Question

If $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$, then find the value of ${A^{40}}$.
A. $\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  0&0
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
  { - 1}&1 \\
  0&{ - 1}
\end{array}} \right]$

Answer 1

Hint: In this question, in order to find ${A^{40}}$,we need to find ${A^2}$ as ${({A^2})^{20}} = {A^{40}}$. To solve these types of questions, we must know about the identity matrix, complex numbers, and basic operations on matrices, such as addition, subtraction, multiplication, and transpose of the matrix.

Complete step by step Solution:
Given: $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$
Step 1: To find ${A^2}$, we have to multiply $A$ with $A$. In order to do that, multiply and add each element of the first row of matrix $A$ with the respective elements of the first column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$
Result for this will be, $R1 = \left[ {(0 \times 0) + (i \times ( - i))} \right]$
Step 2: Multiply and add each element of the first row of matrix $A$ with the respective elements of the second column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(0 \times i) + (i \times 0)} \right]$
Step 3: Multiply and add each element of the second row of matrix $A$ with the respective elements of the first column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(( - i) \times 0) + (0 \times ( - i))} \right]$
Step 4: Multiply and add each element of the second row of matrix $A$ with the respective elements of the second column of matrix $A$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$ $A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(( - i) \times i) + (0 \times 0)} \right]$
Step 5: Matrix multiplication of matrix $A$ and $A$, that is,${A^2}$ will be,
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and $R4,$ we get:
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
  {(0 \times 0) + (i \times ( - i))}&{(0 \times i) + (i \times 0)} \\
  {(( - i) \times 0) + (0 \times ( - i))}&{(( - i) \times i) + (0 \times 0)}
\end{array}} \right]\]
After solving the above expressions, we get:
${A^2} = \left[ {\begin{array}{*{20}{c}}
  {0 + \left( { - {i^2}} \right)}&{\;\;\;\;\;\;0 + 0} \\
  {0 + 0}&{( - {i^2}) + 0}
\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}
  { - {i^2}}&0 \\
  0&{ - {i^2}}
\end{array}} \right]$ $(\because A = \;\left[ {\begin{array}{*{20}{c}}
  0&i \\
  { - i}&0
\end{array}} \right])$
Step 6: Taking $ - {i^2}$ common, as shown below:
${A^2} = - {i^2}\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
Step 7: Since $i$ is an imaginary part of complex number and ${i^2} = - 1$, we replace ${i^2}$ with -1, as shown below:
\[{A^2} \Rightarrow - ( - 1)\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\] \[(\because - ( - 1) = 1)\]
Step 8: Since, an identity matrix is a square matrix having 1s on the main diagonal and 0s everywhere else, so \[\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\] is an identity matrix of order\[(2 \times 2)\].
\[{A^2} = I\] (\[\because \]Identity matrix is denoted by \[I\])
Step 9: Taking power on both sides,
\[{({A^2})^{20}} = {I^{20}}\]
\[{A^{40}} = I\] \[(\because {I^n} = I)\]
\[{A^{40}} = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\]

Therefore, the correct option is (B).

Additional Information: In this type of question, where we have to multiply two given matrices, whose order is different, like a matrix A of order \[\left( {m \times n} \right)\], where ‘m’ is the number of rows of matrix A and ‘n’ is the number of columns of matrix A.
Another matrix B of order \[\left( {n \times p} \right)\], where ‘n’ is the number of rows of matrix B and ‘p’ is the number of columns of matrix B.
Then, the order of the resultant matrix should be\[\left( {m \times p} \right)\].

Note:In this type of question, where we have to multiply two given matrices, if the number of columns of the first matrix is equal to the number of rows of the second matrix, then multiplication is possible. Otherwise, multiplication is not possible.