Defactoring algorithms: Difference between revisions

ArrowHead294 (talk | contribs)
m Inversion by hand: Change to array format
ArrowHead294 (talk | contribs)
Line 441: Line 441:
Okay, now let's target the bottom-right entry. How can we make that 3 into a 0? Let's subtract the 1st row from the 3rd row 3 times:
Okay, now let's target the bottom-right entry. How can we make that 3 into a 0? Let's subtract the 1st row from the 3rd row 3 times:


<math>\left[ \begin{array} {ccc|ccc}
<math>
\left[ \begin{array} {ccc|ccc}
\color{blue}1 & \color{blue}0 & \color{blue}2  &  \color{blue}0 & \color{blue}1 & \color{blue}0 \\
\color{blue}1 & \color{blue}0 & \color{blue}2  &  \color{blue}0 & \color{blue}1 & \color{blue}0 \\
0 & 1 & 0  &  0 & 0 & 1 \\
0 & 1 & 0  &  0 & 0 & 1 \\
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<math>
<math>
\left[ \begin{array} {l} \begin{matrix}
\left[ \begin{array} {ccc|ccc}
1 & 0 & 2  &  0 & 1 & 0 \\
1 & 0 & 2  &  0 & 1 & 0 \\
\color{blue}0 & \color{blue}1 & \color{blue}0  &  \color{blue}0 & \color{blue}0 & \color{blue}1 \\
\color{blue}0 & \color{blue}1 & \color{blue}0  &  \color{blue}0 & \color{blue}0 & \color{blue}1 \\
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<math>
<math>
\left[ \begin{array} {l} \begin{matrix}
\left[ \begin{array} {ccc|ccc}
\color{red}1 & \color{red}0 & \color{red}0  &  \color{red}1 & \color{red}-2 & \color{red}2 \\
\color{red}1 & \color{red}0 & \color{red}0  &  \color{red}1 & \color{red}-2 & \color{red}2 \\
0 & 1 & 0  &  0 & 0 & 1 \\
0 & 1 & 0  &  0 & 0 & 1 \\
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<math>
<math>
\left[ \begin{array} {l} \begin{matrix}
\left[ \begin{array} {ccc|ccc}
1 & 0 & 0  &  1 & -2 & 2 \\
1 & 0 & 0  &  1 & -2 & 2 \\
0 & 1 & 0  &  0 & 0 & 1 \\
0 & 1 & 0  &  0 & 0 & 1 \\