Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{cc}\alpha & 2\\ 2& \alpha \end{array}\right]$ and $|A^3|=125$ then the value of $\alpha$ is

Welcome, future engineers! Today, we are going to unravel a problem that might look like a simple algebraic exercise, but it is actually a gateway into the beautiful, structured world of linear algebra.

We are given a matrix $A=\left[\begin{array}{cc}\alpha & 2\\ 2& \alpha \end{array}\right]$ and a condition $|A^3|=125$.

At first glance, you might be tempted to jump straight into calculating $A^3$. But stop! Take a breath. In the JEE Advanced, time is your most precious resource, and brute force is rarely the intended path. Instead, let us look for the hidden structure.

The key to this problem lies in a fundamental property of determinants: $|A^n|=|A{|}^n$. This property is a lifesaver.

It tells us that the determinant of a matrix raised to a power is simply the determinant of the original matrix, raised to that same power. By applying this, our condition $|A^3|=125$ transforms into:

Suddenly, the problem is no longer about matrix multiplication; it is about simple arithmetic. We know that $125$ is $5^3$. Therefore, $|A{|}^3=5^3$, which implies $|A|=5$. See how the complexity just melted away?

Now that we have the numerical value of the determinant, let us look at the matrix $A$ itself. For any $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, the determinant is defined as $ad-bc$.

Applying this to our matrix $A=\left[\begin{array}{cc}\alpha & 2\\ 2& \alpha \end{array}\right]$, we get:

Now we have two expressions for the same determinant: one from the given condition ($|A|=5$) and one from the matrix definition ($|A|={\alpha }^2-4$).

Since both expressions represent the same value, we can equate them:

This leads us to ${\alpha }^2=9$. Here is where many students stumble.

When you solve ${\alpha }^2=9$, you must remember that both $3$ and $-3$ are valid solutions because $(-3{)}^2$ is also $9$. Thus, $\alpha =\pm 3$.

This is the beauty of the problem—it tests not just your knowledge of matrix properties, but also your attention to detail in solving quadratic equations. You have successfully navigated the trap, applied the property, and arrived at the correct answer. Keep this mindset, and no matrix problem will ever be too daunting for you!