>>9760098View it as a linear map:
The image of this map is the span of f(1,0)=(1,2,-1) and f(0,1)=(-1,1,-5).
That span is 2-dimensional since the above vectors are linearly independent (if they were dependent, one would have to be a multiple of the other). It's some plane in R^3.
Now, the system has a solution iff (4,9,3) is in the image of f.
If it is in the image, then f(1,0) , f(0,1) and (4,9.3) would be linearly dependent. That happens iff the matrix with columns f(1,0) , f(0,1) and (4,9.3) has a zero determinant. *
If it is not in the image, then f(1,0) , f(0,1) and (4,9.3) would be linearly independent. That happens iff the matrix with columns f(1,0) , f(0,1) and (4,9.3) has a non-zero determinant.
The determinant turns out to be 27, so there's no solution.
* If (4,9,3) is outside of the plane, then f(0,1) , f(1,0) and (4,9,3) span a parallelepiped who volume can be measured by that determinant, while if (4,9,3) lies in the plane then the "parallelepiped" has 0 volume (0 deteminant).
