Solve the following for $D$: $ABDB^{-1} = I$

Question

So, here's what I know, 'I' is typically:

\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}

But, from that I really have no idea where to go with this question. I know basic matrix inversion rules, but I am unsure how to solve.

Answer 1

Hint: How would you solve this equation for $d$?

$$abdc=1$$ assuming $a,b,c \neq 0$. Let's also assume that $a,b,c,d$ do not commute. Secondly, the $I$ you mentioned above is often denoted $I_3$, as it is the $3\times 3$ identity matrix. You are working with $I_n$, which has a diagonal containing $n$ $1$'s.

Answer 2

$ ABDB^{-1} = I \quad\leadsto\quad D = B^{-1}A^{-1}IB $