Exercises 27 and 28 prove special cases of the facts about elementary matrices stated in the box following Example 5. Here Ais a \({\bf{3}} \times {\bf{3}}\) matrix and \(I = {I_{\bf{3}}}\). (A general proof would require slightly more notation.)

28. Show that if row 3 of Ais replaced by \(ro{w_3}\left( A \right) - 4 \cdot ro{w_1}\left( A \right)\) the result is EA, where Eis formed from Iby replacing \(ro{w_3}\left( I \right) - 4 \cdot ro{w_{\bf{1}}}\left( I \right)\).

Suppose for\(i = 1,2,3\), matrix A is\(A = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right)}\end{aligned}} \right)\).

Use row one to obtain\({\rm{ro}}{{\rm{w}}_3}\left( A \right)\)in terms of rows three and four. Add \( - 4\) times \({\rm{ro}}{{\rm{w}}_1}\left( A \right)\) to \({\rm{ro}}{{\rm{w}}_3}\left( A \right)\).

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right)}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)}\end{aligned}} \right)\)

Apply the equation\({\rm{ro}}{{\rm{w}}_i}\left( A \right) = {\rm{ro}}{{\rm{w}}_i}\left( I \right) \cdot A\).

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( I \right) \cdot A}\\{{\rm{ro}}{{\rm{w}}_2}\left( I \right) \cdot A}\\{{\rm{ro}}{{\rm{w}}_3}\left( I \right) \cdot A - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( I \right) \cdot A}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( I \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( I \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( I \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( I \right)}\end{aligned}} \right)A\\ = EA\end{aligned}\)

Hence, proved.

Therefore, if row three of Ais replaced by\({\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)\), the result becomes EA.