Question

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

Answer 1

Hint:
we have given the value of $A = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
Then, we will find the value of ${A^{4n + 1}}$
First, We will find the value of ${A^2}$ then after finding the value of ${A^2}$ we will find the value of ${A^4}$
After that using the property ${A^{4n + 1}}={A^{4n}}$.

Complete step by step solution:
Here, we have given the value of $A = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
So, we have to find the value of ${A^{4n + 1}}$
Now,
 ${A^2} = A.A$
 $\therefore {A^2} = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$
 $\therefore $ By solving above equation, we get
 ${A^2} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
Now, for ${A^4}$
 ${A^4} = {A^2}.{A^2}$
 $\therefore {A^4} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
$\therefore $By solving above equation, we get
 ${A^4} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = I$
We can also write ${A^{4n + 1}}$ as ${A^{4n}}$ .A
 $\therefore {A^{4n}}.A = {\left( {{A^4}} \right)^n}.A$
$\because $We have proven above that ${A^4} = I$
 $\therefore {A^{4n}}.A = {\left( I \right)^n}.A$
As we know that ${I^n} = I$
 $\therefore {A^{4n}}.A = I.A$
 $\therefore {A^{4n + 1}} = A$
 $\therefore {A^{4n + 1}} = \left[ {\begin{array}{*{20}{c}}
  i&0&0 \\
  0&i&0 \\
  0&0&i
\end{array}} \right]$

Note:
Additional Information:
Some different types of Matrices:
Symmetric Matrix: A square matrix A $ = \left[ {{a_{ij}}} \right]$ is called a symmetric matrix if ${a_{ij}} = {a_{ij}},$ for all i, j.
Skew-Symmetric Matrix: When ${a_{ij}} = - {a_{ij}}$
Orthogonal Matrix: If $A{A^T} = {I_n} = {A^T}.A$
Involuntary Matrix: ${A^2} = I$ or ${A^{ - 1}} = A$
Idempotent Matrix: If ${A^2} = A$