If $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right]$ , then ${A^n} = {2^k}A$ , where k= ?
A. ${2^{n - 1}}$
B. n+1
C. n-1
D. $2\left( {n - 1} \right)$

Hint: In this question to find the value of k we will find the value of ${A^2}$, ${A^3}$ and, \[{A^4}\] with the help of the property of matrix i.e.
 $\left[ {\begin{array}{*{20}{c}}
  {ak}&{bk} \\
  {ck}&{dk}
\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$
Now we will find the value of ${A^2}$ , ${A^3}$ and, \[{A^4}\] in terms A so that we can compare these equations with the given equation i.e. ${A^n} = {2^k}A$ so that we can find the relation between n and k to get the required answer.

Complete step-by-step answer:
Given data: $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right]$
We know that, $\left[ {\begin{array}{*{20}{c}}
  {ak}&{bk} \\
  {ck}&{dk}
\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$
Therefore using this property on matrix A
  $ \Rightarrow A = 2\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]...........(i)$
Multiplying both sides by matrix A
 $ \Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]$
On multiplication of matrix and simplification we get,
 $ \Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}}
  {1 + 1}&{ - 1 - 1} \\
  { - 1 - 1}&{1 + 1}
\end{array}} \right]$
On simplifying the elements of the matrix we get,
 $ \Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right]..................(ii)$
Now, we know that $\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right] = A$
 $ \Rightarrow {A^2} = {2^2}A.............(iii)$
Now using the property of a matrix i.e. $\left[ {\begin{array}{*{20}{c}}
  {ak}&{bk} \\
  {ck}&{dk}
\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ in equation(ii)
 $ \Rightarrow {A^2} = {2^3}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]$
On multiplying equation(i) and equation(ii)
 $ \Rightarrow {A^3} = {2^4}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]$
On multiplication of matrix and simplification
 $ \Rightarrow {A^3} = {2^4}\left[ {\begin{array}{*{20}{c}}
  {1 + 1}&{ - 1 - 1} \\
  { - 1 - 1}&{1 + 1}
\end{array}} \right]$
On simplifying the elements of the matrix
 \[ \Rightarrow {A^3} = {2^4}\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right].................(iv)\]
Now, we know that $\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right] = A$
 $ \Rightarrow {A^3} = {2^4}A............(v)$
Now using the property of a matrix i.e. $\left[ {\begin{array}{*{20}{c}}
  {ak}&{bk} \\
  {ck}&{dk}
\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ in equation(iv)
 \[ \Rightarrow {A^3} = {2^5}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\]
Now multiplying equation(i) and equation(iv)
 \[ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\]
On multiplication of matrix and simplification
 $ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}}
  {1 + 1}&{ - 1 - 1} \\
  { - 1 - 1}&{1 + 1}
\end{array}} \right]$
On simplifying the elements of the matrix
 \[ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right]\]
Now, we know that $\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right] = A$
 $ \Rightarrow {A^4} = {2^6}A...............(vi)$
Now on concluding from equation (iii), equation(v) and, equation(vi)
We can say that, $k = 2\left( {n - 1} \right)$
Hence, Option (D) is correct.

Note: While taking common any scalar elements from the square matrix some of the students apply the property of the determinant
i.e. $\left| {\begin{array}{*{20}{c}}
  {ka}&{kb} \\
  {tc}&{td}
\end{array}} \right| = kt\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|$ or $\left| {\begin{array}{*{20}{c}}
  {ka}&{kb} \\
  {kc}&{kd}
\end{array}} \right| = {k^2}\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right|$
which is not applicable for the matrix as the matrix follow the property
i.e. $\left[ {\begin{array}{*{20}{c}}
  {ak}&{bk} \\
  {ck}&{dk}
\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$
so most of the students make mistakes while taking common like while finding the ${A^2}$ using the equation (ii)
i.e. ${A^2} = {2^2}\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 2}&2
\end{array}} \right]$
 $ \Rightarrow {A^2} = {2^2} \times 2 \times 2\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]$ , which is not correct and will take us to the wrong answer, so avoid making mistakes like this to get the correct answer.