Consider the system of linear equations x + y + z = 5, x + 2y + λ²z = 9, x + 3y + λz = μ, where λ, μ

Q: Why do we care so much about $\Delta = 0$?

Think of $\Delta$ as the 'uniqueness factor'. When $\Delta \neq 0$, the inverse of the matrix exists, which mathematically guarantees a unique solution. When $\Delta = 0$, the matrix is singular—it collapses dimensions. It is like trying to find a unique intersection of three planes that might be parallel or overlapping. We must check $\Delta_x, \Delta_y, \Delta_z$ to see if the planes intersect at a line (infinite solutions) or never meet at all (inconsistent).

Q: Wait, why is Option 4 wrong? It says $\lambda \neq 1$ and $\mu \neq 13$.

This is the classic JEE trap! You correctly identified $\lambda = 1$ as a critical value, but you missed the second root of the quadratic, $\lambda = -\frac{1}{2}$. If $\lambda = -\frac{1}{2}$, then $\Delta = 0$, meaning the system cannot have a unique solution regardless of $\mu$. Option 4 claims a unique solution exists for all $\lambda \neq 1$, which is false because it fails at $\lambda = -\frac{1}{2}$.

Q: Do I really need to check $\Delta_z$ for every case?

In an exam, you want to be efficient. Once you identify the incorrect option (Option 4) by finding the missing root, you do not *need* to check the others to solve the question. However, for deep understanding, checking $\Delta_z$ confirms the consistency for $\lambda = 1$. It is a great way to verify your logic, but do not waste time if you have already found the definitive error.

Question

Consider the system of linear equations $x + y + z = 5, x + 2y + \lambda^2 z = 9, x + 3y + \lambda z = \mu$, where $\lambda, \mu \in \mathbb{R}$. Then, which of the following statement is NOT correct?

Options:

(A) System has infinite number of solution if $\lambda = 1$ and $\mu = 13$

(B) System is inconsistent if $\lambda = 1$ and $\mu 
eq 13$

(C) System is consistent if $\lambda 
eq 1$ and $\mu = 13$

(D) System has unique solution if $\lambda 
eq 1$ and $\mu 
eq 13$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a classic JEE Advanced problem that tests not just your algebraic skills, but your ability to see the 'soul' of a system of linear equations.

We are given three equations:
$x + y + z = 5$
$x + 2y + \lambda^2 z = 9$
$x + 3y + \lambda z = \mu$

At first glance, it looks like a standard system, but the presence of parameters $\lambda$ and $\mu$ turns this into a detective story. We need to find the statement that is NOT correct.

### The Gatekeeper of Uniqueness

Before we touch any algebra, let us visualize what is happening. We have three planes in 3D space. A unique solution means these three planes intersect at exactly one point.

This happens when the determinant of the coefficient matrix, $\Delta$, is non-zero. If $\Delta = 0$, the planes are either parallel, or they intersect in a line, or they are coincident.

Let us construct the determinant:

$$
\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & \lambda^2 \ 1 & 3 & \lambda \end{vmatrix}
$$

This matrix holds the DNA of our system. If this determinant is zero, the system loses its 'uniqueness' property.

### The Algebraic Dance

Now, let us expand this determinant along the first row. Expanding along the first row, we get:

$$
\Delta = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2)
$$

Let us simplify this carefully. We have $\Delta = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1$. Combining the like terms, we arrive at:

$$
\Delta = -2\lambda^2 + \lambda + 1
$$

This is a quadratic expression. To find the critical values, we factorize it:

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

This is the moment of truth! Setting $\Delta = 0$ gives us two critical values: $\lambda = 1$ and $\lambda = -\frac{1}{2}$. These are the points where the system's behavior changes drastically.

### The Trap

Here is where many students stumble. Look at the statement: "System has a unique solution if $\lambda 
eq 1$ and $\mu 
eq 13$."

It sounds plausible, right? But wait! We just found that $\Delta = 0$ at $\lambda = 1$ AND $\lambda = -\frac{1}{2}$.

