7.16
Note that a basic solution here must satisfy the 3 constraints uniquely for a set of valid basic variables. As
there are 5 variables and 3 equality constraints, a basic solution must have 2 nonbasic variables and 3 basic
variables. We assume that vector solutions have form
 
x1
x2 
 

x=
x3  .
x4 
x5
(a) No, because the solution does not satisfy the last constraint.
(b) No, because the solution is not a unique solution. If the solution was basic, then x1 and x4 would be
nonbasic. Thus, we must check to see if the system

   
3 1 0
x2
1
1 0 1 x3  = 1
6 1 3
x5
4
has a unique solution. However, the columns of the matrix are linearly dependent because
     
 
1
0
0
3
1 − 3 0 − 1 = 0 .
1
3
0
6
Thus, the system does not have a unique solution. Indeed, this linear combination reveals that
(0, 3, −8, 0, −2) is also a solution.
(c) Here, we have 3 possible nonbasic variables. In this case, if there are a pair of nonbasic variables that
indicate the solution is basic, then the solution is basic. Here, this is true: using x1 and x2 as nonbasic
variables, we are left with checking whether

   
1 0 0
x3
1
0 2 1 x4  = 1
1 0 3
4
x5
has a unique solution. Since the columns are linear independent (note how elementary row operations
can be used to turn the matrix into an identity matrix), the system does have a unique solution, so
(0, 0, 1, 0, 1) is basic.
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