Let \( \alpha, \beta, \) and \( \gamma \) be real numbers. Consider the following system of linear equations:
\( x + 2y + z = 7 \)
\( x + \alpha z = 11 \)
\( 2x - 3y + \beta z = \gamma \)
Match each entry in List I to the correct entries in List II
List I List II (P) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has (1) a unique solution (Q) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has (2) no solution (R) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has (3) infinitely many solutions (S) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) 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
Let \( \alpha, \beta, \) and \( \gamma \) be real numbers. Consider the following system of linear equations:
\( x + 2y + z = 7 \)
\( x + \alpha z = 11 \)
\( 2x - 3y + \beta z = \gamma \)
Match each entry in List I to the correct entries in List II
| List I | List II | ||
| (P) | If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has | (1) | a unique solution |
| (Q) | If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has | (2) | no solution |
| (R) | If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has | (3) | infinitely many solutions |
| (S) | If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) 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 | ||
(P) →(3), (Q)→ (2), (R) →(1), (S) →(4)
(P) →(3), (Q)→ (2), (R) →(5), (S) →(4)
(P) →(2), (Q)→ (1), (R) →(4), (S) →(5)
(P) →(2), (Q)→ (1), (R) →(1), (S) →(3)
The Correct Option is A
Approach Solution - 1
We have the following system of linear equations:
\(x + 2y + z = 7\)
\(x + \alpha z = 11\)
\(2x - 3y + \beta z = \gamma\)
Let's analyze the scenarios provided in List I and match them with List II:
(P) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has
When \(\beta = \frac{1}{2}(7\alpha - 3)\) and \(\gamma = 28\), the third equation becomes \((7\alpha-3)/2 = 28\). This implies the system is consistent and has a special form of \(\beta\) and \(\gamma\), leading to infinitely many solutions as it matches a dependency condition. Thus, it associates with (1) a unique solution, but due to the specific \(\beta\) form, it aligns more with (3).
(Q) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has
Here, with the condition that \(\gamma \neq 28\), the system becomes inconsistent if derived equations contradict each other, resulting in no solution. Thus, it matches with (2).
(R) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has
With \(\alpha=1\), \(\beta\neq(7\alpha-3)/2 = 2\) and \(\gamma \neq 28\), the system is consistent and independent, leading to a unique solution. Thus it matches with (1).
(S) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \(\gamma = 28\), then the system has
With this condition, we substitute and verify one of the solution propositions whereby, assuming a direct solution trial reveals: \(x=11\), \(y=-2\), \(z=0\) satisfies all equations. Thus, matches (4).
Hence, the correct matching is: (P) →(3), (Q)→ (2), (R) →(1), (S) →(4)
Approach Solution -2
Solving the System of Equations
We are given the system of equations:
\[ x + 2y + z = 7 \] \[ x + \alpha z = 11 \] \[ 2x - 3y + \beta z = \gamma \]
We can express the determinant as:
\[ \triangle = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix} \] and from the equation, we get: \[ 3\alpha - 2(\beta - 2\alpha) - 3 = 0 \]
Solving the above equation, we find:
\[ 7\alpha - 2\beta = 3 \quad \Rightarrow \quad \beta = \frac{1}{2} (7\alpha - 3) \]
Next, we compute the determinants \( \triangle_1, \triangle_2, \) and \( \triangle_3 \):
\[ \triangle_1 = \begin{vmatrix} 7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta \end{vmatrix}, \quad \triangle_2 = \begin{vmatrix} 1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta \end{vmatrix}, \quad \triangle_3 = \begin{vmatrix} 1 & 7 & 2 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma \end{vmatrix} \]
We solve for \( \gamma \) by simplifying \( \triangle_3 \):
\[ 33 - 2(\gamma - 22) + 7(-3) = 0 \quad \Rightarrow \quad \gamma = 28 \]
Next, we simplify \( \triangle_1 \):
\[ \triangle_1 = 21\alpha - 2(11\beta - \alpha \beta) - 33 \] \[ = 21\alpha - 22\beta + 22\beta + 2\alpha \beta - 33 \]
Next, we simplify \( \triangle_2 \):
\[ \triangle_2 = 11\beta - \alpha\beta - 7(\beta - 2\alpha) + \gamma - 22 \] \[ = 14\alpha + 4\beta + \gamma - \alpha\gamma - 22 \]
Case (P):
If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then:
\[ \triangle = 0, \quad \triangle_1 = 0, \quad \triangle_2 = 0, \quad \triangle_3 = 0 \] There are infinitely many solutions, and \( x = 11, y = -2, z = 0 \) satisfy all the three equations.
Case (Q):
If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then:
\[ \triangle = 0, \quad \text{but} \quad \triangle_3 \neq 0 \quad \text{so no solution} \]
Case (R):
If \( \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1 \) and \( \gamma \neq 28 \), then:
\[ \triangle \neq 0, \quad \triangle_3 \neq 0 \quad \text{so a unique solution} \]
Case (S):
If \( \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1 \) and \( \gamma = 28 \), then:
\[ \triangle \neq 0, \quad \triangle_3 = 0, \quad \triangle_1 \neq 0, \quad \triangle_2 \neq 0 \quad \text{so a unique solution} \]
Conclusion:
Thus, the correct option is (A) \( \text{(P)} \to (3), \, \text{(Q)} \to (2), \, \text{(R)} \to (1), \, \text{(S)} \to (4) \).
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Questions Asked in JEE Advanced exam
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Concepts Used:
Linear Inequalities in One Variable
The graph of a linear inequality in one variable is a number line. We can use an open circle for < and > and a closed circle for ≤ and ≥.
Inequalities that have the same solution are commonly known as equivalents. There are several properties of inequalities as well as the properties of equality. All the properties below are also true for inequalities including ≥ and ≤.
The addition property of inequality says that adding the same number to each side of the inequality gives an equivalent inequality.
If x>y, then x+z>y+z If x>y, then x+z>y+z
If x<y, then x+z<y+z If x<y, then x+z<y+z
The subtraction property of inequality tells us that subtracting the same number from both sides of an inequality produces an equivalent inequality.
If x>y, then x−z>y−z If x>y, then x−z>y−z
If x<y, then x−z<y−z Ifx<y, then x−z<y−z
The multiplication property of inequality tells us that multiplication on both sides of an inequality with a positive number gives an equivalent inequality.
If x>y and z>0, then xz>yz If x>y and z>0, then xz>yz
If x<y and z>0, then xz<yz If x<y and z>0,then xz<yz