Solving Homogeneous System of 3 Equations & 4 Variables

[scientifico]

Hi,
I'm solving out a homogeneous 3 equations and 4 variables system so I considered one variable as known term but the determinant of the matrix is 0, how do I use Cramer in this case ?
these are the 3 equations

2x + 3y - z - 2v = 0
4x - 3y - 5z + 5v = 0
8x + 3y - 7z + v = 0

determinant of {{2,3,-1},{4,-3,-5},{8,3,-7}} is 0

thanks

 

[LCKurtz]

Hi,
I'm solving out a homogeneous 3 equations and 4 variables system so I considered one variable as known term but the determinant of the matrix is 0, how do I use Cramer in this case ?
these are the 3 equations

2x + 3y - z - 2v = 0
4x - 3y - 5z + 5v = 0
8x + 3y - 7z + v = 0

determinant of {{2,3,-1},{4,-3,-5},{8,3,-7}} is 0

thanks

The normal way to solve a system like this would be to use row reduction.

 

[scientifico]

but how can I solve it with Cramer ?

 

[Mark44]

Hi,
I'm solving out a homogeneous 3 equations and 4 variables system so I considered one variable as known term but the determinant of the matrix is 0, how do I use Cramer in this case ?
these are the 3 equations

I don't think there is any reason that you should assume that one variable is known. This is apparently a system of three equations in four unknowns. The matrix of coefficients isn't square, so the concept of the determinant doesn't apply, and you can't use Cramer's Rule.

2x + 3y - z - 2v = 0
4x - 3y - 5z + 5v = 0
8x + 3y - 7z + v = 0

determinant of {{2,3,-1},{4,-3,-5},{8,3,-7}} is 0

thanks

 

[Mark44]

but how can I solve it with Cramer ?

As already noted, Cramer's Rule doesn't apply here. Follow the advice that LCKurtz gave.

 

[LCKurtz]

Also note that it isn't obvious just looking at the system whether there might be 1 or more free variables. You really need row reduction to answer that.

 

[Ray Vickson]

Hi,
I'm solving out a homogeneous 3 equations and 4 variables system so I considered one variable as known term but the determinant of the matrix is 0, how do I use Cramer in this case ?
these are the 3 equations

2x + 3y - z - 2v = 0
4x - 3y - 5z + 5v = 0
8x + 3y - 7z + v = 0

determinant of {{2,3,-1},{4,-3,-5},{8,3,-7}} is 0

thanks

Cramer's rule only applies when the determinant of coefficients is nonzero. No matter which column you omit, the resulting 3x3 matrix has zero determinant, so in this question Cramer's Rule fails in every single case you try.

As others have suggested, just use row reduction; but an equivalent description would be: just use the variable-elimination method as taught in high school.