Let ${\mathrm{D}}_1=\left|\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}+\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}+\mathrm{d}\\ \mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}-\mathrm{b}\end{array}\right|\text{ and }{\mathrm{D}}_2=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{b}& \mathrm{d}& \mathrm{b}+\mathrm{d}\\ \mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{b}+\mathrm{c}\end{array}\right|$ then the value of $\left|\frac{{\mathrm{D}}_1}{{\mathrm{D}}_2}\right|,$ Where $\mathrm{b}\ne 0\text{ and }\mathrm{ad}\ne \mathrm{bc}$, is ____________.

 $\begin{array}{r}{D}_1=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{a}+\mathrm{b}\\ \mathrm{c}& \mathrm{d}& \mathrm{c}+\mathrm{d}\\ \mathrm{a}& \mathrm{b}& \mathrm{a}-\mathrm{b}\end{array}\right|\\ {\mathrm{C}}_2\to {\mathrm{C}}_1+{\mathrm{C}}_2\end{array}$

$\begin{array}{r}{\mathrm{C}}_3\to {\mathrm{C}}_2-{\mathrm{C}}_3+{\mathrm{C}}_1\\ {\mathrm{D}}_1=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{a}+\mathrm{b}& 0\\ \mathrm{c}& \mathrm{c}+\mathrm{d}& 0\\ \mathrm{a}& \mathrm{a}+\mathrm{b}& 2\mathrm{b}\end{array}\right|\end{array}$

$\begin{array}{r}=2\mathrm{b}(\mathrm{ac}+\mathrm{ad}-\mathrm{ac}-\mathrm{cb})\\ =2\mathrm{b}(\mathrm{ad}-\mathrm{bc})\end{array}$

$\begin{array}{r}{\mathrm{D}}_2=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{a}+\mathrm{c}& 0\\ \mathrm{b}& \mathrm{b}+\mathrm{d}& 0\\ \mathrm{a}& \mathrm{a}+\mathrm{c}& -\mathrm{b}\end{array}\right|\\ =-\mathrm{b}(\mathrm{ab}+\mathrm{ad}-\mathrm{ab}-\mathrm{ac})\\ =-\mathrm{b}(\mathrm{ad}-\mathrm{bc})\end{array}$

we have $\frac{{\mathrm{D}}_1}{{\mathrm{D}}_2}=\frac{-2\mathrm{b}(\mathrm{ad}-\mathrm{bc})}{\mathrm{b}(\mathrm{ad}-\mathrm{bc})}=-2$