Question

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Answer 1

Answer:

Inverse of A is $\left[\begin{array}{cc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

Step-by-step explanation:

Given: $A = \left[\begin{array}{cc}a+ib&c+id\\-c+id&a-ib\end{array}\right]$ and a²+b²+c²+d² = 1.

⇒Determinant of A = |A| = (a+ib)(a-ib)-(-c+id)(c+id) = (a+ib)(a-ib)-{-(c-id)(c+id)}

or |A| = a²-i²b²+c²-i²d² = a²+b²+c²+d²         {∵ i² = -1}

⇒ |A| = 1

⇒ Inverse of matrix A exists.

Now, adjoint of A = adjA = $\left[\begin{array}{cc}a-ib&-(c+id)\\-(-c+id)&a+ib\end{array}\right]$

Note: Adjoint of a 2×2 matrix can be obtained by interchanging the diagonal elements and changing the sign of off-diagonal elements.

⇒ adjA = $\left[\begin{array}{cc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

w.k.t. Inverse of a matrix A = A⁻¹ = $\frac{adjA}{|A|}$

⇒ A⁻¹ = $\frac{1}{|A|}\left[\begin{array}{cc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

or A⁻¹ = $\left[\begin{array}{cc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$           {∵|A| = 1}

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