Question: If \(A = \left[ {\begin{array}{*{20}{c}} i&0&0 \\\ 0&i;&0 \\\ 0&0&i; \end{array}} \...

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

Solution

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 ${A^2} = A$