Let A =
1 |
1 |
3 |
2 |
1 |
2 |
4 |
3 |
1 |
3 |
a |
b |
To answer the questions, we will first reduce A to its RREf as under:
Add -1 times the 1st row to the 2nd row
Add -1 times the 1st row to the 3rd row
Add -2 times the 2nd row to the 3rd row
Add -1 times the 2nd row to the 1st row
Then the RREf of A is
1 |
0 |
2 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
a-5 |
b-4 |
a). Now, if a = 5 and b = 4, the solutions to the given system will be x+2z = 1 or x = 1-2z and y+z = 1 or, y = 1-z so that (x,y,z)T = (1-2z,1-z,z)T= (1,1,0)T +z(-2,-1,1)T. Thus, the given system will have infinite solutions if a = 5 and b = 4.
b). If a = 5 and b ≠4, then the last row of the augmented matrix will be (0,0,0,1) which means 0 = 1. Then the given system will be inconsistent. Thus, the given system will be inconsistent if a = 5 and b ≠4.
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