The number of diagonal matrix, `A` or order`n` which `A^3=A` is

Options

A. 1
B. 0
C. `2^(n)`
D. `3^(n)`

Answer

D

Explanation

A= diag `(d_(1), d_(2), .... ,d_(n))`
Given, `A^(3)=A`
`implies` diag `(d_(1)^(3), d_(2)^(3), ...., b_(n)^(3))=` diag `(d_(1), d_(2), ... , d_(n))`
`implies d_(1)^(3)=d_(1), d_(2)^(3)=d_(2),..., d_(n)^(3)=d_(n)`
Hence, all `d_(1), d_(2), d_(3), ..., 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 `3^(n)`.