Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}5& 5\alpha & \alpha \\ 0& \alpha & 5\alpha \\ 0& 0& 5\end{array}\right]$. If $|A^2|=25$, then $|\alpha |$ equals

Imagine you are standing before a complex grid of numbers, a $3\times 3$ matrix:

At first glance, it might look like a daunting collection of variables and constants. But in the world of linear algebra, structure is everything.

The first step to mastering any matrix problem is to observe its anatomy. Look at the elements below the main diagonal; they are all zero. This is the hallmark of an Upper Triangular Matrix.

Why is being an upper triangular matrix so helpful? Because the determinant of such a matrix is simply the product of its main diagonal elements.

You do not need to perform the long, tedious expansion process involving minors and cofactors. The determinant $|A|$ is just the product of the diagonal entries:

Just like that, we have reduced a matrix of nine elements down to a single, elegant expression. This is the beauty of recognizing patterns in mathematics.

Now, the problem introduces a condition: $|A^2|=25$. We could try to multiply matrix $A$ by itself, but that would be a path filled with potential for arithmetic errors.

Instead, we reach into our toolkit for a fundamental property of determinants: $|A^n|=|A{|}^n$. This means the determinant of $A$ squared is simply the square of the determinant of $A$.

So, we can write:

We are now left with a simple algebraic equation: $(25\alpha {)}^2=25$. Let us expand this carefully.

Squaring $25$ gives us $625$, and squaring $\alpha$ gives us ${\alpha }^2$. Our equation becomes:

To isolate ${\alpha }^2$, we divide both sides by $625$:

Simplifying this fraction, we get ${\alpha }^2=\frac{1}{25}$. Finally, to find the absolute value of $\alpha$, we take the square root of both sides:

And there it is! Through the power of observation and the application of fundamental properties, we have navigated the problem with precision. Remember, in JEE Advanced, the goal is not just to calculate, but to see the structure.