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# Problems on Inhomogeneous Systems

1.
Row reducing the augmented matrix with Octave gives
octave:5> A=[1,2,3,4;2,4,8,10;3,6,11,14];
octave:6> B=[1;6;7];
octave:7> rref([A,B])
ans =

1.00000   2.00000   0.00000   1.00000  -5.00000
0.00000   0.00000   1.00000   1.00000   2.00000
0.00000   0.00000   0.00000   0.00000   0.00000


Assigning the values s and t to the variables x2 and x4, and solving for the leading variables, gives

2.
Row reduce the augmented matrix:
octave:8> A=[1,3,3;2,6,9;-1,-3,3];
octave:9> B=[1;5;5];
octave:10> rref([A,B])
ans =

1   3   0  -2
0   0   1   1
0   0   0   0

Solving the reduced system for x1 and x3 gives

Assigning the value s to x2 gives

3.
Row reducing the augmented matrix gives:
octave:11> A=[1,3,1,2;2,6,4,8;0,0,2,4];
octave:12> B=[1;3;1];
octave:13> rref([A,B])
ans =

1.00000  3.00000  0.00000  0.00000  0.50000
0.00000  0.00000  1.00000  2.00000  0.50000
0.00000  0.00000  0.00000  0.00000  0.00000

Solving for x1 and x3 gives

Setting x2=s and x3=t gives

4.
Row reducing the augmented matrix gives
octave:14> A=[1,2,2,4,6;1,2,3,6,9;0,0,1,2,3];
octave:15> B=[0;0;0];
octave:16> rref([A,B])
ans =

1  2  0  0  0  0
0  0  1  2  3  0
0  0  0  0  0  0

Solving for x1 and x3 yields

Setting x2=s, x4=t, and x5=u gives

5.
Row reducing the augmented matrix gives
octave:17> A=[1,2,-2;2,5,-4;4,9,-8];
octave:18> B=[1;0;0];
octave:19> rref([A,B])
ans =

1   0  -2   0
0   1   0   0
0   0   0   1

This gives the equivalent system of equations

Since no value values for x1, x2, and x3 can make the last equation true, the system has no solutions.

Up: Solutions to Supplementary Homework Previous: Problems on Row Reduction
Frank Beatrous
1998-09-21