Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{cc}2& 3\\ 3& 5\end{array}\right]$, then the determinant of the matrix $(A^{2025}-3A^{2024}+A^{2023})$ is

Mastering Matrix Polynomials

Imagine you are sitting in the examination hall, the clock is ticking, and you see a matrix $A=\left[\begin{array}{cc}2& 3\\ 3& 5\end{array}\right]$ raised to the power of $2025$. Your heart might skip a beat; it is designed to do that.

The JEE Advanced examiners love to present you with massive exponents to test your composure. But here is the secret: in the world of linear algebra, high powers are rarely meant to be calculated directly.

They are almost always a mask, hiding a simpler, more elegant structure waiting to be revealed. Let us embark on this journey to dismantle this problem, not with brute force, but with the precision of a surgeon.

The Art of Factoring

When you look at the expression $|A^{2025}-3A^{2024}+A^{2023}|$, your first instinct might be to panic. Resist that. Instead, look for the common thread.

In algebra, we factor out the lowest power to simplify expressions, and the same rule applies to matrices. We can pull out $A^{2023}$ from the entire expression.

This transforms our terrifying polynomial into the following form:

Suddenly, the problem has changed its shape. We are no longer dealing with a gargantuan power; we are dealing with a product of two matrices: $A^{2023}$ and the quadratic matrix $(A^2-3A+I)$. This is the first victory.

The Power of Determinants

Now, we invoke one of the most powerful tools in our arsenal: the multiplicative property of determinants. We know that for any two square matrices $X$ and $Y$, the determinant of their product is the product of their determinants: $|XY|=|X|\cdot |Y|$.

Applying this to our expression, we get:

Furthermore, we know that $|A^n|=|A{|}^n$. This is a massive simplification! We have effectively decoupled the problem into two independent tasks: finding $|A{|}^{2023}$ and finding the determinant of the quadratic matrix.

The Arithmetic Grind

Let us handle the first task. We calculate the determinant of $A=\left[\begin{array}{cc}2& 3\\ 3& 5\end{array}\right]$:

Since $|A|=1$, then $|A{|}^{2023}=1^{2023}=1$. The first half of our problem has vanished into thin air!

Now, for the second task: evaluating the quadratic matrix $M=A^2-3A+I$. First, we compute $A^2$ by multiplying $A$ with itself:

Next, we compute $3A$ by scalar multiplication:

Finally, we assemble the matrix $M=A^2-3A+I$:

The Final Victory

We are at the finish line. We need the determinant of $M=\left[\begin{array}{cc}8& 12\\ 12& 20\end{array}\right]$:

Combining our results, the total determinant is $|A{|}^{2023}\cdot |M|=1\cdot 16=16$.

Look at that. We started with a problem that seemed to require calculating powers up to $2025$, and we finished with a simple integer. This is the beauty of mathematics.

It is not about brute force; it is about finding the elegant path through the complexity. You have successfully navigated the trap, performed the matrix operations with precision, and arrived at the final answer of 16.