Consider the system of equation \( x+y+3z=8,\ 2x+y+bz=24,\ x+ay+3z=10 \), then which of the following statements are True?
(A) The system of equations has a unique solution if \( a\ =\ 2\ ,b\ =\ 3 \)
(B) The system of equations is inconsistent if \( a\ =1,b\in\mathrm{R} \) be any real number.
(C) The system of equations is inconsistent if \( a\ \ne 1 \), and \( b\ \ne 6 \)
(D) The system of equations has a unique solution if \( a\ =4 \) and \( b=3 \).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
The system has a unique solution exactly when the coefficient determinant is non-zero.
\( \displaystyle \begin{vmatrix}1 & 1 & 3\\2 & 1 & b\\1 & a & 3\end{vmatrix}=1(3-ab)-1(6-b)+3(2a-1) \)
\( =-ab+b+6a-6 \)
\( =(a-1)(6-b) \)
With a = 1, the first and third equations become x + y + 3z = 8 and x + y + 3z = 10, which contradict each other.
So the determinant is (a − 1)(6 − b): the solution is unique when a ≠ 1 and b ≠ 6, and the system is inconsistent when a = 1.
Checking the statements
(A) The system of equations has a unique solution if \( a\ =\ 2\ ,b\ =\ 3 \) — Correct. The determinant is (2 − 1)(6 − 3) = 3 ≠ 0.
(B) The system of equations is inconsistent if \( a\ =1,b\in\mathrm{R} \) be any real number. — Correct. With a = 1, x + y + 3z cannot be both 8 and 10.
(C) The system of equations is inconsistent if \( a\ \ne 1 \), and \( b\ \ne 6 \) — Incorrect. Then the determinant is non-zero, so the system has a unique solution.
(D) The system of equations has a unique solution if \( a\ =4 \) and \( b=3 \). — Correct. The determinant is (4 − 1)(6 − 3) = 9 ≠ 0.