Consider the following system of linear equations x+y+z+w = 2 x- y+2z+3w =-10ZS

Havlishkq

Havlishkq

Answered question

2022-02-24

Consider the following system of linear equations
x+y+z+w=2
xy+2z+3w=10
x3y+3z+5w=7
x+3yw=1
If we add the first two (solve for y) or the last two (solve for y), then both are same: 2x+3z+4w=8
Does it mean that all the points (a, y, b, c) are the solutions for the system of equations? Here (a,b,c) is the point lying on the plane 2x+3z+4w=8 and y is arbitrary.
Whereas, if we compute the row echelon form it says the rank of the coefficients matrix is 2 and the augmented matrix is 3. This implies the solution does not exist.
Any help will be good!

Answer & Explanation

Judith Couch

Judith Couch

Beginner2022-02-25Added 3 answers

Here is a smaller toy example which, I believe, captures the essence of your problem (it is a system of four equations in two unknowns which is clearly inconsistent):
x+y=1
x+y=2
x+y=1
x+y=2
Adding either the first or the last pair of equations together yields the line 2x+2y=3. But you can't conclude anything about the solution space from this single equation. Really all you are saying is that if (x,y) solves a pair of original equations, then it solves the new equation, but the converse is not true at all. What you can do with such a combination of two previous equations is substitute it in the original system (this is one of the elementary row operations), but then you still have to solve the system.

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