If the following system of linear equations
2x + y + z = 5
x - y + z = 3
x + y + az = b
has no solution, then :

Solution

Maths sir · 5 steps

No solution needs Δ = 0 first, and then a right-hand side that refuses to fit.

1. Coefficient determinant. If Δ ≠ 0, the system always has exactly one solution. So for no solution we need Δ = 0 first.

   Δ = |2, 1, 1; 1, -1, 1; 1, 1, a| = 2(-a-1) - 1(a-1) + 1(1+1) = 1 - 3a

2. Find a. 1 - 3a = 0 gives a = 1/3. With this value the left side of the third equation is built from the first two.
3. Match the left sides. Take 2/3 of the first equation minus 1/3 of the second. That gives x + y + 1/3z = 10/3 - 1 = 7/3, which has the same left side as the third equation.
4. Check the right side. The third equation says the same left side equals b. If b = 7/3 there are infinitely many solutions. If b ≠ 7/3 the equations contradict each other, so there is no solution. Checking Δₓ = 2b - 14/3 gives the same condition.
5. So. No solution means a = 1/3 and b ≠ 7/3, which is option (d).

The idea to keep. For no solution, first make Δ = 0. Then check the right-hand side: if it does not match the same combination of equations, the system is inconsistent.

Answer from the NTA final answer key. Solution by PrepWiser.