
If\[A = \left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\], then \[{A^{100}} = \]
A. \[{2^{100}}A\]
B. \[{2^{99}}A\]
C. \[{2^{101}}A\]
D. None of these
Answer
162.9k+ views
Hint: In the given problem we first find the value of \[{A^2}\]. We then write \[{A^2}\] in terms of matrix\[A\]. Similarly, we find the value of \[{A^3}\] and write it in terms of matrix \[A\]. We then find a general term \[{A^n}\]whose value is in terms of matrix \[A\]. From this general term we can find the value of \[{A^{100}}\].
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,
\[A = \left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&2 \\
2&2
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = 2A\]
\[{A^3} = {A^2}.A = 2\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^2}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^2}A\]
In general,
\[{A^n} = {2^{n - 1}}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^{n - 1}}A\]
Therefore \[{A^{100}} = {2^{99}}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^{99}}A\]
Option B. 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 and hence find the required term from the general term.
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,
\[A = \left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&2 \\
2&2
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = 2A\]
\[{A^3} = {A^2}.A = 2\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^2}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^2}A\]
In general,
\[{A^n} = {2^{n - 1}}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^{n - 1}}A\]
Therefore \[{A^{100}} = {2^{99}}\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&1
\end{array}} \right] = {2^{99}}A\]
Option B. 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 and hence find the required term from the general term.
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