Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $x-ky-z=0$, $kx-y-z=0$, $x+y-z=0$ has a non-zero solution, then the possible values of $k$ are

Imagine you are standing in a 3D room. You have three flat sheets of paper, each representing a linear equation. Because each equation is equal to zero, these planes are not floating randomly; they are all anchored, passing perfectly through the origin $(0,0,0)$.

This is the essence of a homogeneous system $AX=0$. The origin is our trivial meeting point—a place where all three planes are guaranteed to intersect.

However, the question asks for a non-zero solution. This means the planes must share more than just a single point; they must intersect along an entire line.

How do we translate this geometric requirement into algebra? If the determinant of the coefficient matrix $A$ were non-zero, the matrix would be invertible, and the only solution would be $X=A^{-1}0$, which is just the origin.

To escape this "trivial" trap, we need the matrix to be singular. We need the determinant to vanish: $|A|=0$. This is the gateway to infinitely many solutions.

Let us extract the coefficients from our equations: $x-ky-z=0$, $kx-y-z=0$, and $x+y-z=0$. The matrix $A$ is formed by these coefficients:

Now, we expand along the first row. We take the first element, $1$, and multiply it by the minor:

Next, we handle the middle term, $-(-k)$, which becomes $+k$, multiplied by the minor:

Finally, we take the last term, $-1$, multiplied by the minor:

Putting it all together, we have:

As we expand, we see:

Look closely at the terms involving $k$. We have a $+k$ and a $-k$! They cancel out perfectly, leaving us with:

This is the beauty of mathematics—what started as a complex system of planes has collapsed into the elegant, simple equation $k^2=1$.

Solving this, we find $k=\pm 1$. These are the values that force our planes to align in such a way that they share a common line, providing us with the non-zero solutions we were hunting for.