Gaussian elimination
Solving a linear system by Gaussian elimination: augmented matrix, row operations, row echelon form, back-substitution, and the three possible outcomes (one solution, none, infinitely many).
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A few cards from the deck
Augmented matrix of a system: what is it?
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The array that gathers the coefficients of the unknowns of each equation, one row per equation, with the column of right-hand sides alongside on the right.
It holds the whole system in a form where the unknowns are no longer rewritten: x + 2y = 5 and 3x + 4y = 6 become the rows (1 2 | 5) and (3 4 | 6).
Which of these row operations can change the solution set of the system?
- L2 ← 0 × L2
- Swap L1 and L3
- L2 ← L2 + 5 × L1
- L2 ← 3 × L2
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L2 ← 0 × L2
Multiplying by 0 erases the equation: it cannot be recovered. The other three operations can be undone (swap again, subtract 5 L1, divide by 3).
To cancel the coefficient a of row Li with the pivot p of row Lj, you perform Li ← Li − … × Lj.
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To cancel the coefficient a of row Li with the pivot p of row Lj, you perform Li ← Li − (a/p) × Lj.
With this multiplier, the new coefficient is a − (a/p) × p = 0. Only a nonzero pivot allows this quotient to be formed.
Elementary row operation: what is it?
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Swapping two rows, multiplying a row by a nonzero real number, or adding to a row a multiple of another row.
These three operations can be undone, so they leave the set of solutions unchanged. Multiplying by 0 cannot be undone: it erases an equation.
Solving a system by Gaussian elimination: which step comes after "Write the augmented matrix of the system"?
- Cancel the coefficients below the pivot
- Choose a nonzero pivot (swapping rows if needed)
- Repeat on the remaining rows until you reach row echelon form
- Work back up from the last equation to the first
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Choose a nonzero pivot (swapping rows if needed)
Elimination goes down and builds the echelon form; back-substitution then uses each equation to find one more unknown, starting with the last.