
Which of the following is true for the matrix \[A = \left[ {\begin{array}{*{20}{c}}i&{1 - 2i}\\{ - 1 - 2i}&0\end{array}} \right]\].
A. Symmetric
B. Skew- Symmetric
C. Hermitian
D. Skew-Hermitian
Answer
162.3k+ views
Hint:We will find the transpose of the given matrix. Then check whether the transpose matrix is equal to A or -A, to determine whether the matrix is symmetric or skew-symmetric. Then We will find the conjugate of the transpose matrix, to check whether the given matrix is a Hermitian or Skew-Hermitian
Formula used:
Symmetric matrix: \[{A^T} = A\]
Skew Symmetric matrix: \[{A^T} = - A\]
Hermitian: \[{\left( {\overline A } \right)^T} = A\]
Skew Hermitian: \[{\left( {\overline A } \right)^T} = - A\]
Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}i&{1 - 2i}\\{ - 1 - 2i}&0\end{array}} \right]\].
Now we will find the transpose of the matrix by converting the row into a column.
\[{A^T} = \left[ {\begin{array}{*{20}{c}}i&{ - 1 - 2i}\\{1 - 2i}&0\end{array}} \right]\]
Since \[{A^T} \ne A\] and \[{A^T} \ne - A\], thus A is neither a symmetric matrix nor a skew-symmetric matrix.
To find the conjugate of \[{A^T}\], we will change the sign of the imaginary part of all elements of the matrix.
\[{\left( {\overline A } \right)^T} = \left[ {\begin{array}{*{20}{c}}{ - i}&{ - 1 + 2i}\\{1 + 2i}&0\end{array}} \right]\]
\[{\left( {\overline A } \right)^T} = \left[ {\begin{array}{*{20}{c}}{ - i}&{ - \left( {1 - 2i} \right)}\\{ - \left( { - 1 - 2i} \right)}&0\end{array}} \right]\]
\[ \Rightarrow {\left( {\overline A } \right)^T} = - \left[ {\begin{array}{*{20}{c}}i&{\left( {1 - 2i} \right)}\\{\left( { - 1 - 2i} \right)}&0\end{array}} \right]\]
\[ \Rightarrow {\left( {\overline A } \right)^T} = - A\]
Thus, the given matrix is skew Hermitian.
Hence option D is the correct option
Note: In the hermitian, we find the conjugate of the transpose matrix of the given matrix. Hermitian is applicable when at least one of the elements of the matrix complex. To solve the given question first we find the transpose of the given matrix and identify the complex number. Then write the conjugate of the complex numbers in place of complex numbers.
Formula used:
Symmetric matrix: \[{A^T} = A\]
Skew Symmetric matrix: \[{A^T} = - A\]
Hermitian: \[{\left( {\overline A } \right)^T} = A\]
Skew Hermitian: \[{\left( {\overline A } \right)^T} = - A\]
Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}i&{1 - 2i}\\{ - 1 - 2i}&0\end{array}} \right]\].
Now we will find the transpose of the matrix by converting the row into a column.
\[{A^T} = \left[ {\begin{array}{*{20}{c}}i&{ - 1 - 2i}\\{1 - 2i}&0\end{array}} \right]\]
Since \[{A^T} \ne A\] and \[{A^T} \ne - A\], thus A is neither a symmetric matrix nor a skew-symmetric matrix.
To find the conjugate of \[{A^T}\], we will change the sign of the imaginary part of all elements of the matrix.
\[{\left( {\overline A } \right)^T} = \left[ {\begin{array}{*{20}{c}}{ - i}&{ - 1 + 2i}\\{1 + 2i}&0\end{array}} \right]\]
\[{\left( {\overline A } \right)^T} = \left[ {\begin{array}{*{20}{c}}{ - i}&{ - \left( {1 - 2i} \right)}\\{ - \left( { - 1 - 2i} \right)}&0\end{array}} \right]\]
\[ \Rightarrow {\left( {\overline A } \right)^T} = - \left[ {\begin{array}{*{20}{c}}i&{\left( {1 - 2i} \right)}\\{\left( { - 1 - 2i} \right)}&0\end{array}} \right]\]
\[ \Rightarrow {\left( {\overline A } \right)^T} = - A\]
Thus, the given matrix is skew Hermitian.
Hence option D is the correct option
Note: In the hermitian, we find the conjugate of the transpose matrix of the given matrix. Hermitian is applicable when at least one of the elements of the matrix complex. To solve the given question first we find the transpose of the given matrix and identify the complex number. Then write the conjugate of the complex numbers in place of complex numbers.
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