If we choose $\lambda = -\frac{1}{2}$, the determinant is zero, and the system **cannot** have a unique solution, regardless of what $\mu$ is. This statement ignores the second critical value, making it the **incorrect statement**.

### Verifying the Consistency

For the sake of completeness, let us look at the case $\lambda = 1$. We know $\Delta = 0$. To check for consistency, we calculate $\Delta_z$ by replacing the third column with the constants $5, 9, \mu$:

$$
\Delta_z = \begin{vmatrix} 1 & 1 & 5 \ 1 & 2 & 9 \ 1 & 3 & \mu \end{vmatrix}
$$

Expanding this, we get:

$$
\Delta_z = 1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2) = \mu - 13
$$

If $\mu = 13$, then $\Delta_z = 0$, leading to infinite solutions. If $\mu 
eq 13$, then $\Delta_z 
eq 0$, leading to no solution (inconsistent).

### Conclusion

The beauty of this problem lies in the realization that a system's behavior is defined by its parameters. By finding the roots of the determinant, we mapped out the 'danger zones' for $\lambda$.

Always remember: in JEE Advanced, **precision is your greatest weapon**. You did not just solve an equation; you analyzed the entire landscape of the system. Keep that curiosity alive, and keep pushing boundaries!

Animated Solution for Mathematics - Matrices and Determinants: Consider the system of linear equations $x + y + z = 5, x + 2 y + λ^{2} z = 9, x + 3 y + λ z = μ$ , where $λ, μ \in R$ . Then, which of the following statement is NOT correct?

Select Answer:

Visualized Solution

System of Linear Equations

Cramer's Rule and Determinants

Setting up the Main Determinant $Δ$

Expanding the Determinant $Δ$

Factorizing $Δ$

Finding Critical Values of $λ$

Condition for Unique Solution

Analyzing Option 4

Verifying Other Options: Case $λ = 1$

Evaluating $Δ_{z}$

Consistency Conditions for $λ = 1$

Final Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Gatekeeper of Uniqueness

The Algebraic Dance

The Trap

Verifying the Consistency

Conclusion

Similar Questions

For the system of linear equations $2 x - y + 3 z = 5$ , $3 x + 2 y - z = 7$ , $4 x + 5 y + a z = β$ , which of the following is NOT correct?

The system of linear equations $λ x + 2 y + 2 z = 5$ , $2 λ x + 3 y + 5 z = 8$ , $4 x + λ y + 6 z = 10$ has:

The following system of linear equations $3 x + 3 y + 2 z = 9, 3 x + 2 y + 2 z = 9, x - y + 4 z = 8$ has:

For the system of linear equations $a x + y + z = 1$ , $x + a y + z = 1$ , $x + y + a z = β$ , which one of the following statements is NOT correct ?

Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

For the system of equations $x + y + z = 6$ , $x + 2 y + a z = 10$ , $x + 3 y + 5 z = β$ , which one of the following is NOT true?

The system of linear equations $x + y + z = 2$ , $2 x + 3 y + 2 z = 5$ , $2 x + 3 y + (a^{2} - 1) z = a + 1$

The system of linear equations $λ x + 2 y + 2 z = 5 2 x + 3 y + 5 z = 8 4 x + λ y + 6 z = 10$ has

For the system of linear equations $x + y + z = 6$ , $α x + β y + 7 z = 3$ , $x + 2 y + 3 z = 14$ , which of the following is NOT true ?

For the system of linear equations $2 x + 4 y + 2 a z = b$ , $x + 2 y + 3 z = 4$ , $2 x + 5 y + 2 z = 8$ which of the following is NOT correct?

Select Answer:

Visualized Solution

System of Linear Equations

Cramer's Rule and Determinants

Setting up the Main Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Expanding the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Factorizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Finding Critical Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Condition for Unique Solution

Analyzing Option 4

Verifying Other Options: Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=1

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δz​

Consistency Conditions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=1