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SOLVING A SYSTEM OF LINEAR EQUATIONS 91
which denotes a system of linear equations having an identity matrix as the
coefficient matrix
[] [3] [2] [1] T
I x = b = [ b b b ]
1 2 3
[]
and, consequently, take the RHS vector b as the final solution.
Note that we don’t have to distinguish the two steps, the forward/backward
elimination. In other words, during the forward elimination, we do the pivot-
ing operations in such a way that the pivot becomes one and other elements
above/below the pivot in the same column become zeros.
Consider the following system of linear equations:
−1 −2 2 x 1 −1
1 1 −1 x 2 = 1 (2.2.28)
1 2 −1 x 3 2
We construct the augmented matrix by combining the coefficient matrix and the
RHS vector to write
a 11 a 12 a 13 b 1 −1 −2 2 −1 : r 1
a 21 a 22 a 23 b 2 = 1 1 −1 1 : r 2 (2.2.29)
a 31 a 32 a 33 b 3 1 2 −1 2 : r 3
and apply the Gauss–Jordan elimination procedure.
(1)
First, we divide the first row r 1 by a 11 =−1 to make the new first row r
1
(1) (1)
have the pivot a = 1 and subtract a m1 × r (m = 2, 3) from the second and
11 1
third row r 2 and r 3 to get
(1) (1) (1) (1) (1)
r 1 ÷ (−1) → a a a b 1 2 −21 : r
11 12 13 1 1
(1) (1) (1) (1) (1)
r 2 − 1 × r → a a a b (1) = 0 −1 1 0 : r
1 21 22 23 2 2
(1) (1) (1) (1) (1) (1)
r 3 − 1 × r → a a a b 0 0 1 1 : r
1 31 32 33 3 3
(2.2.30a)
(1) (1)
Then, we divide the second row r by a =−1 to make the new second row
2 22
(2) (2) (1) (2)
r have the pivot a = 1 and subtract a × r (m = 1, 3) from the first and
2 22 m2 2
(1) (1)
third row r and r to get
1 3
r (1) − 2 × r (2) → a (2) a (2) a (2) b (2) : r (2)
1 2 11 12 13 1 10 0 1 1
(1) (2) (2) (2) (2)
r ÷ (−1) → a a a b (2) = 01 −1 0 : r
2 21 22 23 2 2
(1) (2) (2) (2) (2) (2) (2)
r − 0 × r → a a a b 00 1 1 : r
3 2 31 32 33 3 3
(2.2.30b)
(2) (2)
Lastly, we divide the third row r by a = 1 to make the new third row
3 33
(3) (3) (2) (3)
r have the pivot a = 1 and subtract a × r (m = 1, 2) from the first and
3 33 m3 3
