【高代】Matrices:row reduction

05-10

In this chapter, how to reduce a matrix to a row echelon matrix and how to use this to determine the solution of a system of equations will be shown.

some kind of short.

To begin with, elementary row operations are introduced( in case 2x2matrices) :

(i) $egin{pmatrix} 1&a\ 0&1 end{pmatrix} or egin{pmatrix} 1&0\ a&1 end{pmatrix}$ (ii) $egin{pmatrix} 0&1\ 1&0 end{pmatrix}$ (iii) $egin{pmatrix} c&0\ 0&1 end{pmatrix} or egin{pmatrix} 1&0\ 0&c end{pmatrix}$

These matrices are called elementary matrices, which are invertible, as well as their inverses. For a matrix M, by a sequence of operations, it will be simplified, and we called this procedureas row reduction:

$M=E_k...E_2E_1M$

Then, use this method, we can reduce an arbitary matrix to a row echelon matrix,which has the following properties:

(a)If (row i)of M is zero, then (row j) is zero for all j>i.

(b)If (row i) isnt zero, its first nonzero entry is 1, which is called a pivot(主元).

(c)If (row (i+1)) isnt zero, the pivot in (row (i+1)) is to the right of the pivot in (row i).

(d)The entries above a pivot are zero.

The (d) should be paid more attention to.

Now we mainly use this to solve two problems: how to sovle square systems and how to find a inverse of M.

We must first know the following coonditons are equivalent:

A is invertible; A is a product of elementary matrices; A can be reduced to the identity by a sequence of elementary row operations.

So $E_k...E_1A=I ightarrow E_k...E_1I=A^{-1}$ .We can do this by reduce [A|I] to [I|A].

We have mentioned augmented matrix before. Let [A|B] be the result of a sequence of row operations. The observation is that [A|B] and [A|B] has the same solution, which is of great help to determine the case of its solution.

Its straightforward that every homogneous linear equation AX=0 has the trivial solution X=0. Using the row echelon form, we conclude that if there are more unknowns than equations then the homogeneous equation AX=0 has a nontrivial solution.

Thus we handle the square system like this:

these conditions are equivalent:

(a) A is invertible.

(b)The system AX=B has a unique solution for every column vector B.

(c)The system AX=0 has only the trivial solution X=0.

Containing what has been said above, we can solve the square system more easily.

T.B.C.