Animated Solution for Mathematics - Matrices and Determinants: The value of $k\in \mathbb{R}$, for which the following system of linear equations $3x-y+4z=3$, $x+2y-3z=-2$, $6x+5y+kz=-3$ has infinitely many solutions, is :

Imagine standing in a room where three walls meet. Usually, they meet at a single corner—a point. But what if those walls were shifted?

What if they all met along a single, shared edge? That is the geometric reality of a system with infinitely many solutions. We are looking for the value of $k$ that forces these three planes to align perfectly along a common line.

To solve this, we turn to the elegance of Cramer's Rule. We are given the system: $3x-y+4z=3$ $x+2y-3z=-2$ $6x+5y+kz=-3$

For a system to have infinitely many solutions, the main determinant of the coefficient matrix, $\Delta$, must be zero. If $\Delta$ is non-zero, the system has a unique solution. We need the system to be singular, which means $\Delta =0$.

Let us construct our determinant:

Expanding along the first row, we get:

Simplifying this, we find:

This expands to:

Combining the terms, we get $7k+35=0$, which leads us directly to $k=-5$.

We must be careful. As I always tell my students, $\Delta =0$ is a necessary condition, but not sufficient. It could also imply 'No Solution'.

To be certain, we must check ${\Delta }_x$. Replacing the first column with the constants $3,-2,-3$, we get:

Expanding this, we get:

Since ${\Delta }_x=0$, the system is indeed consistent and has infinitely many solutions. You have successfully navigated the trap! Keep this rigor in your toolkit, and you will conquer any JEE problem.