Question

The number of diagonal matrix ${\text{A}}$ of order n for which {{\text{A}}^{\text{3}}}$$$${\text{ = A}}is
A. 1
B. 0
C. 2
D. 3

Answer 1

First, we take $A =$diag(${d_1},{d_2},$……${d_n}$) and find ${A^3}$, and we apply the given condition, and then solve to get the required answer.

Complete step-by-step answer:
We have,$A =$diag(${d_1},{d_2},$……${d_n}$)
As we know for a diagonal matrix, ${A^n} = diag(d_1^n,d_2^n,........,d_n^n)$
So now, ${A^3}$= diag($d_1^3,d_2^3,........,d_n^3$)
As given,{A^3}$$$$ = A
So we have,
diag(${d_1},{d_2},$……${d_n}$) = diag($d_1^3,d_2^3,........,d_n^3$)
so, ${d_1}$= $d_1^3$,$d_2^{}$ =$d_2^3$,…………., ${d_n}$= $d_n^3$
for all $i = 1,2,3,...,n$,
we have, ${d_i}$=$d_i^3$
\Rightarrow $$$${d_i}- $d_i^3$= 0
\Rightarrow $$$${d_i}( $1 - d_i^2$) = 0
So, we have ${d_i}$= 0 or $1 - d_i^2$=0 \Rightarrow $$$$d_i^2= 1 \Rightarrow $$$${d_i}= $\pm$1
Then we get, ${d_i}$= 0, $\pm$1
So, The number of diagonal matrices $A$ of order n for which {A^3}$$$$ = Ais, 3.

Note: If all the ${d_i}$’s are not equal then by changing the terms between 0 and $\pm$1 we can have total ${3^n}$ number of matrices for which The number of diagonal matrix $A$ of order n for which {A^3}$$$$ = A.