If \[A\] is an unitary matrix then \[\left| A \right|\] is equal to:
A) $1$
B) $ - 1$
C) $ \pm 1$
D) $2$
Answer
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Hint: A square matrix A is said to be unitary if its transpose is its own inverse and all its entries should belong to complex numbers. A unitary matrix is a matrix whose inverse equals its conjugate transpose. Unitary matrices are the complex analog of real orthogonal matrices.
Complete step-by-step answer:
In mathematics, a complex square matrix A is unitary if its conjugate transpose \[{A^ * }\]is also its inverse.
A unitary matrix can be defined as a square complex matrix A for which,
\[A{A^*} = {A^*}A = I\]
\[{A^*}\]= Conjugate transpose of A
\[I\]= Identity matrix
When we are working with square matrices we are mapping a finite dimensional space to itself whenever we multiply.
Now let's take a situation where we are finding the determinant of the complete equation mentioned above.
\[A{A^*} = {A^*}A = I\]
Taking determinant of complete equation.
\[ \Rightarrow \left| {A{A^*}} \right| = \left| {{A^*}A} \right| = \left| I \right|\]
Separating the determinant of each term in the equation.
\[ \Rightarrow \left| {\left| A \right| \times \left| {{A^*}} \right|} \right| = \left| {\left| {{A^*}} \right| \times \left| A \right|} \right| = \left| I \right|\]
Removing the determinant above the whole equation of both sides.
\[ \Rightarrow \left| A \right| \times \left| {{A^*}} \right| = \left| {{A^*}} \right| \times \left| A \right| = 1\]
Now cancelling\[\left| {{A^*}} \right|\]from the equation we get,
\[ \Rightarrow \left| A \right| = \left| A \right| = 1\]
\[|A|\]can be a complex number with modulus/magnitude 1.
So, option (A) is the correct answer.
Note: If matrix A is called Unitary matrix then it satisfy this condition \[A{A^*} = {A^*}A = I\] where \[{A^*}\]= Transpose Conjugate of A = \[{\left( {A\prime } \right)^T}\] (first you Conjugate and then Transpose , you will get Unitary matrix)
Properties of Unitary matrix:
1) If A is a Unitary matrix then\[{A^{ - 1}}\]is also a Unitary matrix.
2) If A is a Unitary matrix then \[{A^*}\] is also a Unitary matrix.
3) If A&B are Unitary matrices, then A.B is a Unitary matrix.
4) If A is Unitary matrix then \[{A^{ - 1}} = {A^*}\]
5) If A is Unitary matrix then it's determinant is of Modulus Unity (always1).
Complete step-by-step answer:
In mathematics, a complex square matrix A is unitary if its conjugate transpose \[{A^ * }\]is also its inverse.
A unitary matrix can be defined as a square complex matrix A for which,
\[A{A^*} = {A^*}A = I\]
\[{A^*}\]= Conjugate transpose of A
\[I\]= Identity matrix
When we are working with square matrices we are mapping a finite dimensional space to itself whenever we multiply.
Now let's take a situation where we are finding the determinant of the complete equation mentioned above.
\[A{A^*} = {A^*}A = I\]
Taking determinant of complete equation.
\[ \Rightarrow \left| {A{A^*}} \right| = \left| {{A^*}A} \right| = \left| I \right|\]
Separating the determinant of each term in the equation.
\[ \Rightarrow \left| {\left| A \right| \times \left| {{A^*}} \right|} \right| = \left| {\left| {{A^*}} \right| \times \left| A \right|} \right| = \left| I \right|\]
Removing the determinant above the whole equation of both sides.
\[ \Rightarrow \left| A \right| \times \left| {{A^*}} \right| = \left| {{A^*}} \right| \times \left| A \right| = 1\]
Now cancelling\[\left| {{A^*}} \right|\]from the equation we get,
\[ \Rightarrow \left| A \right| = \left| A \right| = 1\]
\[|A|\]can be a complex number with modulus/magnitude 1.
So, option (A) is the correct answer.
Note: If matrix A is called Unitary matrix then it satisfy this condition \[A{A^*} = {A^*}A = I\] where \[{A^*}\]= Transpose Conjugate of A = \[{\left( {A\prime } \right)^T}\] (first you Conjugate and then Transpose , you will get Unitary matrix)
Properties of Unitary matrix:
1) If A is a Unitary matrix then\[{A^{ - 1}}\]is also a Unitary matrix.
2) If A is a Unitary matrix then \[{A^*}\] is also a Unitary matrix.
3) If A&B are Unitary matrices, then A.B is a Unitary matrix.
4) If A is Unitary matrix then \[{A^{ - 1}} = {A^*}\]
5) If A is Unitary matrix then it's determinant is of Modulus Unity (always1).
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