Question

Let \(k\) be a constant. The equations \(kx + y = 3\) and \(4x + ky = 4\) have a unique solution if and only if

• A. \(|k|≠2\)
• B. \(|k|=2\)
• C. \(k≠2\)
• D. \(k=2\)

Answer 1

The Correct Option is A

To find when the system of equations \(kx + y = 3\) and \(4x + ky = 4\) has a unique solution, we rely on the condition that the determinant of the coefficient matrix is non-zero.

The coefficient matrix is:  

\[\begin{bmatrix} k & 1 \\ 4 & k \end{bmatrix}\]

The determinant of this matrix is computed as: 

\[\text{det} = k \cdot k - 1 \cdot 4 = k^2 - 4\]

For a unique solution, the determinant must be non-zero: 

\[k^2-4 \neq 0 \]\]

Solving the inequality: 

\[k^2 \neq 4 \]\]\[|k| \neq 2 \]\]

Thus, the system has a unique solution if and only if \(|k|≠2\)\\)

Answer 2

Consider two linear equations of the form: \[ ax + by = c \quad \text{and} \quad dx + ey = f \] 

These two equations have a unique solution if the coefficients satisfy: \[ \frac{a}{d} \ne \frac{b}{e} \]

Given:

Let's assume: \[ \frac{k}{4} \ne \frac{1}{k} \]

Solving the inequality:

Cross-multiplying: \[ k^2 \ne 4 \Rightarrow |k| \ne 2 \]

Conclusion:

The condition for the system to have a unique solution is: \[ \boxed{|k| \ne 2} \]

Final Answer:

The correct option is (A): \[ \boxed{|k| \ne 2} \]