One motivation is that it makes solving equations very easy! If A= LU where, of course, L is a "lower diagonal" and U is "upper diagonal", then we can solve Ax= b by writing it as LUx= b so that Ux= y= L-1b, which can be done by 'back substitution" and then solving x= U-1y again by back substitution.
For example, if
[tex]A= LU= \begin{bmatrix}2 & 0 \\ 1 & 3\end{bmatrix}\begin{bmatrix}1 & 3 \\ 0 & 2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix}2 \\ 1\end{bmatrix}[/tex]
We can let Ux= y so the equation becomes
[tex]Ly= \begin{bmatrix}2 & 0 \\ 1 & 3\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix}= \begin{bmatrix}2 & 1 \end{bmatrix}[/tex]
The first row is equivalent to the equation [itex]2y_1= 2[/itex] so we have immediately [itex]y_1= 1[/itex]. With that value, the second equation, [itex]y_1+ 3y_2= 1[/itex] becomes [itex]1+ 3y_2= 1[/itex] so that [itex]3y_2= 0[/itex] and [itex]y_2= 0[/itex].
Then, since we defined y to be Ux, we have
[tex]x= \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix}1 \\ 0 \end{bmatrix}[/tex]
Now, the second row gives the equation [itex]2x_2= 0[/itex] so that [itex]x_2= 0[/itex] and then the top row becomes [itex]x_1+ 3x_2= x_1= 1[/itex].