The system of equations in 3 variables \( x,y \) and z as \( x+y=5,y+az=3 \) , \( x+by=8 \) has a unique solution if the value of \( a \) and \( b \) are
(A) \( a=1,b=1 \)
(B) \( a=0,b=1 \)
(C) \( a\in(-\infty,0) \) and \( b\ne 1 \)
(D) \( a\ne 0 \) and \( b\in(3,4) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C) and (D) only
The system has a unique solution exactly when the coefficient determinant is non-zero.
\( \displaystyle \begin{vmatrix}1 & 1 & 0\\0 & 1 & a\\1 & b & 0\end{vmatrix} \)
\( \displaystyle =1(0-ab)-1(0-a)+0 \)
\( \displaystyle =a-ab \)
\( \displaystyle =a(1-b) \)
So the determinant is a(1 − b), and the solution is unique exactly when a ≠ 0 and b ≠ 1.
Checking the statements
(A) \( a=1,b=1 \) — Incorrect. b = 1 makes a(1 − b) = 0.
(B) \( a=0,b=1 \) — Incorrect. a = 0 makes a(1 − b) = 0.
(C) \( a\in(-\infty,0) \) and \( b\ne 1 \) — Correct. Here a ≠ 0 and b ≠ 1.
(D) \( a\ne 0 \) and \( b\in(3,4) \) — Correct. Here a ≠ 0, and b ≠ 1 since b lies between 3 and 4.