1. Key Concept: Consistency of a System of Linear Equations (Cramer's Rule)

For a system of linear equations $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we use determinants to determine its consistency. Let $\Delta = \det(A)$ be the determinant of the coefficient matrix. Let $\Delta_x, \Delta_y, \Delta_z$ be the determinants formed by replacing the respective column of the coefficient matrix $A$ with the constant terms from matrix $B$.

The conditions for consistency are as follows:

• Unique Solution (Consistent): The system has a unique solution if and only if $\Delta \ne 0$. In this case, $x = \frac{\Delta_x}{\Delta}$, $y = \frac{\Delta_y}{\Delta}$, $z = \frac{\Delta_z}{\Delta}$.
• No Solution (Inconsistent): The system has no solution if and only if $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
• Infinitely Many Solutions (Consistent): The system has infinitely many solutions if and only if $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.

Our goal is to find the conditions for which the system is inconsistent, i.e., has no solution.

2. Setting up the System

The given system of linear equations is:

1. $3x - 2y - kz = 10$
2. $2x - 4y - 2z = 6$
3. $x + 2y - z = 5m$

From these equations, we can identify the coefficient matrix $A$ and the constant vector $B$: $A = \begin{pmatrix} 3 & -2 & -k \\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 6 \\ 5m \end{pmatrix}$

3. Calculating the Determinant of the Coefficient Matrix ($\Delta$)

First, we calculate $\Delta = \det(A)$. This step is crucial because the value of $\Delta$ dictates whether a unique solution exists or if we need to investigate further for inconsistency or infinitely many solutions. $\Delta = \begin{vmatrix} 3 & -2 & -k \\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{vmatrix}$ Expanding along the first row: $\Delta = 3 \left( (-4)(-1) - (-2)(2) \right) - (-2) \left( (2)(-1) - (-2)(1) \right) + (-k) \left( (2)(2) - (-4)(1) \right)$ $\Delta = 3 (4 + 4) + 2 (-2 + 2) - k (4 + 4)$ $\Delta = 3(8) + 2(0) - k(8)$ $\Delta = 24 - 8k$

For the system to be inconsistent (or have infinitely many solutions), we must have $\Delta = 0$. Setting $\Delta = 0$: $24 - 8k = 0$ $8k = 24$ $k = 3$ So, the system can only be inconsistent (or have infinitely many solutions) if $k=3$. If $k \ne 3$, then $\Delta \ne 0$, which means the system has a unique solution and is therefore consistent.

4. Calculating Other Determinants ($\Delta_x, \Delta_y, \Delta_z$) for $k=3$

Now that we know $k=3$ is the condition for $\Delta = 0$, we substitute $k=3$ into the coefficient matrix and calculate $\Delta_x, \Delta_y, \Delta_z$. These determinants help us distinguish between no solution (inconsistent) and infinitely many solutions (consistent).

Substituting $k=3$, the coefficient matrix becomes: $A = \begin{pmatrix} 3 & -2 & -3 \\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{pmatrix}$

Calculation of $\Delta_x$: Replace the first column of $A$ with the constants from $B$: $\Delta_x = \begin{vmatrix} 10 & -2 & -3 \\ 6 & -4 & -2 \\ 5m & 2 & -1 \end{vmatrix}$ Expanding along the first row: $\Delta_x = 10 \left( (-4)(-1) - (-2)(2) \right) - (-2) \left( (6)(-1) - (-2)(5m) \right) + (-3) \left( (6)(2) - (-4)(5m) \right)$ $\Delta_x = 10 (4 + 4) + 2 (-6 + 10m) - 3 (12 + 20m)$ $\Delta_x = 10(8) - 12 + 20m - 36 - 60m$ $\Delta_x = 80 - 48 - 40m$ $\Delta_x = 32 - 40m$

Calculation of $\Delta_y$: Replace the second column of $A$ with the constants from $B$: $\Delta_y = \begin{vmatrix} 3 & 10 & -3 \\ 2 & 6 & -2 \\ 1 & 5m & -1 \end{vmatrix}$ Expanding along the first row: $\Delta_y = 3 \left( (6)(-1) - (-2)(5m) \right) - 10 \left( (2)(-1) - (-2)(1) \right) + (-3) \left( (2)(5m) - (6)(1) \right)$ $\Delta_y = 3 (-6 + 10m) - 10 (-2 + 2) - 3 (10m - 6)$ $\Delta_y = -18 + 30m - 10(0) - 30m + 18$ $\Delta_y = 0$

