Animated Solution for Mathematics - Matrices and Determinants: If $x,y,z$ are in arithmetic progression with common difference $d,x\ne 3d$, and the determinant of the matrix $\left[\begin{array}{ccc}3& 4\sqrt{2}& x\\ 4& 5\sqrt{2}& y\\ 5& k& z\end{array}\right]$ is zero, then the value of $k^2$ is

Imagine you are standing before a complex $3\times 3$ matrix. It looks intimidating, but as an elite aspirant, you know that every matrix in a JEE Advanced problem is a puzzle waiting to be solved with elegance, not brute force.

Let us embark on this journey to solve for $k^2$ by uncovering the hidden symmetry within.

The problem begins with a statement: $x,y,z$ are in an Arithmetic Progression (A.P.) with common difference $d$. This means the middle term $y$ is the arithmetic mean of its neighbors.

Mathematically, this gives us the master key:

Keep this equation etched in your mind. It is the lever that will move the entire matrix.

Now, consider the matrix:

Our goal is to find the determinant, which is given as zero. Instead of expanding, we observe the third column. Since $x+z-2y=0$, we perform the row operation $R_2\to R_1+R_3-2R_2$.

Applying this to the first column:

Applying this to the second column:

Applying this to the third column:

Our matrix now simplifies to:

The determinant is now a breeze to calculate. Expanding along the second row, which contains two zeros, is the most efficient path:

This simplifies to:

We are left with a product of two terms equaling zero. This implies either $(k-6\sqrt{2})=0$ or $(3z-5x)=0$.

Here is where the JEE Advanced examiner tests your vigilance. Could $(3z-5x)$ be zero?

If $3z-5x=0$, we substitute $z=x+2d$:

However, the problem explicitly states $x eq 3d$. This is the trap, and we must reject this case.

Therefore, the only remaining possibility is:

Finally, we calculate the required value:

You have successfully navigated the trap and found the solution. Always look for the symmetry, respect the constraints, and never fear the variables.