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7.3 Reduced Row Echelon Form 205

In this example, we started with C and obtained a reduced matrix. If we had used a different
sequence of elementary row operations, could we have reached a different reduced matrix? The
answer is no.

THEOREM 7.7

Every matrix is row equivalent to a matrix in reduced form. Further, the reduced form of a matrix
is completely determined by the matrix itself, not by the reduction process. That is, no matter
what sequence of elementary row operations is used to produce a reduced matrix equivalent to
A, the same reduced matrix will result.

Proof Every matrix can be manipulated to reduced form by following the idea of Example 7.13.
First, move any zero rows to the bottom of the matrix by row interchanges. Then start at the
upper left leading entry, and by multiplying this row by a scalar, obtain a matrix having a 1 in
this position. Add multiples of this row to the other rows to obtain zeros in this column below
this leading entry. Then move to the second row and carry out the same procedure starting with
its leading entry. After each nonzero row has been treated, a reduced matrix results.

In view of the uniqueness of the reduced form of a given matrix, we will denote the reduced
form of A as A R . The process of determining A R , by any sequence of elementary row operations,
is referred to as reducing A.

EXAMPLE 7.14
Let
⎛ ⎞
0000 0
0020 0
⎜ ⎟
A = ⎜ ⎟ .
⎝ 0101
1 ⎠
0030 −4
We will reduce this matrix. First interchange rows to move the zero row to the bottom of the
matrix:
⎛ ⎞ ⎛ ⎞
0000 0 0020 0
0020 0 0101 1
⎜ ⎟ ⎜ ⎟
A = ⎜ ⎟ → ⎜ ⎟ .
⎝ 0101
1 ⎠ ⎝ 0030 −4 ⎠
0030 −4 0000 0
The leading entry of row one is in the 1,3 position, and the leading entry of row three is in the 2,2
position. We want the leading entries to move down the matrix from left to right, so interchange
rows one and two to obtain
⎛ ⎞ ⎛ ⎞
0020 0 0101 1
0101 1 0020 0
⎜ ⎟ ⎜ ⎟
⎜ ⎟ → ⎜ ⎟ .
⎝ 0030 −4 ⎠ ⎝ 0030 −4 ⎠
0000 0 0000 0
The leading entry of (the new) row one is 1 and already has zeros below it, so move to the second
row and find its leading entry, which is 2. Multiply row two by 1/2:
⎛ ⎞ ⎛ ⎞
0101 1 0101 1
⎜ 0020 0 ⎟ ⎜ 0010 0 ⎟
⎜ ⎟ → ⎜ ⎟ .
⎝ 0030 −4 ⎠ ⎝ 0030 −4 ⎠
0000 0 0000 0

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October 14, 2010 14:23 THM/NEIL Page-205 27410_07_ch07_p187-246