If $A=\left[\begin{array}{cc}a+i b & c+i d \ -c+i d & a-i b\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}a+i b & -c-i d \ -c+i d & a-i b\end{array}\right]$. Find $\left(a^2+b^2+c^2+d^2\right)$.

Question

If $A=\left[\begin{array}{cc}a+i b & c+i d \ -c+i d & a-i b\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}a+i b & -c-i d \ -c+i d & a-i b\end{array}\right]$. Find $\left(a^2+b^2+c^2+d^2\right)$., 1, -1, $i$, $-i$

MARKS App · Accepted Answer

Given matrix $A=\left[\begin{array}{cc} a+i b & c+i d \ -c+i d & a-i b \end{array}\right]$ So, $\quad A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}a-i b & -c-i d \ c-i d & a+i b\end{array}\right]$ $\begin{aligned} & =\frac{1}{a^2+b^2+c^2+d^2}\left[\begin{array}{cc} a-i b & -c-i d \ c-i d & a+i b \end{array}\right] \ & =\left[\begin{array}{cc} a+i b & -c-i d \ -c+i d & a-i b \end{array}\right] \quad \text{(given)} \end{aligned}$ $\therefore \quad b=0$ and $c=d=0$ and $a^2+b^2+c^2+d^2=1$

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