[tex] \sf{if \: matrix \: A~ = \: \frac{1}{ \sqrt{2} } } \sf{\bigg[ \begin{matrix}1& i \\ - i& a\end{matrix} \bigg],i = \sqrt{ - 1} }[/tex]
[tex] \sf{is \: unitary \: matrix, \: a \: is \: equal \: to \: ? }[/tex]
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[tex] \sf{if \: matrix \: A~ = \: \frac{1}{ \sqrt{2} } } \sf{\bigg[ \begin{matrix}1& i \\ - i& a\end{matrix} \bigg],i = \sqrt{ - 1} }[/tex]
[tex] \sf{is \: unitary \: matrix, \: a \: is \: equal \: to \: ? }[/tex]
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[ Note: Kindly see my answer from brainly app. ]
At first,
[tex]i\:=\:\sqrt{-1}\\=>\:i^2\:=\:-1 [/tex]
[tex]\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ A = \frac{1}{ \sqrt{2} } \left [ \pink{ { \begin{array}{cc} 1&i \\ - i&a\end{array}}}\right] \\ \\ \sf \: where \: \: A \sf \: \: i s\: a \: unitary \: matrix. \\ \\ \bf \: as \: \: A \: \: is \: a \: unitary \: matrix, \: we \: \\ \bf \: know \: that \: : \\ \\ \blue{ \boxed{ \bf \: A {A}^{ \theta} = I }}\: \\ \\ \sf \: here, \\ \\ \bf \: {A}^{ \theta} = conjugate \: transpose \: of \: A \\ \\ \sf \: and \\ \\ \bf \: {A}^{ \theta} = { {( \bar{A}})^{ T} }^{} \\ \\ \end{array}}}}[/tex]
[tex]\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: now \\ \\ \bf \bar{A} = \frac{1}{ \sqrt{2} } \left [ \pink{ { \begin{array}{cc}1& - i \\ i&a \end{array}}}\right]\\ \\ \bf \: \therefore \: {A}^{ \theta} = ( { \bar{A}})^{ T} = \frac{1}{ \sqrt{2} }\left [ \pink{ { \begin{array}{cc}1&i \\ - i&a \end{array}}}\right] \\ \\ \end{array}}}}[/tex]
[tex]\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: according \: to \: the \: question : \\ \\ A {A}^{ \theta} = I \\ \\ \small{ \rm \implies \: \frac{1}{ \sqrt{2} } \left [ \pink{ { \begin{array}{cc} 1&i \\ - i&a\end{array}}}\right]. \frac{1}{ \sqrt{2} }\left [ \pink{ { \begin{array}{cc} 1&i \\ - i&a \end{array}}}\right] = \left [ \orange{ { \begin{array}{cc}1&0 \\ 0&1 \end{array}}}\right]} \: \: \\ \\ \rm \implies \: \frac{1}{2} \left [ \orange{ { \begin{array}{cc} 1 - {i}^{2}&&i + ai \\ - i - ai&& - {i}^{2} + {a}^{2} \end{array}}}\right] = \left [ \orange{ { \begin{array}{cc} 1&0 \\ 0&1\end{array}}}\right] \\ \\ \rm \implies \: \frac{1}{2} \left [ \orange{ { \begin{array}{cc} 1 + 1&&i + ai \\ - i - ai&& {a}^{2} \end{array}}}\right] = \left [ \orange{ { \begin{array}{cc} 1&0 \\ 0&1 \end{array}}}\right] \\ \\ \rm \implies \:\left [ \orange{ { \begin{array}{cc} 1& \frac{i + ai}{2} \\ \\ \frac{ - i - ai}{2} & {a}^{2} \end{array}}}\right] = \left [ \orange{ { \begin{array}{cc} 1&0 \\ 0&1 \end{array}}}\right] \\ \\ \rm \: comparing \: both \: sides : \\ \\ \frac{i + ai}{2} = 0 \\ \\ \rm \implies \:i(1 + a) = 2 \times 0 = 0 \\ \\ \rm \implies \:1 + a = 0 \\ \\ \rm \therefore \: a = - 1 \\ \\ \\ \blue{ \boxed{\rm \: \therefore \: value \: of \: \: a = - 1}}\end{array}}}}[/tex]
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