For the matrix \(A\) of Exercise 2, find elementary matrices \(E_1, E_2, \ldots, E_k\) such that \[ E_k \cdots E_2E_1A = I. \]
Solution
For \[ A=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}, \] perform the row operations \[ R_2\leftarrow R_2-2R_1,\quad R_3\leftarrow R_3-3R_1,\quad R_2\leftarrow \tfrac12R_2,\quad R_3\leftarrow R_3-3R_2, \] \[ R_3\leftarrow -2R_3,\quad R_2\leftarrow R_2+\tfrac12R_3,\quad R_1\leftarrow R_1-R_3,\quad R_1\leftarrow R_1+R_2. \] They reduce \(A\) to \(I_3\). The corresponding elementary matrices are \[ E_1=\begin{bmatrix}1&0&0\\-2&1&0\\0&0&1\end{bmatrix},\quad E_2=\begin{bmatrix}1&0&0\\0&1&0\\-3&0&1\end{bmatrix},\quad E_3=\begin{bmatrix}1&0&0\\0&\frac12&0\\0&0&1\end{bmatrix}, \] \[ E_4=\begin{bmatrix}1&0&0\\0&1&0\\0&-3&1\end{bmatrix},\quad E_5=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix},\quad E_6=\begin{bmatrix}1&0&0\\0&1&\frac12\\0&0&1\end{bmatrix}, \] \[ E_7=\begin{bmatrix}1&0&-1\\0&1&0\\0&0&1\end{bmatrix},\qquad E_8=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}. \] Therefore \[ E_8E_7E_6E_5E_4E_3E_2E_1A=I_3. \]