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The concept of elementary row operations, which include row swapping, scalar multiplication, and row sum. These operations preserve the linear system in question and are used to reduce a matrix to Reduced Row Echelon Form (RREF) through Gauss-Jordan Elimination. The document also covers the uniqueness of RREF and provides examples and exercises.
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Our goal is to begin with an arbitrary matrix and apply operations that respect row equivalence until we have a matrix in Reduced Row Echelon Form (RREF). The three elementary row operations are:
Why do these preserve the linear system in question? Swapping rows is just changing the order of the equations begin considered, which certainly should not alter the solutions. Scalar multiplication is just multiplying the equation by the same number on both sides, which does not change the solution(s) of the equation. Likewise, if two equations share a common solution, adding one to the other preserves the solution. There is a very simple process for row reducing a matrix, working column by column. This system is called Gauss-Jordan Elimination.
Example 3 x 3 = 9 x 1 +5x 2 − 2 x 3 = 2 1 3 x^1 +2x^2 = 3 First we write the system as an augmented matrix:
1 3 2 0 3
R 1 ↔ ∼R 2
1 3 2 0 3 1 5 − 2 2 0 0 3 9
R 1 =R 1 − 6 R 2 ∼
(^13) R 3 ∼
Now we’re in RREF and can see that the solution to the system is given by x 1 = 1, x 2 = 3, and x 3 = 1; it happens to be a unique solution. Notice that we kept track of the steps we were taking; this is important for checking work!
removing columns does not affect row equivalence. Call the new, smaller,
matrices Aˆ and Bˆ. The new matrices should look this: Aˆ =
( IdN a 0 0
) and
( IdN b 0 0
) , where IdN is an N xN identity matrix and a and b are
vectors. Now if Aˆ and Bˆ have the same solution, then we must have a = b. But this is a contradiction! Then A = B.
Hefferon, Chapter One, Section 1.1 and 1. Wikipedia, Systems of Linear Equations
For which values of k does the system below have a solution?
x − 3 y = 6 x +3z = − 3 2 x +ky +(3 − k)z = 1