If a-b-c and d are real numbers such that and if A- a-ib c-id -c-id a-ib - then A -1 -

The correct option is A | a+ib c id c id a ib | A= a+ib amp; c+id c+id amp; a ib | A | = a ^ 2 + b ^ 2 + c ^ 2 + d ...

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Inverse of a Matrix

If a,b,c an...

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If $a, b, c$ and $d$ are real numbers such that and if $A = ∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣$ then $A^{- 1} =$

∣ ∣ ∣ \begin{matrix} a + i b & - c - i d c - i d & a - i b \end{matrix} ∣ ∣ ∣

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∣ ∣ ∣ \begin{matrix} a - i b & c + i d - c + i d & a + i b \end{matrix} ∣ ∣ ∣

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∣ ∣ ∣ \begin{matrix} a - i b & - c - i d c - i d & a + i b \end{matrix} ∣ ∣ ∣

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∣ ∣ ∣ \begin{matrix} a + i b & c + i d c - i d & a - i b \end{matrix} ∣ ∣ ∣

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∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣

|\begin{matrix} a + i b & c + i d \\ - c + i d & a - i b \end{matrix}| .

a, b, c

d

a^{2} + b^{2} + c^{2} + d^{2} = 1

I f A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

A^{- 1} =

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

A^{- 1}

a, b, c

d

a^{2} + b^{2} + c^{2} + d^{2} = 1

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}],

i^{2} = - 1,

A^{- 1} =

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