文档介绍：Solutions for Chapter 1
Solutions for exercises in section 1. 2
. (1, 0, 0)
. (1, 2, 3)
. (1, 0, −1)
. (−1/2, 1/2, 0, 1)
2 −4 3
. 4 −7 4
5 −8 4
. Every row operation is reversible. In particular the “inverse” of any row operation
is again a row operation of the same type.
π
. 2 ,π,0
. The third equation in the triangularized form is 0x3 =1, which is impossible
to solve.
. The third equation in the triangularized form is 0x3 =0, and all numbers are
solutions. This means that you can start the back substitution with any value
whatsoever and consequently produce infinitely many solutions for the system.
− 11 − 3
. α= 3,β= 2 , and γ= 2
. (a) If xi = the number initially in chamber #i, then
.4x1 +0x2 +0x3 + .2x4 =12
0x1 + .4x2 + .3x3 + .2x4 =25
0x1 + .3x2 + .4x3 + .2x4 =26
.6x1 + .3x2 + .3x3 + .4x4 =37
and the solution is x1 =10,x2 =20,x3 =30, and x4 =40.
(b) 16, 22, 22, 40
. To interchange rows i and j, perform the following sequence of Type II and
Type III operations.
Rj ← Rj + Ri (replace row j by the sum of row j and i)
Ri ← Ri − Rj (replace row i by the difference of row i and j)
Rj ← Rj + Ri (replace row j by the sum of row j and i)
Ri ←−Ri (replace row i by its negative)
. (a) This has the effect of interchanging the order of the unknowns— xj and
xk are permuted. (b) The solution to the new system is the same as the
2 Solutions
solution to the old system except that the solution for the jth unknown of the
1 th
new system isx ˆj = α xj. This has the effect of “changing the units” of the j
unknown. (c) The solution to the new system is the same as the solution for
the old system except that the solution for the kth unknown in the new system
isx ˆk = xk −αxj.
. h = 1
ij i+j−1
x1 y1
x2 y2
. If x = . and y = . are two different solutions, then
. .
xm ym
x1+y1
2