If the system of linear equationsx+ky+3z=03x+ky–2z=02x+4y–3z=0has a non-zero solution (x,y,z), then xzy2 is equal to:(1)30(2)−30(3)−10(4)10In the solution given, the determinant of matrix formed from the coefficients of x,y,z in the three equations is equated to 0 and the corresponding value for k is found.However, as far as I know the condition for non-zero solutions is that the determinant must not equal 0. Is there anything I am missing? I am not asking for the solution. I only need some clarity in the concept. Any help is greatly appreciated.

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Jowiszowy9zb · Accepted Answer

The zero vector is always a solution to a homogeneos system. It is the unique solution when the determinant of A is non-zero and the matrix is non-singular.  To have a non-zero (non-trivial) solution, the determinant must be zero. In fact, we are going to have infinitely many solutions. The corresponding matrix is singular.  For example, in one dimensional case, ax=0. If a≠0, then x=0 is the unique solution. If a=0, then any non-zero x is a solution.

vefibiongedogn7z · Accepted Answer

You got it the other way round. The correct statement is:  det(A)=0⇔ker(A)  non-trivial(⋅)  You assume the existence of a non-zero solution (x,y,z)T and you want to compute xzy2. For the computation you essentially need some kind of equation relating x,y,z to each other. For this, you use (x,y,z)∈ker(A). Unfortunately, you do neither know ker(A) nor can you compute it without first eliminating k. Fortunately, though, we can apply equation (∗) since we assumed the existence of a nonzero solution, i.e. a non-trivial kernel. Equating det(A)=0 implies k=??? (a specific real number) in your case. Now apply the ideas above.

If the system of linear equations x+ky+3z=0 3x+ky–2z=0 2x+4y–3z=0

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