Exercise 1.4.3

Notice that when we apply ‘elimination’ process on a matrix $A$, at the end of the process, it becomes an upper triangular matrix. So we can search for matrices that when left multiply the $A$, they reduce the $A$ to $U$. We look for one matrix for each step in an ‘elimination’ process.

Suppose $A$ is $n$ by $n$ matrix. The first step is to transform the first elements of all rows except the first row to be zeros. We achieve this by multiply a number to the first row and subtract from each row below. Represented by a left multiply matrix, we have

$E_1=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill -l_{21}\hfill & \hfill 1\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill -l_{31}\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \dots \hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill -l_{n1}\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 1\hfill \end{array}\right]$

Now apply step 2 of ‘elimination’ to change all the second elements for rows below the second row to zero, we have $E_2=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -l_{32}\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill -l_{n2}\hfill & \hfill \dots \hfill & \hfill 1\hfill \end{array}\right]$

Continue this until we have $E_{n-1}=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill \dots \hfill & \hfill -l_{n,n-1}\hfill & \hfill 1\hfill \end{array}\right]$

Now it’s clear that $E=E_{n-1}\dots E_2{E}_1$.

For $A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 2\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right]$, we have $n=3$, so $E_1=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -l_{21}\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -l_{31}\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ and $E_2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -l_{32}\hfill & \hfill 1\hfill \end{array}\right]$

so $E=E_2{E}_1=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -l_{21}\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -l_{31}+l_{21}{l}_{32}\hfill & \hfill -l_{32}\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

And $L=E^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 2\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill \end{array}\right]$