Question

Prove that the determinant of a unitary matrix has absolute value 1

Answer 1

SOLUTION

TO PROVE

The determinant of a unitary matrix has absolute value 1

CONCEPT TO BE IMPLEMENTED

Unitary Matrix

A complex n × n matrix is said to be Unitary Matrix if

$\sf{A {A}^{ \circ} = I_n}$

$\sf{where \: \: {A}^{ \circ} \: is \: conjugate \: transpose \: of \: \: A}$

EVALUATION

Let A be the Unitary Matrix

Then

$\sf{A {A}^{ \circ} = I_n}$

$\implies \: \sf{A { \overline{A}}^{t} = I_n}$

$\implies \: \sf{det(A { \overline{A}}^{t} )=det( I_n)}$

$\implies \: \sf{det(A) \: det( { \overline{A}}^{t} )=1}$

$\implies \: \sf{det(A) \: det( \overline{A} )=1}$

$\implies \: \sf{det(A) \: \overline{ det(A )}=1}$

$\implies \: \sf{ { | \: det(A) \: | }^{2} =1}$

$\implies \: \sf{ { | \: det(A) \: | } =1}$

Hence the determinant of a unitary matrix has absolute value 1

Hence proved

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