Question

\[\text{If }A=\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]\text{ then }{{A}^{64}}\text{ is:}\]
(a) \[\left[ \begin{matrix}
   1 & 32 \\
   32 & 1 \\
\end{matrix} \right]\]
(b) \[\left[ \begin{matrix}
   1 & 0 \\
   32 & 1 \\
\end{matrix} \right]\]
(c) \[\left[ \begin{matrix}
   1 & 32 \\
   0 & 1 \\
\end{matrix} \right]\]
(d) None of these

Answer 1

Hint: To solve this question, we will first find matrices \[{{A}^{2}},{{A}^{3}},{{A}^{4}}.\] Then we will develop a relationship between the matrix A and the calculated matrices \[{{A}^{2}},{{A}^{3}},{{A}^{4}}.\] With the help of this relation, we will determine what will be matrix \[{{A}^{64}}.\]

Complete step-by-step answer:
In our question, it is given that
\[A=\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]\]
We will consider that
\[{{A}^{n}}=\left[ \begin{matrix}
   {{a}_{11n}} & {{a}_{12n}} \\
   {{a}_{21n}} & {{a}_{22n}} \\
\end{matrix} \right]\]
Thus,
\[{{a}_{111}}=1,{{a}_{121}}=\dfrac{1}{2},{{a}_{211}}=0,{{a}_{221}}=1\]
Now, to solve this question, we will first find the matrix \[{{A}^{2}}\]. The matrix \[{{A}^{2}}\] is given by:
\[{{A}^{2}}=A\times A\]
\[{{A}^{2}}=\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]......\left( i \right)\]
The product of any two square matrices A and B is given by:
\[AB=\left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} \\
   {{a}_{3}} & {{a}_{4}} \\
\end{matrix} \right]\left[ \begin{matrix}
   {{b}_{1}} & {{b}_{2}} \\
   {{b}_{3}} & {{b}_{4}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{a }_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{3}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{4}} \\
   {{a}_{3}}{{b}_{1}}+{{a}_{4}}{{b}_{3}} & {{a}_{3}}{{b}_{2}}+{{a}_{4}}{{b}_{4}} \\
\end{matrix} \right]\]
With the help of the above formula, we will find the matrix \[{{A}^{2}}.\] Thus, we have the following equation:
\[{{A}^{2}}=\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   \left( 1\times 1 \right)+\left( \dfrac{1}{2}\times 0 \right) & \left( 1\times \dfrac{1}{2} \right)+\left( \dfrac{1}{2}\times 1 \right) \\
   \left( 0\times 1 \right)+\left( 1\times 0 \right) & \left( 0\times \dfrac{1}{2} \right)+\left( 1\times 1 \right) \\
\end{matrix} \right]\]
\[{{A}^{2}}=\left[ \begin{matrix}
   1+0 & \dfrac{1}{2}+\dfrac{1}{2} \\
   0+0 & 0+1 \\
\end{matrix} \right]\]
\[{{A}^{2}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, we will find matrix \[{{A}^{3}}.\] The matrix \[{{A}^{3}}\] is given by:
\[{{A}^{3}}=A\times A\times A\]
\[{{A}^{3}}=\left( A\times A \right)\times A\]
\[{{A}^{3}}={{A}^{2}}\times A.....\left( ii \right)\]
Now, we will put the value of \[{{A}^{2}}\] and A in the second equation. Thus, we get the following equation:
\[{{A}^{3}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   \left( 1\times 1 \right)+\left( 1\times 0 \right) & \left( 1\times \dfrac{1}{2} \right)+\left( 1\times 1 \right) \\
   \left( 0\times 1 \right)+\left( 1\times 0 \right) & \left( 0\times \dfrac{1}{2} \right)+\left( 1\times 1 \right) \\
\end{matrix} \right]\]
\[{{A}^{3}}=\left[ \begin{matrix}
   1+0 & \dfrac{1}{2}+1 \\
   0+0 & 0+1 \\
\end{matrix} \right]\]
\[{{A}^{3}}=\left[ \begin{matrix}
   1 & \dfrac{3}{2} \\
   0 & 1 \\
\end{matrix} \right]\]
