \mathrm{cos}2\theta \end{matrix}][\begin{matrix}\mathrm{cos}\theta i\mathrm{sin}\theta \\ i\mathrm{sin}\theta \mathrm{cos}\theta \end{matrix}]$$A3=[cos2θisin2θisin2θcos2θ​][cosθisinθisinθcosθ​]

$$=[\begin{matrix}\mathrm{cos}3\theta i\mathrm{sin}3\theta \\ i\mathrm{sin}3\theta \mathrm{cos}3\theta \end{matrix}]$$=[cos3θisin3θisin3θcos3θ​]

Similarly, ${A}^{5}=[\begin{matrix}\mathrm{cos}5\theta i\mathrm{sin}5\theta \\ i\mathrm{sin}5\theta \mathrm{cos}5\theta \end{matrix}]=[\begin{matrix}a b \\ c d\end{matrix}]$A5=[cos5θisin5θisin5θcos5θ​]=[abcd​]

$a=\mathrm{cos}5\theta$a=cos5θ, $b=i\mathrm{sin}5\theta$b=isin5θ, $c=i\mathrm{sin}5\theta$c=isin5θ

$$d=\mathrm{cos}5\theta$$d=cos5θ

$$a=d,b=c$$a=d,b=c

$(1)$(1) ${a}^{2}+{b}^{2}={\mathrm{cos}}^{2}5\theta -{\mathrm{sin}}^{2}5\theta =\mathrm{cos}10\theta \in (0,1)$a2+b2=cos25θ−sin25θ=cos10θ∈(0,1) as $\theta =\frac{\pi }{24}$θ=24π​

$(2)$(2) ${a}^{2}-{d}^{2}=0$a2−d2=0

$(3)$(3) ${a}^{2}-{b}^{2}={\mathrm{cos}}^{2}5\theta +{\mathrm{sin}}^{2}5\theta =1$a2−b2=cos25θ+sin25θ=1

$(4)$(4) ${a}^{2}-{c}^{2}={a}^{2}-{b}^{2}=1$a2−c2=a2−b2=1