Animated Solution for Mathematics - Matrices and Determinants: The number of values of $k$ for which the linear equations $4x+ky+2z=0$, $kx+4y+z=0$ and $2x+2y+z=0$ possess a non-zero solution is

Imagine you are standing before a set of linear equations. At first glance, they look like any other system, but look closer at the right-hand side. Every single equation equals zero.

This is not a coincidence; it is a defining characteristic of a homogeneous system of linear equations. In the world of JEE Advanced, recognizing this structure is your first step toward victory.

We represent this system as $AX=0$, where $A$ is our coefficient matrix and $X$ is the column vector of variables.

A homogeneous system is unique because it always possesses the trivial solution, where $x=0,y=0,z=0$. But the problem asks for something more: a non-zero solution.

For this to happen, the system must be singular. Geometrically, this means the planes represented by these equations do not intersect at a single point, but rather along a line or a plane, creating infinitely many solutions.

Algebraically, this requires the determinant of the coefficient matrix to be exactly zero: $\det (A)=0$. This is the key that unlocks the puzzle.

Now, let us set up our determinant:

We expand this along the first row. Remember the sign convention: plus, minus, plus.

For the first element, $4$, we block its row and column to get $4(4\cdot 1-1\cdot 2)$, which simplifies to $4(2)$.

For the second element, $k$, we apply the negative sign: $-k(k\cdot 1-1\cdot 2)$, which simplifies to $-k(k-2)$.

Finally, for the third element, $2$, we have $+2(k\cdot 2-4\cdot 2)$, which simplifies to $2(2k-8)$.

Putting it all together, we have the equation:

Now, we simplify this expression step by step. We have $8-k^2+2k+4k-16=0$.

Combining like terms, we arrive at $-k^2+6k-8=0$. To make this quadratic easier to handle, we multiply by $-1$ to get:

Factorizing this, we find $(k-4)(k-2)=0$. This gives us two possible values for $k$: $k=4$ and $k=2$.

The question asks for the number of values of $k$, and since we have found two distinct values, the answer is 2.

You have successfully navigated the matrix, mastered the determinant, and solved the quadratic. This is the beauty of mathematics—taking a complex system and reducing it to its core essence.