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

Hint: 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 \[{A^3}\]\[ = A\]is, 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 \[{A^3}\]\[ = A\].