Question

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

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.