Animated Solution for Mathematics - Matrices and Determinants: Given $2x-y+2z=2$, $x-2y+z=-4$, $x+y+\lambda z=4$ then the value of $\lambda$ such that the given system of equation has NO solution, is

A Journey into Linear Systems

My dear student, welcome to the fascinating world of linear algebra. Today, we are not just solving a system of equations; we are exploring the very architecture of 3D space.

Imagine you are standing in a room, and each of these three equations represents a flat plane extending infinitely in all directions. Our goal is to find the specific value of $\lambda$ that makes these three planes refuse to meet at a single point. This is the essence of an inconsistent system.

Cramer's Rule

When we face a $3\times 3$ system, we have a powerful compass: Cramer's Rule. It tells us that the behavior of the system is dictated by the determinant of the coefficient matrix, which we call $\Delta$.

Think of $\Delta$ as the 'volume' or 'scaling factor' of the system. If $\Delta eq 0$, the planes intersect at exactly one point—a unique solution.

But we want the opposite. We want the planes to be 'out of sync'. For the system to have no solution, we need $\Delta =0$ and at least one of the auxiliary determinants (${\Delta }_x,{\Delta }_y,{\Delta }_z$) to be non-zero. This is our roadmap.

The Determinant Expansion

Let us construct our main determinant $\Delta$ using the coefficients of $x,y,$ and $z$:

Take a deep breath. Expanding this is just a matter of careful bookkeeping. We expand along the first row:

Let's simplify this step-by-step. The first term becomes $2(-2\lambda -1)=-4\lambda -2$. The second term is $+1(\lambda -1)=\lambda -1$. The third term is $2(1+2)=2(3)=6$.

Combining these, we get:

For the system to be inconsistent, we set $\Delta =0$:

The Crucial Verification

Many students stop here, but we are not finished! We have found the candidate $\lambda =1$, but we must ensure this doesn't lead to infinitely many solutions.

We need to check ${\Delta }_z$ (replacing the $z$-column with the constants $2,-4,4$):

Expanding this along the first row:

Since ${\Delta }_z=6$, which is clearly not zero, we have confirmed our result. The system is indeed inconsistent when $\lambda =1$. You have successfully navigated the trap!