Consider the system of equations x - 2y + 3z = -1, -x + y - 2z = k, x - 3y + 4z = 1. STATEMENT - 1 :

Q: Wait, why does $D=0$ mean no solution? Can't it be infinite solutions?

That is a fantastic question! You are absolutely right that $D=0$ is not enough to guarantee 'no solution'. When $D=0$, the system is either inconsistent (no solution) or dependent (infinite solutions). To distinguish between them, we must check the other determinants like $D_x, D_y,$ or $D_z$. If any of these are non-zero, the system is inconsistent. If all are zero, we have to look deeper, but in this specific problem, we found $D_y = k-3$, which directly tells us when the system becomes inconsistent.

Q: How do I know which determinant to calculate? Why $D_y$?

Great intuition! You don't necessarily have to calculate $D_y$ first. You could calculate $D_x$ or $D_z$. However, in this problem, the structure of the equations makes $D_y$ very easy to compute. The key is that for 'no solution', you only need *one* of the determinants to be non-zero. Once you find that $D_y \neq 0$, you have already proven the system is inconsistent.

Q: Is Statement 2 just a random determinant? How does it relate to the system?

It is not random at all! In JEE Advanced problems, you will often see statements that seem disconnected. The trick is to evaluate the determinant in Statement 2 and compare it to the ones you calculated for the system. Here, we discovered that the determinant $\Delta$ is exactly $-D_y$. This means Statement 2 is just a clever, disguised way of stating the condition $D_y \neq 0$.

Question

Consider the system of equations $x - 2y + 3z = -1$, $-x + y - 2z = k$, $x - 3y + 4z = 1$.
STATEMENT - 1 : The system of equations has no solution for $k 
eq 3$ and
STATEMENT - 2 : The determinant $\begin{vmatrix} 1 & 3 & -1 \ -1 & -2 & k \ 1 & 4 & 1 \end{vmatrix} 
eq 0$, for $k 
eq 3$.

Options:

(A) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C) STATEMENT - 1 is True, STATEMENT - 2 is False

(D) STATEMENT - 1 is False, STATEMENT - 2 is True

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space surrounded by three infinite planes defined by the following equations:
$x - 2y + 3z = -1$
$-x + y - 2z = k$
$x - 3y + 4z = 1$

Our goal is to determine the nature of their intersection. We must decide if they meet at a single point, along a line, or if they form a prism with no common intersection.

### The First Step: The Determinant $D$

To understand the fundamental nature of the system, we examine the coefficient matrix:
$$A = \begin{pmatrix} 1 & -2 & 3 \ -1 & 1 & -2 \ 1 & -3 & 4 \end{pmatrix}$$

We calculate the determinant $D$ by expanding along the first row:
$$D = 1(4 - 6) - (-2)(-4 + 2) + 3(3 - 1)$$

Simplifying this expression, we find:
$$D = 1(-2) + 2(-2) + 3(2) = -2 - 4 + 6 = 0$$

The fact that $D = 0$ is a critical observation. It implies that the normal vectors of these planes are linearly dependent, meaning the system cannot have a unique solution.

### The Mystery of the Parameter $k$

We now investigate the condition for the system to be inconsistent. According to Cramer's Rule, for a system to have no solution when $D = 0$, at least one of the modified determinants (such as $D_y$) must be non-zero.

We calculate $D_y$ by replacing the $y$-column with the constants from the right-hand side of the equations:
$$D_y = \begin{vmatrix} 1 & -1 & 3 \ -1 & k & -2 \ 1 & 1 & 4 \end{vmatrix}$$

Expanding this determinant yields:
$$D_y = 1(4k + 2) - (-1)(-4 + 2) + 3(-1 - k)$$

Simplifying further, we obtain:
$$D_y = 4k + 2 - 2 - 3 - 3k = k - 3$$

The system is inconsistent if $D_y 
eq 0$, which implies $k - 3 
eq 0$, or **$k 
eq 3$**. This confirms that Statement 1 is true.

### The Hidden Connection

Next, we evaluate the determinant $\Delta$ provided in Statement 2:
$$\Delta = \begin{vmatrix} 1 & 3 & -1 \ -1 & -2 & k \ 1 & 4 & 1 \end{vmatrix}$$

Expanding along the first row, we get:
$$\Delta = 1(-2 - 4k) - 3(-1 - k) - 1(-4 + 2)$$

Simplifying this expression results in:
$$\Delta = -2 - 4k + 3 + 3k + 2 = 3 - k$$

We observe that $\Delta = -(k - 3) = -D_y$. Statement 2 claims that $\Delta 
eq 0$ for $k 
eq 3$.

Since $\Delta = -D_y$, this condition is mathematically identical to requiring $D_y 
eq 0$. Therefore, Statement 2 is true and serves as the correct logical explanation for Statement 1.

Select Answer:

Visualized Solution

The System of Equations

Defining the Determinant $D$

Evaluating $D$

Geometric Meaning of $D = 0$

Calculating $D_{y}$

Expanding $D_{y}$

Analyzing Statement 1

What if $k = 3$ ?

Analyzing Statement 2

Connecting the Statements

Final Conclusion

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

Analyzing the Setup

The First Step

The Mystery of the Parameter $k$

The Hidden Connection

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Select Answer:

Visualized Solution

The System of Equations

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

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

Geometric Meaning of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=0

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

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

Analyzing Statement 1

What if .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=3?