System of Linear Equations
Chapter Eight Rectangular Arrays of Jiuzhang suanshu provided an algorithm for solving System of linear equations by method of elimination:
Problem 8-1: Suppose we have 3 bundles of top quality cereals, 2 bundles of medium quality cereals, and a bundle of low quality cereal with accumulative weight of 39 dou. We also have 2, 3 and 1 bundles of respective cereals amounting to 34 dou; we also have 1,2 and 3 bundles of respective cereals, totaling 26 dou.
Find the quantity of top, medium, and poor quality cereals. In algebra, this problem can be expressed in three system equations with three unknowns.
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-
-
- 3x+2y+z=39
- 2x+3y+z=34
- x+2y+3z=26
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-
This problem was solved in Jiuzhang suanshu with counting rods laid out on a counting board in a tabular format similar to a 3x4 matrix:
| quality | left column | center column | right column |
| top | |||
| medium | |||
| low | |||
| shi |
Algorithm:
- Multiply the center column with right column top quality number.
- Repeatedly subtract right column from center column, until the top number of center column =0
- multiply the left column with the value of top row of right column
- Repeatedly subtract right column from left column, until the top number of left column=0
- After applying above elimination algorithm to the reduced center column and left column, the matrix was reduced to triangular shape:
| quality | left column | center column | right column |
| top | |||
| medium | |||
| low | |||
| shi |
The amount of on bundle of low quality cereal =
From which the amount of one bundle of top and medium quality cereals can be found easily:
One bundle of top quality cereals=9 dou
One bundle of medium cereal=4 dou >
Read more about this topic: Rod Calculus
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