Question

If the system of equations \[ 2x+3y-3z=3, \qquad x+2y+\alpha z=1, \qquad 2x-y+z=\beta \] has infinitely many solutions, then \[ \frac{\alpha}{\beta}-\frac{\beta}{\alpha} = \]

• A. \(\dfrac{53}{14}\)
• B. \(\dfrac{45}{14}\)
• C. \(-\dfrac{53}{14}\)
• D. \(-\dfrac{45}{14}\)

Hint:

For infinitely many solutions in a system of linear equations, one equation must be expressible as a linear combination of the others. A quick method is to compare coefficients directly after assuming: \[ E_3=\lambda E_1+\mu E_2 \]

Answer 1

The Correct Option is B

Concept: A system of linear equations has infinitely many solutions when: \[ \text{Rank}(A)=\text{Rank}(A|B)<\text{number of variables} \] This means one equation must be a linear combination of the others. Hence, for infinitely many solutions, the third equation must be dependent on the first two equations.

Step 1: Write the given equations. \[ 2x+3y-3z=3 \qquad \cdots (1) \] \[ x+2y+\alpha z=1 \qquad \cdots (2) \] \[ 2x-y+z=\beta \qquad \cdots (3) \] Since infinitely many solutions exist, equation (3) must be obtainable from equations (1) and (2). Assume: \[ (3)=\lambda(1)+\mu(2) \]

Step 2: Compare coefficients of \(x\), \(y\), and constants. From coefficient of \(x\): \[ 2\lambda+\mu=2 \qquad \cdots (4) \] From coefficient of \(y\): \[ 3\lambda+2\mu=-1 \qquad \cdots (5) \] Solving (4) and (5): From (4), \[ \mu=2-2\lambda \] Substitute into (5): \[ 3\lambda+2(2-2\lambda)=-1 \] \[ 3\lambda+4-4\lambda=-1 \] \[ -\lambda=-5 \] \[ \lambda=5 \] Then, \[ \mu=2-2(5) \] \[ \mu=-8 \]

Step 3: Find the value of \(\alpha\). Compare coefficients of \(z\): From equation (3), \[ 1=\lambda(-3)+\mu(\alpha) \] Substitute \(\lambda=5,\ \mu=-8\): \[ 1=5(-3)-8\alpha \] \[ 1=-15-8\alpha \] \[ 16=-8\alpha \] \[ \alpha=-2 \]

Step 4: Find the value of \(\beta\). Compare constants: \[ \beta=3\lambda+1\mu \] Substitute values: \[ \beta=3(5)+(-8) \] \[ \beta=15-8 \] \[ \beta=7 \]

Step 5: Compute \[ \frac{\alpha}{\beta}-\frac{\beta}{\alpha} \] Substitute: \[ \frac{-2}{7}-\frac{7}{-2} \] \[ =-\frac{2}{7}+\frac{7}{2} \] Taking LCM \(=14\): \[ =\frac{-4+49}{14} \] \[ =\frac{45}{14} \] Thus, \[ \boxed{\frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{45}{14}} \] Hence, the correct option is: \[ \boxed{(B)\ \frac{45}{14}} \]