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Elementary Row Operations and Gauss-Jordan Elimination, Schemes and Mind Maps of Linear Algebra

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.

What you will learn

  • What are the three elementary row operations?
  • Why do these operations preserve the linear system?
  • How is Gauss-Jordan Elimination used to reduce a matrix to RREF?

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/27/2022

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Elementary Row Operations
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:
(Row Swap) Exchange any two rows.
(Scalar Multiplication) Multiply any row by a constant.
(Row Sum) Add a multiple of one row to another row.
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.
1. If all entries in a given column are zero, then the associated variable is
undetermined; make a note of the undetermined variable(s) and then
ignore all such columns.
2. Swap rows so that the first entry in the first column is non-zero.
3. Multiply the first row by λso that the pivot is 1.
4. Add multiples of the first row to each other row so that the first entry
of every other row is zero.
5. Now ignore the first row and first column and repeat steps 1-5 until
the matrix is in RREF.
Example
3x3= 9
x1+5x22x3= 2
1
3x1+2x2= 3
First we write the system as an augmented matrix:
1
pf3
pf4
pf5

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Elementary Row Operations

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:

  • (Row Swap) Exchange any two rows.
  • (Scalar Multiplication) Multiply any row by a constant.
  • (Row Sum) Add a multiple of one row to another row.

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.

  1. If all entries in a given column are zero, then the associated variable is undetermined; make a note of the undetermined variable(s) and then ignore all such columns.
  2. Swap rows so that the first entry in the first column is non-zero.
  3. Multiply the first row by λ so that the pivot is 1.
  4. Add multiples of the first row to each other row so that the first entry of every other row is zero.
  5. Now ignore the first row and first column and repeat steps 1-5 until the matrix is in RREF.

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

 

3 ∼R 1

  

  

R 2 =R ∼ 2 −R 1

  

  

− ∼R 2

 

 

R 1 =R 1 − 6 R 2 ∼

 

 

(^13) R 3 ∼

 

 

R 1 =R ∼ 1 +12R 3

 

 

R 2 =R ∼ 2 − 2 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

Bˆ =

( 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.

References

Hefferon, Chapter One, Section 1.1 and 1. Wikipedia, Systems of Linear Equations

Review Problems

  1. Explain why row equivalence is not affected by removing columns. Is row equivalence affected by removing rows? Prove or give a counter- example.
  2. (Gaussian Elimination) Another method for solving linear systems is to use row operations to bring the augmented matrix to row-echelon form. In row echelon form, the pivots are not necessarily set to one, and we only require that all entries left of the pivots are zero, not necessarily entries above a pivot. Provide a counterexample to show that row-echelon form is not unique. Once a system is in row echelon form, it can be solved by “back sub- stitution.” Write the following row echelon matrix as a system of equations, then solve the system using back-substitution.  

 

  1. Explain why the linear system has no solutions:   

  

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