Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}1& 2\\ -1& 4\end{array}\right]$. If $A^{-1}=\alpha I+\beta A,\alpha ,\beta \in \mathbb{R},I$ is a $2\times 2$ identity matrix, then $4(\alpha -\beta )$ is equal to :

A Journey into Cayley-Hamilton

Welcome, my dear student. Today, we are not just solving a matrix problem; we are uncovering the hidden structure of linear transformations. When you look at a matrix like $A=\left[\begin{array}{cc}1& 2\\ -1& 4\end{array}\right]$, what do you see?

Do you see a grid of numbers, or do you see a transformation waiting to be understood? Many students approach this problem by immediately reaching for the adjoint formula to find $A^{-1}$. While that is a valid path, it is the path of the laborer. We, however, are aiming for the path of the architect.

We are given the equation $A^{-1}=\alpha I+\beta A$. The temptation is to calculate $A^{-1}$ directly, find the determinant, find the adjoint, and then solve for $\alpha$ and $\beta$. But stop!

In the heat of a JEE Advanced exam, every second counts. If you calculate the inverse directly, you are doing unnecessary work. The problem is practically screaming at you to use the relationship between the matrix and its characteristic polynomial.

The Matrix's Signature

Every square matrix has a 'signature'—a characteristic polynomial that defines its eigenvalues. We find this by solving the determinant equation:

Let us substitute our matrix $A$ into this expression. We are subtracting $\lambda$ from the diagonal elements:

Now, we expand this $2\times 2$ determinant using the rule $(ad-bc)$:

Let us expand this carefully. I want you to be meticulous here; a single sign error can derail the entire process. Expanding the brackets, we get:

Combining the like terms, we arrive at the characteristic polynomial:

This quadratic equation is the DNA of our matrix. It tells us how the matrix behaves.

The Bridge

Here is the beautiful, powerful theorem that will save you minutes of calculation: The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. This means we can replace the scalar $\lambda$ with the matrix $A$, and the constant term $6$ with $6I$ (where $I$ is the identity matrix).

Thus, we can write:

Look at this equation. It is a direct link between $A^2$, $A$, and $I$. We are so close to finding $A^{-1}$!

We need an expression for $A^{-1}$. How do we get it from $A^2-5A+6I=0$? We multiply the entire equation by $A^{-1}$:

Distributing $A^{-1}$ across the terms, we get:

Since $A^{-1}A=I$ and $A^{-1}I=A^{-1}$, this simplifies beautifully to:

Now, let us isolate $6A^{-1}$:

Dividing by $6$, we finally have the form we were looking for:

Comparing this to our target equation $A^{-1}=\alpha I+\beta A$, we can immediately identify our constants:

The problem asks us to calculate $4(\alpha -\beta )$. Let us plug in our values:

And there we have it! The final answer is 4.