Page 274 - Advanced engineering mathematics

254 CHAPTER 8 Determinants

There are many ways to evaluate |A|. One way to begin is to exploit the 1 in the 1,3 position to
get zeros in the other locations in column 3. Add −2 times row one to row three, −5 times row
one to row four, and 3 times row one to row five to get
⎛ ⎞
−6 0 1 3 2
⎜ −1 5 0 1 7 ⎟
⎜ ⎟
B = ⎜ 20 3 0 −5 3 ⎟ .
⎜ ⎟
⎝ 30 1 0 −18 −8 ⎠
−17 15 0 18 10
Adding a multiple of one row to another does not change the value of the determinant, so
|A|=|B|.
Furthermore, by equation (8.3),
|B|= (−1) 1+3 (1)|C|=|C|,
where C is the 4 × 4 matrix formed by deleting row one and column three of B:
⎛ ⎞
−1 5 1 7
20 3 −5 3
⎜ ⎟
C = ⎜ ⎟ .
⎝ 30 1 −18 −8 ⎠
−17 15 18 10
Now work on C. Again, there are many ways to proceed. We will use the −1inthe 1,1 position
to get zeros in row one. Add 5 times column one to column two, add column one to column three
and add 7 times column one to column four of C to get
⎛ ⎞
−1 0 0 0
⎜ 20 103 15 143 ⎟
D = ⎜ ⎟ .
⎝ 30 151 12 202 ⎠
−17 70 1 −109
Because we added a multiple of one column to another,
|C|=|D|.
And, using equation (8.3) again,

|D|= (−1) 1+1 (−1)|E|=−|E|,
where E is the 3 × 3 matrix formed from D by deleting row one and column one:
⎛ ⎞
103 15 143
E = ⎝ 151 12 202 ⎠ .
−70 1 −109
To evaluate E wewill usethe1inthe3,2 place. Add −1 times row three to row one and −12
times row three to row two to get
⎛ ⎞
1153 0 1778
F = ⎝ 991 0 1510 ⎠ .
−70 1 −109
Then
|E|=|F|.
Furthermore

|F|= (−1) 3+2 (1)|G|=−|G|

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October 14, 2010 14:26 THM/NEIL Page-254 27410_08_ch08_p247-266