For what value of k, the system of linear equations has no solution ?
(3k + 1)x + 3y - 2 = 0
(k^2 + 1)x + (k - 2)y − 5 = 0

Question

For what value of k, the system of linear equations has no solution ? 
(3k + 1)x + 3y - 2 = 0 
(k^2 + 1)x + (k - 2)y − 5 = 0

Accepted Answer

To determine the value of $ k $ for which the given system of linear equations has no solution, we need to analyze the conditions under which two linear equations are inconsistent, meaning they represent parallel lines that never intersect.

The given system of equations is:

1. $(3k + 1)x + 3y - 2 = 0$
2. $(k^2 + 1)x + (k - 2)y - 5 = 0$

For the system to have no solution, the lines must be parallel. This occurs when the ratios of the coefficients of $ x $ and $ y $ are equal, but the ratio of the constant terms is different. Specifically, for the lines to be parallel, the following condition must hold:

$$
\frac{3k + 1}{k^2 + 1} = \frac{3}{k - 2} \neq \frac{-2}{-5}
$$

Let's solve the equation $\frac{3k + 1}{k^2 + 1} = \frac{3}{k - 2}$:

Cross-multiplying gives:

$$
(3k + 1)(k - 2) = 3(k^2 + 1)
$$

Expanding both sides:

$$
3k^2 - 6k + k - 2 = 3k^2 + 3
$$

Simplifying:

$$
3k^2 - 5k - 2 = 3k^2 + 3
$$

Subtract $3k^2$ from both sides:

$$
-5k - 2 = 3
$$

Add 2 to both sides:

$$
-5k = 5
$$

Divide by -5:

$$
k = -1
$$

Thus, when $ k = -1 $, the lines are parallel, and the system of equations has no solution. This confirms that Option 2 is correct.

Now, let's explain why the other options are incorrect:

1. **Option 1 (k = 1):** Substituting $ k = 1 $ into the equations does not satisfy the condition for parallel lines. The ratios of the coefficients do not equate, meaning the lines are not parallel, and the system may have a solution.

2. **Option 3 (k = 2):** Substituting $ k = 2 $ results in undefined expressions due to division by zero in the ratio $\frac{3}{k - 2}$, which is not a valid scenario for determining parallelism.

3. **Option 4 (k = 6):** Substituting $ k = 6 $ into the equations also does not satisfy the condition for parallel lines. The ratios of the coefficients do not equate, meaning the lines are not parallel, and the system may have a solution.

Therefore, the correct value of $ k $ that makes the system of equations have no solution is indeed $ k = -1 $, supporting Option 2 as the correct answer.