Question

If $A = \left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]$, then ${A^4}$ is equal to
A $\left[ {\begin{array}{*{20}{c}}
  1&{{a^4}} \\
  0&1
\end{array}} \right]$
B $\left[ {\begin{array}{*{20}{c}}
  4&{4a} \\
  0&4
\end{array}} \right]$
C $\left[ {\begin{array}{*{20}{c}}
  4&{{a^2}} \\
  0&4
\end{array}} \right]$
D $\left[ {\begin{array}{*{20}{c}}
  1&{4a} \\
  0&1
\end{array}} \right]$

Answer 1

Hint: First we will find ${A^2}$ by multiplying $A$ with $A$. Then will find ${A^3}$ by multiplying ${A^2}$ with $A$. Then will find the ${A^4}$ by multiplying ${A^3}$ with $A$.

Complete step by step Solution:
 Given, $A = \left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]$
${A^2} = A.A$
${A^2} = \left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]$
After multiplying we will get
$ = \left[ {\begin{array}{*{20}{c}}
  1&{a + a} \\
  0&1
\end{array}} \right]$
After solving will get
$ = \left[ {\begin{array}{*{20}{c}}
  1&{2a} \\
  0&1
\end{array}} \right]$
${A^3} = {A^2}.A$
${A^3} = \left[ {\begin{array}{*{20}{c}}
  1&{2a} \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]$
After multiplying, we will get
$ = \left[ {\begin{array}{*{20}{c}}
  1&{a + 2a} \\
  0&1
\end{array}} \right]$
After solving, we get
\[ = \left[ {\begin{array}{*{20}{c}}
  1&{3a} \\
  0&1
\end{array}} \right]\]
${A^4} = {A^3}.A$
${A^4} = \left[ {\begin{array}{*{20}{c}}
  1&{3a} \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&a \\
  0&1
\end{array}} \right]$
After multiplying, we will get
$ = \left[ {\begin{array}{*{20}{c}}
  1&{a + 3a} \\
  0&1
\end{array}} \right]$
After solving it, will get
$ = \left[ {\begin{array}{*{20}{c}}
  1&{4a} \\
  0&1
\end{array}} \right]$

Therefore, the correct option is (D).

Note: Students should do each multiplication correctly with the correct logic of multiplication of matrices to get the required answer.