Page 224 - Advanced engineering mathematics

204 CHAPTER 7 Matrices and Linear Systems

EXAMPLE 7.12
The following matrices are all reduced:
⎛ ⎞
0130

1 −4 1 0
⎝
0 0 0 1 , 0001 ⎠
0000
⎛ ⎞ ⎛ ⎞
01200 1 0 0 2 1
⎜ 00010 ⎟ ⎜ 010 −24 ⎟
,
⎜ ⎟ ⎜ ⎟ .
⎝ 00000 ⎠ ⎝ 0 0 1 0 1 ⎠
00000 0 0 0 0 0
EXAMPLE 7.13
⎛ ⎞
2 0 1
C = 0 −46 ⎠
⎝
2 −25
is not reduced. However, we claim that, by a sequence of elementary row operations, we can
transform C to a reduced matrix.
First, if the matrix had one or more zero rows, we would interchange rows to place these at
the bottom of the new matrix. In this example C has no zero rows.
In view of condition (4) of the definition, start at the upper left corner. Multiply row one by
1/2 to obtain a matrix having a leading entry of 1 in row one:
⎛ ⎞
1 0 1/2
C → 0 −4 6 ⎠ .
⎝
2 −2 5
To get zeros below the 1 in the 1,1− position, add −2 times row one to row three:
⎛ ⎞ ⎛ ⎞
1 0 1/2 1 0 1/2
⎝ 0 −4 6 ⎠ → 0 −4 6 ⎠ .
⎝
2 −2 5 0 −2 4
Now look across row two of the last matrix. The leading entry is −4, so divide this row by −4:
⎛ ⎞ ⎛ ⎞
1 0 1/2 1 0 1/2
⎝ 0 −4 6 ⎠ → 0 1 −3/2 ⎠ .
⎝
0 −2 4 0 −2 4
Asidefromthe1inthe2,2 position of the last matrix, we want zeros in column two. Add 2 times
row two to row three:
⎛ ⎞ ⎛ ⎞
1 0 1/2 10 1/2
⎝ 0 1 −3/2 ⎠ → 01 −3/2 ⎠ .
⎝
0 −2 4 00 1
It happens that the leading entry of row three of the last matrix is 1. To get zeros in column
three above this leading entry, add 3/2 times row three to row one, then −1/2 times row three to
row one:
⎛ ⎞ ⎛ ⎞
1 0 1/2 100
⎝ 0
1 −3/2 ⎠ → 010 ⎠ .
⎝
0 0 1 001
This is a reduced matrix that is row equivalent to C, having been obtained from it by a sequence
of elementary row operations.

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October 14, 2010 14:23 THM/NEIL Page-204 27410_07_ch07_p187-246