Back Substitution

Pronunciation: /bæk ˈsʌb stɪˌtut ʃən/ ?

Back substitution is an algorithm for solving a linear system that has been reduced to row-eschelon form, which is an upper triangular matrix:

\\left[\\array{1 & a_{12} & a_{13} & a_{14} \\\\ 0 & 1 & a_{23} & a_{24} \\\\ 0 & 0 & 1 & a_{34}}\\right]
Since the rows in a matrix represent equations, and the columns represent variables in those equations, the matrix
\\left[\\array{1 & 3 & -2 & -1 \\\\ 0 & 1 & 2 & -1 \\\\ 0 & 0 & 1 & 3}\\right]
can be expressed as the linear system
\\array{ x+3y-2x=-1 \\\\ y+2z=-1 \\\\ z=3 }
The value of z is know, so 3 can be substituted into the equation just above it:
y+2\\cdot3\\;=\\;-1\\;\\Rightarrow\\;y+6\\;=\\;-1\\;\\Rightarrow\\;y\\;=\\;-7
Now that the values of y and z are known, these two values can be substituted into the first equation:
x+3\\cdot(-7)-2\\cdot3\\;=\\;-1\\;\\Rightarrow\\;x-21-6\\;=\\;-1\\;\\Rightarrow\\;x\\;=\\;26
So the solution to the linear system is:
x\\;=\\;26,\\quad y\\;=\\;-7,\\quad z\\;=\\;3

Cite this article as:


Back Substitution. 2008-11-20. All Math Words Encyclopedia. Life is a Story Problem LLC. http://www.allmathwords.org/en/b/backsubstitution.html.

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2008-11-20: Initial version (McAdams, David.)

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