Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $2x+3y-z=-2$, $x+y+z=4$, $x-y+|\lambda |z=4\lambda -4$ where $\lambda \in \mathbb{R}$, has no solution, then

Imagine you are standing in a vast, three-dimensional room. You have three planes, each defined by a linear equation. Usually, these three planes intersect at a single, beautiful point—a unique solution.

But today, we are looking for a scenario where these planes refuse to meet. We are hunting for the value of $\lambda$ that makes the system inconsistent.

The Determinant

Our first step is to identify the 'gatekeeper' of the system: the determinant of the coefficient matrix, $\Delta$. If $\Delta eq 0$, the planes intersect at a unique point.

To ensure the system has no solution, we must force $\Delta =0$. Let us construct our matrix:

Expanding this along the first row, we get:

Simplifying this, we find:

This leads us to $-|\lambda |+7=0$, or simply $|\lambda |=7$. This is our critical threshold.

Now, we have two candidates: $\lambda =7$ and $\lambda =-7$. Many students stop here, but this is where the real algebra begins. We must test these candidates to ensure the system is truly inconsistent.

Let us look at $\lambda =7$ first. Substituting $\lambda =7$ into our third equation, $x-y+|\lambda |z=4\lambda -4$, we get $x-y+7z=24$.

If we perform elimination on the system, we find that the equations are consistent, leading to infinitely many solutions. That is not what we want.

Now, let us test $\lambda =-7$. Substituting this into the third equation, we get $x-y+7z=-32$.

When we eliminate $y$ using the first two equations, we get $x+4z=14$. When we eliminate $y$ using the second and third equations, we get $x+4z=-14$.

Look at that! We have $x+4z=14$ and $x+4z=-14$. This is a blatant mathematical contradiction, as $14 eq -14$.

This contradiction proves that the planes are positioned such that they never share a common intersection. The system is inconsistent.

We have found our answer: $\lambda =-7$.