- #1
chwala
Gold Member
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- TL;DR Summary
- Looking at this now- refreshing on rref kindly see attached.
This is fine but i have my own approach as follows;
##\begin{pmatrix}
-7 & -6 & -12 & -33 \\
5 & 5 & 7& 24 \\
1 & 0 & 4 & 5
\end{pmatrix} ## → Row 1 times 5 and row 2 times 7...
##\begin{pmatrix}
-35 & -30 & -60 & -165 \\
35 & 35 & 49& 168 \\
1 & 0 & 4 & 5
\end{pmatrix}## →
##\begin{pmatrix}
-35 & -30 & -60 & -165 \\
0 & 5 & -11& 3 \\
1 & 0 & 4 & 5
\end{pmatrix}## → Row 1 minus row 2...
##\begin{pmatrix}
-35 & -30 & -60 & -165 \\
1 & 0 & 4& 5 \\
0 & 5 & -11 & 3
\end{pmatrix} ## → R3 and R2 switch...
##\begin{pmatrix}
-35 & -30 & -60 & -165 \\
0 & -30 & 80 & 10 \\
0 & 5 & -11 & 3
\end{pmatrix} ## → R3 times 6 then subtract from R2
##\begin{pmatrix}
-35 & -30 & -60 & -165 \\
0 & -30 & 80 & 10 \\
0 & 0 & 14 & 28
\end{pmatrix} ##
##14x_3=28, x_3=2##
##-30x_2+160=10##
##x_2=5##
##-35x_1-30x_2-60x_3=-165##
##-35x_1-150-120=-165##
##-35x_1=105##
##x_1=-3##.
I fully understand the author's approach of having the leading elements for every row being ##1## conforming with the Row reduced echelon form...
just sharing...cheers