Answer
$AD=I_{3}$ is equivalent to $\left\{\begin{array}{l}
A\mathrm{d}_{1}=\mathrm{e}_{1}, \\
A\mathrm{d}_{2}=\mathrm{e}_{2}, \\
A\mathrm{d}_{3}=\mathrm{e}_{3}
\end{array}\right.$
Select one solution of each equation and place them as columns of $D$.
Then, $AD=I_{3}$.
Work Step by Step
Represent matrices $I_{3}$ and $D$ wth their columns
$I_{3}=[\mathrm{e}_{1}\ \mathrm{e}_{2}\ \mathrm{e}_{3}]$
$D=[\mathrm{d}_{1}\ \mathrm{d}_{2}\ \mathrm{d}_{3}]$.
By definition of $AB=A[\mathrm{b}_{1}\ \mathrm{b}_{2}\ ...\ \mathrm{b}_{p}]=[A\mathrm{b}_{1}\ A\mathrm{b}_{2}\ ...\ A\mathrm{b}_{p}],$
$AD=A[\mathrm{d}_{1}\ \mathrm{d}_{2}\ \mathrm{d}_{3}]=[A\mathrm{d}_{1}\ A\mathrm{d}_{2}\ A\mathrm{d}_{3}]$
$AD=I_{3}$ is equivalent to $\left\{\begin{array}{l}
A\mathrm{d}_{1}=\mathrm{e}_{1}, \\
A\mathrm{d}_{2}=\mathrm{e}_{2}, \\
A\mathrm{d}_{3}=\mathrm{e}_{3}
\end{array}\right.$
Theorem 4 in Section 1.4 states that if A spans $\mathbb{R}^{3},$
then each of these equations has at least one solution
Select one solution of each equation and place them as columns of $D$.
Then, $AD=I_{3}$.
