Elimination.
Simplify, without changing. Then climb.
Predict first
10⁻²⁰ x₁ + x₂ = 1 and x₁ + x₂ = 2. The answer is almost exactly (1, 1). Could a computer, keeping sixteen digits, report x₁ = 0?
Three safe moves
The lines move. The answer can’t.
Swap two equations, scale one, add a multiple of one to another. Then try to cheat, and watch the answer move.
Two pictures
One elimination. Two pictures, one truth.
Subtract half of row 1 from row 2. In one picture the crossing stands still. In the other, the walk stays the same.
Run it
Down, making zeros. Back up, collecting answers.
The augmented matrix, one move at a time. Every move is itself a matrix.
Pivots
The shape leans. Its area never changes.
Elimination slides every point straight down. At the end, base times height is pivot times pivot.
Drag the arrows
Nearly zero
A tiny pivot. A huge mistake.
The truth is almost exactly (1, 1). Shrink the first pivot, and watch a real computer, rounding after every step, lose it.
Without swapping
Swap first
Fragile, however you arrange it: two features that are almost the same measurement
The recipe
Four patients, four scores. Four weights, revealed.
Lecture 3’s risk scores, in real units. Run elimination and recover the recipe behind them.
A new patient walks in
Predicted risk
The clinic is invented, so the numbers come out clean. And four patients fitting four weights exactly isn’t learning; it’s memorisation. That story is Lecture 12’s.
Prove it
Four questions. The reasons matter more than the score.
Question 1 of 4
Why must a pivot never be zero?
Question 2 of 4
Elimination on a 3 × 3 matrix finds pivots 2, 2.5 and 3.2, with no swaps. What is its determinant?
Question 3 of 4
After elimination, the last row reads 0 = −4. What does that tell you?
Question 4 of 4
Why do real solvers swap the largest number into the pivot position, even when the pivot isn’t zero?