Question

Say true or false:
If A is a periodic matrix with period 2, then $ {A^6} = A $
a. True
b. False

Answer 1

Hint: In this question they have mentioned the periodic matrix, the periodic matrix is defined as a square matrix such that $ {A^{k + 1}} = A $ , where k is a its period. Suppose if k = 1 then $ {A^2} = A $ and this is called an idempotent matrix.

Complete step-by-step answer:
Now in this question they have mentioned that the matrix A is a periodic matrix with period 2. So now we will check that by considering the matrix
Let $ A = \left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 6} \\
  { - 3}&2&9 \\
  2&0&{ - 3}
\end{array}} \right] $
First, we prove $ {A^3} = A $
So, consider the matrix A and find $ {A^2} $
 $
  {A^2} = A \times A \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 6} \\
  { - 3}&2&9 \\
  2&0&{ - 3}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 6} \\
  { - 3}&2&9 \\
  2&0&{ - 3}
\end{array}} \right] \\
  $
By multiplying
 $ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 \times 1 - 2 \times - 3 - 6 \times 2}&{1 \times - 2 - 2 \times 2 - 6 \times 0}&{1 \times - 6 - 2 \times 9 - 6 \times - 3} \\
  { - 3 \times 1 + 2 \times - 3 + 9 \times 2}&{ - 3 \times - 2 + 2 \times 2 + 9 \times 0}&{ - 3 \times - 6 + 2 \times 9 + 9 \times - 3} \\
  {2 \times 1 + 0 \times - 3 - 3 \times 2}&{2 \times - 2 + 0 \times 2 - 3 \times 0}&{2 \times - 6 + 0 \times 9 - 3 \times - 3}
\end{array}} \right] $
On simplifying we have
 $
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + 6 - 12}&{ - 2 - 4 + 0}&{ - 6 - 18 + 18} \\
  { - 3 - 6 + 18}&{6 + 4 + 0}&{18 + 18 - 27} \\
  {2 + 0 - 6}&{ - 4 + 0 + 0}&{ - 12 + 0 + 9}
\end{array}} \right] \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  { - 5}&{ - 6}&{ - 6} \\
  9&{10}&9 \\
  { - 4}&{ - 4}&{ - 3}
\end{array}} \right] \\
  $
Now we find the value of $ {A^3} $ that is the product of $ {A^2} $ and A.
 $
  {A^3} = {A^2} \times A \\
   \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 5}&{ - 6}&6 \\
  9&{10}&9 \\
  { - 4}&{ - 4}&{ - 3}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 6} \\
  { - 3}&2&9 \\
  2&0&{ - 3}
\end{array}} \right] \\
  $
By multiplying
 $ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 5 \times 1 - 6 \times - 3 - 6 \times 2}&{ - 5 \times - 2 - 6 \times 2 - 6 \times 0}&{ - 5 \times - 6 - 6 \times 9 - 6 \times - 3} \\
  {9 \times 1 + 10 \times - 3 + 9 \times 2}&{9 \times - 2 + 10 \times 2 + 9 \times 0}&{9 \times - 6 + 10 \times 9 + 9 \times - 3} \\
  { - 4 \times 1 + - 4 \times - 3 - 3 \times 2}&{ - 4 \times - 2 - 4 \times 2 - 3 \times 0}&{ - 4 \times - 6 - 4 \times 9 - 3 \times - 3}
\end{array}} \right] $
On simplifying we have
 $
   \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 5 + 18 - 12}&{10 - 12 + 0}&{30 - 54 + 18} \\
  {9 - 30 + 18}&{ - 18 + 20 + 0}&{ - 54 + 90 - 27} \\
  { - 4 + 12 - 6}&{8 - 8 + 0}&{24 - 36 + 9}
\end{array}} \right] \\
   \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 6} \\
  { - 3}&2&9 \\
  2&0&{ - 3}
\end{array}} \right] \\
  $
Hence, we obtained $ {A^3} = A $ and we can say that the matrix A is a periodic matrix with period 2. We have proved that by considering the matrix. Hence the matrix A is a periodic matrix.
Now we want to prove that $ {A^6} = A $
By the definition of periodic matrix, we have a square matrix such that $ {A^{k + 1}} = A $ , where k is a period. We already proved that $ {A^3} = A $ and its period is 2. If the given matrix is a periodic it satisfies another condition that is $ {A^2} = A $ . Therefore $ {A^6} $ can be written as
 $ {A^6} = {\left( {{A^3}} \right)^2} $
Already we have proved that $ {A^3} = A $ , by substituting that we can write the above inequality as
 $
   \Rightarrow {A^6} = {\left( A \right)^2} \\
   \Rightarrow {A^6} = {A^2} \\
  $
Therefore $ {A^6} \ne A $
The answer for the given question is false. Because $ {A^6} \ne A $ but $ {A^6} = {A^2} $
So, the correct answer is “Option b”.

Note: We must know the definition of the periodic matrix which makes us understand the question properly. We can solve the above question in only two steps . In the question they have already mentioned that a is a periodic matrix with 2 as its period, $ {A^3} = A $ . Now consider $ {A^6} $ can be written as $ {A^6} = {\left( {{A^3}} \right)^2} $ where $ {A^3} = A $ we have $ {A^6} = {A^2} $ and we can say it is false. We did a lengthy process to verify the A is a periodic matrix of period 2.