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If\[A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right]\], then \[{A^n} = \]
A. \[\left[ {\begin{array}{*{20}{c}}
  1&{2n} \\
  0&1
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
  2&n \\
  0&1
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
  1&{2n} \\
  0&{ - 1}
\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}
  1&{2n} \\
  0&1
\end{array}} \right]\]

Answer
VerifiedVerified
162.9k+ views
Hint: In the given problem we first find the value of \[{A^2}\]. We observe that the second element in the matrix \[{A^2}\]is 2 times of 2 which is 4. Similarly, we find the value of \[{A^3}\] and we observe that the second element in matrix \[{A^3}\] is 3 times of 2 which is 6. Hence we write the general term \[{A^n}\] using this observation.

Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]

Complete step by step solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  0&1
\end{array}} \right]\]
\[{A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&6 \\
  0&1
\end{array}} \right]\]
In general, we can say that,
\[{A^n} = \left[ {\begin{array}{*{20}{c}}
  1&{2n} \\
  0&1
\end{array}} \right]\]

Therefore Option A. is the correct answer.

Note: To solve the given problem, one must know to multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix. One must also know to form a general term from a sequence of terms.