Animated Solution for Mathematics - Matrices and Determinants: The system of equations $\lambda x+y+z=0$, $-x+\lambda y+z=0$, $-x-y+\lambda z=0$ will have a non-zero solution if real values of $\lambda$ are given by .........

Imagine you are standing before a system of three linear equations, all whispering the same secret: they are homogeneous. This means that every single equation is set to zero, like $\lambda x+y+z=0$, $-x+\lambda y+z=0$, and $-x-y+\lambda z=0$.

In the world of linear algebra, this is a special state of affairs. A homogeneous system always has a trivial solution—the origin, where $x=0,y=0,z=0$.

We are hunting for the non-trivial, the hidden, the non-zero solutions that exist when these planes intersect in a more complex way. To find them, we must unlock the secret of the determinant.

The condition for a homogeneous system to possess non-trivial solutions is simple yet profound: the determinant of the coefficient matrix must be zero. Think of the determinant as a gatekeeper; if it is non-zero, the gate is locked, and only the trivial solution can pass.

If it is zero, the gate swings open, allowing for a family of non-zero solutions. Let us construct our matrix $A$ from the coefficients:

This is our battlefield.

Now, we expand this determinant along the first row. It is a dance of signs and minors.

The first term is $\lambda$ multiplied by the minor $\left|\begin{array}{cc}\lambda & 1\\ -1& \lambda \end{array}\right|$, which gives $\lambda ({\lambda }^2-(-1))=\lambda ({\lambda }^2+1)$.

Next, we take the second element, $1$, but remember the alternating sign rule: it becomes $-1$ multiplied by the minor $\left|\begin{array}{cc}-1& 1\\ -1& \lambda \end{array}\right|$, resulting in $-1(-\lambda -(-1))=-1(-\lambda +1)$.

Finally, the third element is $1$, multiplied by the minor $\left|\begin{array}{cc}-1& \lambda \\ -1& -1\end{array}\right|$, which gives $1(1-(-\lambda ))=1(1+\lambda )$.

Putting it all together, we have:

This looks like a mess, but watch as the complexity melts away. Expanding the terms, we get:

Notice the magic? The $-1$ and $+1$ cancel out perfectly, leaving us with ${\lambda }^3+3\lambda =0$. This is the beauty of mathematics—what starts as a daunting cubic equation simplifies into something elegant.

Factoring out $\lambda$, we get:

This gives us two paths: $\lambda =0$ or ${\lambda }^2+3=0$. The first path is clear: $\lambda =0$ is a valid real solution.

The second path, ${\lambda }^2=-3$, leads us into the realm of imaginary numbers, which we must reject because the problem demands real values. Thus, we arrive at our destination: the only real value for $\lambda$ that allows for a non-zero solution is $\lambda =0$.