Question

Let S be the set of all column matrices [b₁, b₂, b₃] such that b₁, b₂, b₃ ∈ R and the system of equations (in real variables)

–x + 2y + 5z = b₁

2x – 4y + 3z = b₂

x – 2y + 2z = b₃

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each [b₁, b₂, b₃] ∈ S?

(A) x + 2y + 3z = b₁, 4y + 5z = b₂ and x + 2y + 6z = b₃ 

(B) x + y + 3z = b₁, 5x + 2y + 6z = b₂ and –2x – y – 3z = b₃ 

(C) –x + 2y – 5z = b₁, 2x – 4y + 10z = b₂ and x – 2y + 5z = b₃ 

(D) x + 2y + 5z = b₁, 2x + 3z = b₂ and x + 4y – 5z = b₃

Answer 1

Answer is (A),(D)

The given system of equations is

= 0 ⇒ at least one solution is possible.

Therefore, we have

• Option (A): The given set of equations are

That is, in this case, there is a unique solution. Hence, option (A) is correct. 

• Option (B): The given set of equations are

However, this does not satisfy Eq. (1). Therefore, this system of equations has no solution. Hence, option (B) is incorrect.

 Option (C): The given set of equations are

That is, this has infinitely many solutions. Hence, option (C) is incorrect.

• Option (D): The given set of equations are

That is, in this case, there is a unique solution. Hence, option (D) is correct.