
If \[A = \left( {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right)\] then \[{A^4}\] equals to
A. \[\left( {\begin{array}{*{20}{c}}
1&{ - 4i} \\
0&1
\end{array}} \right)\]
B.\[\left( {\begin{array}{*{20}{c}}
{ - 1}&{ - 4i} \\
0&{ - 1}
\end{array}} \right)\]
C. \[\left( {\begin{array}{*{20}{c}}
{ - i}&4 \\
0&i
\end{array}} \right)\]
D. \[\left( {\begin{array}{*{20}{c}}
1&4 \\
0&1
\end{array}} \right)\]
Answer
161.4k+ views
Hint:We are given a Matrix A in the question. On squaring the matrix A, we get \[{i^2} = - 1\]. On squaring matrix \[{A^2}\] we get matrix \[{A^4}\]. To find \[{A^2}\]we multiply the first row's entries of one matrix by the first column's entries of the second matrix, and so on for all other entries.
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
\[A = \left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right]\]
On squaring the matrix A we get,
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right]\]
On squaring the matrix \[{A^2}\]we get,
\[{A^4} = {A^2}.{A^2} = \left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{ - 4i} \\
0&1
\end{array}} \right]\]
\[ \Rightarrow {A^4} = \left[ {\begin{array}{*{20}{c}}
1&{ - 4i} \\
0&1
\end{array}} \right]\]
Therefore, the correct option is (A).
Note: In order to solve the given question, one must know to multiply two matrices. One must also know the properties of imaginary numbers. Imaginary numbers are the numbers which when squared, gives the negative result i.e., \[{i^2} = - 1\].
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
\[A = \left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right]\]
On squaring the matrix A we get,
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
i&1 \\
0&i
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right]\]
On squaring the matrix \[{A^2}\]we get,
\[{A^4} = {A^2}.{A^2} = \left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}&{2i} \\
0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{ - 4i} \\
0&1
\end{array}} \right]\]
\[ \Rightarrow {A^4} = \left[ {\begin{array}{*{20}{c}}
1&{ - 4i} \\
0&1
\end{array}} \right]\]
Therefore, the correct option is (A).
Note: In order to solve the given question, one must know to multiply two matrices. One must also know the properties of imaginary numbers. Imaginary numbers are the numbers which when squared, gives the negative result i.e., \[{i^2} = - 1\].
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