Back substitution is an algorithm for solving a linear system that has been reduced to row-eschelon form, which is an upper triangular matrix:
Since the rows in a matrix represent equations, and the columns represent variables in those equations, the matrix can be expressed as the linear system The value of z is known, so 3 can be substituted into the equation just above it: Now that the values of y and z are known, these two values can be substituted into the first equation: So the solution to the linear system is:# | A | B | C | D |
E | F | G | H | I |
J | K | L | M | N |
O | P | Q | R | S |
T | U | V | W | X |
Y | Z |
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