Now, we will find the matrix \[{{A}^{4}}.\] The matrix \[{{A}^{4}}\] is given by:
\[{{A}^{4}}=A\times A\times A\times A\]
\[{{A}^{4}}=\left( A\times A \right)\times \left( A\times A \right)\]
\[{{A}^{4}}={{A}^{2}}\times {{A}^{2}}.....\left( iii \right)\]
Now, we will put the values of \[{{A}^{2}}\] in equation (iii). After doing this, we will get the following equation.
\[{{A}^{4}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   \left( 1\times 1 \right)+\left( 1\times 0 \right) & \left( 1\times 1 \right)+\left( 1\times 1 \right) \\
   \left( 0\times 1 \right)+\left( 1\times 0 \right) & \left( 0\times 1 \right)+\left( 1\times 1 \right) \\
\end{matrix} \right]\]
\[{{A}^{4}}=\left[ \begin{matrix}
   1+0 & 1+1 \\
   0+0 & 0+1 \\
\end{matrix} \right]\]
\[{{A}^{4}}=\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, we can see that \[{{a}_{111}}={{a}_{112}}={{a}_{113}}={{a}_{114}}.\]
So, we can say that \[{{a}_{11}}={{a}_{11\left( 64 \right)}}.\]
\[\Rightarrow {{a}_{11\left( 64 \right)}}=1\]
Similarly, we can also see that \[{{a}_{211}}={{a}_{212}}={{a}_{213}}={{a}_{214}}.\]
So, we can say that \[{{a}_{21}}={{a}_{21\left( 64 \right)}}.\]
\[\Rightarrow {{a}_{21\left( 64 \right)}}=0\]
Similarly,
\[\Rightarrow {{a}_{22\left( 64 \right)}}=1\]
But we do not know anything about \[{{a}_{12\left( 64 \right)}}.\] To determine the relation, we have toe following data:
\[{{a}_{121}}=\dfrac{1}{2}\]
\[{{a}_{121}}=1=\dfrac{1}{2}\times 2\]
\[{{a}_{123}}=\dfrac{3}{2}=\dfrac{1}{2}\times 3\]
Similarly,
\[{{a}_{12\left( 64 \right)}}=\dfrac{1}{2}\times 64=32\]
Thus, the matrix \[{{A}^{64}}\] is given by:
\[{{A}^{64}}=\left[ \begin{matrix}
   {{a}_{11\left( 64 \right)}} & {{a}_{12\left( 64 \right)}} \\
   {{a}_{21\left( 64 \right)}} & {{a}_{22\left( 64 \right)}} \\
\end{matrix} \right]\]
On putting the respective values, we will get,
\[{{A}^{64}}=\left[ \begin{matrix}
   1 & 32 \\
   0 & 1 \\
\end{matrix} \right]\]
Hence, the option (c) is the right answer.

Note: We can also solve the above question as follows:
We will calculate \[{{A}^{2}}\] first. We get,
\[{{A}^{2}}=A\times A=\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & \dfrac{1}{2} \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, we will calculate the matrix \[{{A}^{4}}.\] We get,
\[{{A}^{4}}={{A}^{2}}\times {{A}^{2}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, we will calculate the matrix \[{{A}^{8}}.\] We get,
\[{{A}^{8}}={{A}^{4}}\times {{A}^{4}}=\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 4 \\
   0 & 1 \\
\end{matrix} \right]\]
Similarly,
\[{{A}^{16}}={{A}^{8}}\times {{A}^{8}}=\left[ \begin{matrix}
   1 & 4 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 4 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 8 \\
   0 & 1 \\
\end{matrix} \right]\]
\[{{A}^{32}}={{A}^{16}}\times {{A}^{16}}=\left[ \begin{matrix}
   1 & 8 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 8 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 16 \\
   0 & 1 \\
\end{matrix} \right]\]
Hence,
\[{{A}^{64}}={{A}^{32}}\times {{A}^{32}}=\left[ \begin{matrix}
   1 & 16 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 16 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 32 \\
   0 & 1 \\
\end{matrix} \right]\]
Here, we can see that the answer we get here is also the same.