The number of diagonal matrix A of order n for which A³ = A is

A = diag (d₁, d₂, ..., d_n)

Given, A³ = A

$\Rightarrow \mathrm{diag}\left({\mathrm{d}}_1^3,{\mathrm{d}}_2^3,\dots ,{\mathrm{d}}_{\mathrm{n}}^3\right)=\mathrm{diag}\left({\mathrm{d}}_1,{\mathrm{d}}_2,\dots ,{\mathrm{d}}_{\mathrm{n}}\right)$

$\Rightarrow {{\mathrm{d}}_1}^3={\mathrm{d}}_1,{{\mathrm{d}}_2}^3={\mathrm{d}}_2,\dots ,{{\mathrm{d}}_{\mathrm{n}}}^3={\mathrm{d}}_{\mathrm{n}}$

Hence, all d₁, d₂, d₃, ..., d_n have three possible values $\pm 1$, 0.

Each diagonal element can be selected in three ways. Hence, the number of different matrices is 3ⁿ.