Linear Algebra/Gauss-Jordan Reduction/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

Use Gauss-Jordan reduction to solve each system.

Answer

These answers show only the Gauss-Jordan reduction. With it, describing the solution set is easy.

  1. A row swap here makes the arithmetic easier.
This exercise is recommended for all readers.
Problem 2

Find the reduced echelon form of each matrix.

Answer

Use Gauss-Jordan reduction.

This exercise is recommended for all readers.
Problem 3

Find each solution set by using Gauss-Jordan reduction, then reading off the parametrization.

Answer

For the "Gauss" halves, see the answers to Problem I.2.5.

  1. The "Jordan" half goes this way.
    The solution set is this
  2. The second half is
    so the solution is this.
  3. This Jordan half
    gives
    (of course, the zero vector could be omitted from the description).
  4. The "Jordan" half
    ends with this solution set.
Problem 4

Give two distinct echelon form versions of this matrix.

Answer

Routine Gauss' method gives one:

and any cosmetic change, like multiplying the bottom row by ,

gives another.

This exercise is recommended for all readers.
Problem 5

List the reduced echelon forms possible for each size.

Answer

In the cases listed below, we take . Thus, some canonical forms listed below actually include infinitely many cases. In particular, they includes the cases and .

  1. , , ,
  2. , , , , , ,
  3. , , ,
  4. , , , , , ,
This exercise is recommended for all readers.
Problem 6

What results from applying Gauss-Jordan reduction to a nonsingular matrix?

Answer

A nonsingular homogeneous linear system has a unique solution. So a nonsingular matrix must reduce to a (square) matrix that is all 's except for 's down the upper-left to lower-right diagonal, e.g.,

Problem 7

The proof of Lemma 4 contains a reference to the condition on the row pivoting operation.

  1. The definition of row operations has an condition on the swap operation . Show that in this condition is not needed.
  2. Write down a matrix with nonzero entries, and show that the operation is not reversed by .
  3. Expand the proof of that lemma to make explicit exactly where the condition on pivoting is used.
Answer
  1. The operation does not change .
  2. For instance,
    leaves the matrix changed.
  3. If then
    does indeed give back. (Of course, if then the third matrix would have entries of the form .)