JEE Mathematics 2023 Question with Solution

Given: The system is

$$\begin{aligned} \lambda x + y + z &= 1 \\ x + \lambda y + z &= 1 \\ x + y + \lambda z &= 1 \end{aligned}$$

Find: The value of $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|)$ for those real values of $\lambda$ for which the system is inconsistent.

For inconsistency, the coefficient determinant must be zero. So solve

$$\begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0$$

From the given working,

$$(\lambda + 2)\begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0$$

and

$$(\lambda + 2)[1(\lambda^2-1)-1(\lambda-1)+(1-\lambda)] = 0$$

Thus,

$$(\lambda + 2)(\lambda^2-2\lambda+1)=0$$

So the roots are

$$\lambda = -2, \quad \lambda = 1$$

From the solution, at $\lambda = 1$ the system has infinite solutions, and for inconsistency $\lambda = -2$. Therefore,

$$\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) = |-2|^2 + |-2| = 4 + 2 = 6$$

Therefore, the correct option is D.

Given: The coefficient matrix is

$$A = \begin{bmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{bmatrix}$$

with right-hand side vector

$$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$

Find: Which value of $\lambda$ makes the system inconsistent.

First, set the determinant equal to zero:

$$\begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0$$

Using the extracted solution steps, this gives

$$(\lambda + 2)(\lambda - 1)^2 = 0$$

Hence,

$$\lambda = -2 \quad \text{or} \quad \lambda = 1$$

Now classify these values using the solution statement:

• at $\lambda = 1$, the system has infinite solutions,
• for inconsistent system, $\lambda = -2$.

So,

$$S = \{-2\}$$

Therefore,

$$\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) = |{-2}|^2 + |{-2}| = 4 + 2 = 6$$

Hence the required value is $6$, so the correct option is D.