Just do exactly what you are told to do. Use "Gauss Jordan elimination" on the augmented matrix . The last row will have "0" in the first two places, expressions in k in the third and fourth places.

Values of k that make the expression in the third place non-zero give a unique solution since you can just divide through by it to get the solution.

Values of k that make the expression in the third place zero and in the fourth place non-zero give no solution since that corresponds to 0z= non-zero which is not true for any z.

Values of k that make the expressions in both third and fourth places 0 give infinitely many solutions since that corresponds to 0z= 0 which is true for all z.