Back substitution
Back substitution solves a triangular system by substituting known later variables into the equations above them.
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01Here the triangular system has nonzero diagonal coefficients#
Here the triangular system has nonzero diagonal coefficients. Its last equation contains only z: gives . Substitute upward: gives , then gives .
For and , solve the second equation first: . Substitute that value into the first, then divide by 2.
Solve for .
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First , then .
02Back substitution uses solved variables, not raw right-hand sides#
Back substitution uses solved variables, not raw right-hand sides. In , the known value is . Substituting 8 instead would violate that equation.
A learner takes y as 8. Repair . ;
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First , then .
03Check the result in every original equation#
Check the result in every original equation. For and , the pair gives and .
A device reports 10 for twice x plus y. It reports 6 for three times y. Find .
Show answer and explanation
First ; then .
Start at the last equation. Solve, substitute upward, and check every original equation.
- Solve a triangular system by back substitution.
Sources & further reading
- [1]Margalit and Rabinoff: Interactive Linear Algebra, 1.2 Row Reduction ↗Interactive Linear Algebra · Book