Calculation of $\Delta_z$: Replace the third column of $A$ with the constants from $B$: $\Delta_z = \begin{vmatrix} 3 & -2 & 10 \\ 2 & -4 & 6 \\ 1 & 2 & 5m \end{vmatrix}$ Expanding along the first row: $\Delta_z = 3 \left( (-4)(5m) - (6)(2) \right) - (-2) \left( (2)(5m) - (6)(1) \right) + 10 \left( (2)(2) - (-4)(1) \right)$ $\Delta_z = 3 (-20m - 12) + 2 (10m - 6) + 10 (4 + 4)$ $\Delta_z = -60m - 36 + 20m - 12 + 10(8)$ $\Delta_z = -40m - 48 + 80$ $\Delta_z = 32 - 40m$

5. Applying the Condition for Inconsistency

For the system to be inconsistent (no solution), we need two conditions to be met:

1. $\Delta = 0$
2. At least one of $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.

From our calculations:

• Condition 1: $\Delta = 0 \implies k=3$.
• Condition 2: For $k=3$, we have:
  • $\Delta_x = 32 - 40m$
  • $\Delta_y = 0$
  • $\Delta_z = 32 - 40m$

Since $\Delta_y$ is always zero, for condition 2 to be met, we need either $\Delta_x \ne 0$ OR $\Delta_z \ne 0$. Both $\Delta_x$ and $\Delta_z$ are equal to $32 - 40m$. So, we require $32 - 40m \ne 0$. $40m \ne 32$ $m \ne \frac{32}{40}$ $m \ne \frac{4}{5}$

Therefore, the system of linear equations is inconsistent if and only if $k=3$ AND $m \ne \frac{4}{5}$.

6. Analyzing the Results and Conclusion

Let's summarize the conditions for the system based on our analysis:

• If $k \ne 3$: $\Delta \ne 0$, so there is a unique solution (consistent).
• If $k = 3$: $\Delta = 0$.
  • If $m \ne \frac{4}{5}$: Then $\Delta_x \ne 0$ (and $\Delta_z \ne 0$), while $\Delta_y = 0$. Since at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution (inconsistent).
  • If $m = \frac{4}{5}$: Then $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$. Since all determinants are zero, the system has infinitely many solutions (consistent).

The question asks for the condition when the system is inconsistent. Based on our derivation, this occurs when $k=3$ and $m \ne \frac{4}{5}$.

Comparing this with the given options: (A) $k \ne 3, m \in R$ (This leads to a unique solution, hence consistent) (B) $k = 3, m \ne \frac{4}{5}$ (This leads to no solution, hence inconsistent) (C) $k = 3, m = \frac{4}{5}$ (This leads to infinitely many solutions, hence consistent) (D) $k \ne 3, m \ne \frac{4}{5}$ (This leads to a unique solution, hence consistent)

Our derived condition for inconsistency, $k=3$ and $m \ne \frac{4}{5}$, matches option (B).

The final answer is $\boxed{B}$.

7. Tips for JEE Aspirants

• Master Cramer's Rule: It's a fundamental tool for analyzing systems of linear equations. Understand the conditions for unique, no, and infinite solutions thoroughly.
• Systematic Calculation: Determinant calculations can be prone to sign errors. Perform them slowly and methodically, double-checking each step. Row/column operations can simplify determinants before expansion.
• Don't Jump to Conclusions: The condition $\Delta=0$ only tells you there isn't a unique solution. You must calculate $\Delta_x, \Delta_y, \Delta_z$ to distinguish between no solution and infinitely many solutions.
• Gaussian Elimination as an Alternative: For systems with parameters, Gaussian elimination (reducing the augmented matrix to row echelon form) is another powerful method and can often be more intuitive for understanding the geometric interpretation of solutions. It would yield the same results.

8. Key Takeaway

A system of linear equations is inconsistent (has no solution) precisely when the determinant of the coefficient matrix ($\Delta$) is zero, AND at least one of the determinants formed by replacing a column with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) is non-zero. If $\Delta=0$ and all $\Delta_x, \Delta_y, \Delta_z$ are also zero, the system has infinitely many solutions. If $\Delta \ne 0$, there's always a unique solution.