If $A = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$ then ${A^{4n + 1}},n \in N$
$\left( a \right)\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
$\left( b \right)\left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
$\left( c \right)\left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
$\left( d \right)\left[ {\begin{array}{*{20}{c}}
  { - i}&0&0 \\
  0&{ - i}&0 \\
  0&0&{ - i}
\end{array}} \right]$

Hint: In this particular question use the concept that first find out the square of the matrix and then fourth power of the matrix using matrix multiplication and also use that in complex, $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$ so use these concepts to reach the solution of the question.

Complete step by step answer:
Given matrix
$A = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
Now we have to find out the value of ${A^{4n + 1}}$, where $n \in N$.
Now multiply matrix A with matrix A we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
Now apply matrix multiplication rule we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {{i^2} + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 0} \\
  {0 + 0 + 0}&{0 + {i^2} + 0}&{0 + 0 + 0} \\
  {0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + {i^2}}
\end{array}} \right]$
Now as we know that in complex, $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$ s use this property we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
Now multiply ${A^2}$ with ${A^2}$ we have,
$ \Rightarrow {A^4} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
$ \Rightarrow {A^4} = \left[ {\begin{array}{*{20}{c}}
  {{{\left( { - 1} \right)}^2} + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 0} \\
  {0 + 0 + 0}&{0 + {{\left( { - 1} \right)}^2} + 0}&{0 + 0 + 0} \\
  {0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + {{\left( { - 1} \right)}^2}}
\end{array}} \right]$
$ \Rightarrow {A^4} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = I$..................... (1), where I is called as identity matrix.
Now as we know that ${I^m} = 1...........\left( 2 \right),m \in N$
Now we have to find out the value of
${A^{4n + 1}}$
Now the above equation is also written as,
$ \Rightarrow {A^{4n + 1}} = \left( {{A^{4n}}} \right)\left( A \right)$
$ \Rightarrow {A^{4n + 1}} = {\left( {{A^4}} \right)^n}\left( A \right)$
Now from equation (1) we have,
$ \Rightarrow {A^{4n + 1}} = {\left( I \right)^n}\left( A \right)$
Now from equation (2) we have,
$ \Rightarrow {A^{4n + 1}} = \left( 1 \right)\left( A \right)$
$ \Rightarrow {A^{4n + 1}} = \left( A \right)$
$ \Rightarrow {A^{4n + 1}} = A = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
So this is the required answer.

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept we have to remember is that we always recall how to multiply two matrices which is shown above and always recall that the power of identity matrix which belongs to a natural number always remains an identity matrix.