Question

If A = $\left[\begin{array}{ccc}a+ib&c+id\\-c+id&a-ib\end{array}\right]$, a² + b² + c² + d² = 1 then find the inverse of A.

Answer 1

Answer:

$adj.A=A^{-1}=\left[\begin{array}{ccc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

Step-by-step explanation:

As we know that Adjoint of matrix is calculated as Minor × Co-factor of each element and taking transpose of it.

or

$adj.A=[A_{ji}]_{n\times n}$

$A= \left[\begin{array}{ccc}a+ib&c+id\\-c+id&a-ib\end{array}\right] \\\\adj.A=\left[\begin{array}{ccc}a-ib&c-id\\-c-id&a+ib\end{array}\right]^{'}\\\\adj.A= \left[\begin{array}{ccc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

Now

$A^{-1} =\frac{adj.A}{|A|}\\\\|A|=\left[\begin{array}{ccc}a+ib&c+id\\-c+id&a-ib\end{array}\right] \\\\|A|=(a+ib)(a-ib)-(c+id)(-c+id)\\\\a^{2}+b^{2}+c^{2}+d^{2}=1\\\\\\A^{-1} =\frac{1}{1}\left[\begin{array}{ccc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}a-ib&-(c+id)\\c-id&a+ib\end{array}\right]$