Animated Solution for Mathematics - Matrices and Determinants: If $A^2-A+I=0$, then the inverse of $A$ is

Imagine you are standing at the threshold of a new dimension in mathematics. You have encountered the equation $A^2-A+I=0$.

At first glance, it looks like a simple quadratic equation, but you are dealing with matrices—the architects of linear transformations. This is not just arithmetic; it is the language of how space and vectors shift, rotate, and scale.

The Power of Pre-multiplication

In standard algebra, if you had $x^2-x+1=0$, you might reach for the quadratic formula. But here, we have a matrix $A$. We cannot simply 'divide' by $A$ because matrix division is not defined.

Instead, we use a surgical tool: pre-multiplication. We want to introduce $A^{-1}$ into the equation. By multiplying the entire equation $A^2-A+I=0$ from the left by $A^{-1}$, we are essentially applying the 'undo' transformation to the entire system.

We write:

The right side is easy; any matrix multiplied by the zero matrix remains the zero matrix. The left side, however, is where the magic happens.

Distributing the Inverse

Matrix multiplication is distributive. This means we can distribute $A^{-1}$ to every term inside the parentheses:

Now, let us look at each term with the eye of a master. The first term, $A^{-1}{A}^2$, is $A^{-1}(AA)$. By the associative property, this is $(A^{-1}A)A$, which simplifies to $IA$, and finally to $A$.

The second term, $A^{-1}A$, is the very definition of the identity matrix $I$. The third term, $A^{-1}I$, is simply $A^{-1}$, because the identity matrix is the 'unity' of the matrix world—it leaves any matrix unchanged.

Isolating the Inverse

Substituting these back, we get:

We are almost at the finish line. We want to isolate $A^{-1}$. With a simple rearrangement, we move $A$ and $-I$ to the other side:

It is beautiful, isn't it? The inverse of $A$ is simply the identity matrix minus $A$ itself. This result is not just a solution; it is a testament to the internal consistency of linear algebra.

We can even verify it:

Since $A^2-A+I=0$, it follows that $A-A^2=I$. The circle is complete. You have successfully navigated the matrix landscape. Keep this intuition—that matrices are transformations—and you will find that no equation is too daunting to solve.