The system of linear equations x + y + z = 6,
x + 2y + 3z = 10 and x + 2y + az = b has no solution when
b = 2, a = 3
D.
Now, options (a) and (d) satisfy a = 3. If b = 10, then equations (ii) and (iii) become identical and system will have infinite solutions.
Hence,
If x, y,z are all different and not equal to zero and = 0, then the value of x-1 + y-1 + z-1 is equal to
xyz
x-1y-1z-1
- x - y - z
- 1