Question

Let α,β, and γ be real numbers. Consider the following system of linear equations:
x + 2y + z = 7
x + αz = 11
2x - 3y + βz = γ
Match each entry in List I to the correct entries in List II

List I		List II
(P)	If β=\(\frac{1}{2}\)(7α - 3) and \(\gamma\)=28, then the system has	(1)	a unique solution
(Q)	If β=\(\frac{1}{2}\)(7α - 3) and \(\gamma\)\(\neq\)28, then the system has	(2)	no solution
(R)	If β\(\neq\)\(\frac{1}{2}\)(7α - 3)  where \(\alpha\)=1 and \(\gamma\)\(\neq\)28, then the system has	(3)	infinitely many solutions
(S)	If β\(\neq\)\(\frac{1}{2}\)(7α - 3)  where \(\alpha\)=1 and \(\gamma\)=28, then the system has	(4)	x = 11, y = - 2 and z = 0 as a solution
		(5)	x = -15 , y = 4 and z = 0 as a solution

• A. (P) →(3), (Q)→ (2), (R) →(1), (S) →(4)
• B. (P) →(3), (Q)→ (2), (R) →(5), (S) →(4)
• C. (P) →(2), (Q)→ (1), (R) →(4), (S) →(5)
• D. (P) →(2), (Q)→ (1), (R) →(1), (S) →(3)

Answer 1

The Correct Option is A

\(x + 2y + z = 7\)
\(x + αz = 11\)
\(2x - 3y + βz = γ\)

\(\triangle =\) \(\begin{vmatrix}1&2&1 \\ 1 & 0 & \alpha \\ 2 &-3 & \beta \end{vmatrix}\)= 0
\(3\alpha - 2(\beta - 2\alpha) -3 = 0\)
\(7\alpha - 2 \beta = 3\)
⇒ \(\beta = \frac {1}{2} (7 \alpha -3)\)

\(\begin{vmatrix}7&2&1\\ 11&0 &\alpha \\ \gamma &-3 &\beta\end{vmatrix} , \triangle_2 = \begin{vmatrix}1&7&1\\ 1&11 &\alpha \\  2&\gamma &\beta\end{vmatrix} , \triangle_3 = \begin{vmatrix}1&7&2\\ 1&0 &11 \\  2& -3 &\gamma\end{vmatrix}\)

\(\triangle_3 =0\)
⇒ \(33 -2(\gamma -22) + 7(-3) = 0\)
\(\gamma = 28\)
\(\triangle_1 = 21 \alpha - 2(11\beta - \alpha \beta) - 33\)
\(\,\,= 21\alpha -22 \beta + 22 \beta + 2 \alpha \beta -33\)
\(\triangle_2 = 11 \beta - \alpha\beta - 7(\beta - 2 \alpha) + \gamma -22\)
\(\,\, = 14 \alpha + 4 \beta + \gamma - \alpha\gamma -22\)

\(\text{(P) If}\)  \(\beta =  \frac{1}{2}(7\alpha -3) \,\,and \,\,\gamma = 28\)
\(\triangle = 0, \,\triangle_1 = 0, \,\triangle_2 = 0, \,\triangle_3 = 0\)
Infinitely many solutions
x = 11, y = – 2 and z = 0 will satisfy all the three given equations, so it is a solution.

\(\text{(Q) If}\)  \(\beta = \frac {1}{2} (7 \alpha -3) \,\, and \,\, \gamma ≠ 28 \, \,then\)
 \(\triangle =0 , \,but \,\,\triangle_3\neq 0 \text{ so no solution}\)

\(\text{ (R) If }\beta \neq \frac{1}{2}(7\alpha - 3), \alpha = 1 \, and \,\, \gamma\neq 28\)
\(\,\,\triangle \neq 0, \triangle_3 \neq 0 \text{ so a unique solution}\)

\(\text{ (S) \, If }\, \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1, \,\, \gamma = 28\)
\(\,\,\, \triangle \neq 0, \,\triangle_3 = 0 \,, \triangle_1 \neq 0,\, \triangle_2 \neq 0, \text{so a unique solution}\)

x = 11, y = – 2 and z = 0 will satisfy all the three equations.
Hence, The correct option is (A) (P) →(3), (Q)→ (2), (R) →(1), (S) →(4).