# Thread: Gauss reduction with constants a and b

1. ## Gauss reduction with constants a and b

Consider the system of linear equations:

$x + 2y + 3z = 4$
$x + y - 2z = a$
$2x + 3y + bz = 6$

where a and b are constants

For which values of a and b does this system have (i) a unique solution; (ii) no solution; (iii) in finitely many solutions?

Would you first have to gauss reduce this matrix to get 0s below the first non-zero entry of the first row, second row and third row, and then look at what values of a and b that give the particular solutions? Or would you look for the a and b values that give the particular solutions without gauss reduction?

2. Originally Posted by SyNtHeSiS
Consider the system of linear equations:

$x + 2y + 3z = 4$
$x + y - 2z = a$
$2x + 3y + bz = 6$

where a and b are constants

For which values of a and b does this system have (i) a unique solution; (ii) no solution; (iii) infinitely many solutions?

Would you first have to gauss reduce this matrix to get 0s below the first non-zero entry of the first row, second row and third row, and then look at what values of a and b that give the particular solutions? Or would you look for the a and b values that give the particular solutions without gauss reduction?
If the determinant of the matrix of coefficients is non-zero there is a unique solution. So you could start by looking at the determinant.

CB