Answer:
C
Explanation:
If we call [x y z]^T big X and the other matrices A and b
AX=b
A^-1 is the inverse, times both sides by it
A^-1 A X=A^-1 b
A^-1 A X = I X = X where I is the identity matrix
so X = A^-1 b
doing this matrix multiplication gives us c