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The number of diagonal matrix, `A`or order`n`which `A^3=A`is a. is a a. 1 b. 0 c. `2^n`d. `3^n`A. 1B. 0C. `2^(n)`D. `3^(n)`

Answer» Correct Answer - D
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)`.


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