Animated Solution for Mathematics - Matrices and Determinants: Find all values of $\lambda$, such that $x,y,z\ne (0,0,0)$ and $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z=\lambda (x\hat{i}+y\hat{j}+z\hat{k})$ where $\hat{i},\hat{j},\hat{k}$ are unit vectors along the coordinate axes.

Welcome, fellow traveler on the JEE Advanced journey. Today, we are not just solving a problem; we are peeling back the layers of a linear system.

We are given a vector equation:

The condition $(x, y, z) eq (0, 0, 0)$ is our North Star. It tells us that we are not looking for the trivial solution where everything vanishes into nothingness. We are looking for the specific values of $\lambda$ that allow the system to exist in a non-zero state. This is the essence of eigenvalues and eigenvectors, hidden in plain sight.

Translating Vectors to Algebra

To make sense of this, we must speak the language of components. We equate the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides.

For $\hat{i}$, we have $x+3y-4z=\lambda x$, which rearranges to:

For $\hat{j}$, we have $x-3y+5z=\lambda y$, becoming:

For $\hat{k}$, we have $3x+y=\lambda z$, or:

We have successfully transformed a vector equation into a beautiful, homogeneous system of three linear equations. This is the bedrock of our solution.

The Gatekeeper of Solutions

Now, we face the core of the problem. We have a system $AX=0$. In the world of linear algebra, such a system has a non-trivial solution if and only if the determinant of the coefficient matrix $A$ is zero.

If $|A| eq 0$, the matrix is invertible, and we could simply multiply by $A^{-1}$ to find $X=A^{-1}0=0$, which is the trivial solution we are trying to avoid. So, we set the determinant to zero:

This determinant is the gatekeeper. It dictates whether our system collapses or expands.

The Algebraic Dance

Expanding this determinant is where the real work happens. We expand along the first row:

The first term simplifies to $(1-\lambda )({\lambda }^2+3\lambda -5)$. The second term becomes $3\lambda +45$. The third term simplifies to $-40-12\lambda$.

When we combine these, the constants $-5,45,-40$ magically cancel out to zero! We are left with:

Multiplying by $-1$, we get:

We are at the finish line. Factoring out $\lambda$, we get:

The quadratic inside is a perfect square:

This gives us our final values: $\lambda =0$ and $\lambda =-1$. These are the values that allow our system to have non-trivial solutions.

It is elegant, it is precise, and it is the result of careful, step-by-step logic. Remember, in JEE Advanced, the complexity is often just a mask for fundamental principles. Keep your cool, trust your algebra, and the path will always reveal itself.