Key Concept: Definition and Properties of Matrix Inverse

The fundamental concept here is the definition of a matrix inverse. For a square matrix $A$, its inverse, denoted as $A^{-1}$, is a matrix such that when multiplied by $A$, it yields the identity matrix $I$: $AA^{-1} = A^{-1}A = I$ The identity matrix $I$ acts like the number $1$ in scalar multiplication, meaning $AI = IA = A$ for any matrix $A$. The zero matrix $0$ acts like the number $0$, such that $A \cdot 0 = 0 \cdot A = 0$. Our goal is to manipulate the given matrix equation to express $A^{-1}$ in terms of $A$ and $I$.

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Prerequisite: Verifying the Existence of the Inverse

Before attempting to find the inverse, it's crucial to ensure that $A^{-1}$ actually exists. A square matrix $A$ has an inverse if and only if its determinant is non-zero (i.e., $\det(A) \neq 0$).

Let's use the given equation to confirm this: Given: $A^2 - A + I = 0$ Step 1: Rearrange the equation to factor out $A$. To determine if $\det(A) \neq 0$, we can try to express the identity matrix in terms of $A$ and another matrix. $A^2 - A = -I$ Factor out $A$ from the left side. Remember that $A$ can be factored as $A \cdot I$ to maintain matrix dimensions: $A(A - I) = -I$

Step 2: Take the determinant of both sides. For any square matrices $M$ and $N$ of the same size, $\det(MN) = \det(M)\det(N)$. Also, for an $n \times n$ identity matrix $I$, $\det(I) = 1$. For a scalar $k$ and an $n \times n$ matrix $M$, $\det(kM) = k^n \det(M)$. Here, $-I$ is equivalent to $(-1)I$. $\det(A(A - I)) = \det(-I)$ $\det(A) \det(A - I) = (-1)^n \det(I)$ $\det(A) \det(A - I) = (-1)^n \cdot 1$ $\det(A) \det(A - I) = (-1)^n$

Step 3: Conclude invertibility. Since the right-hand side, $(-1)^n$, is either $1$ or $-1$ (and therefore never zero), it implies that neither $\det(A)$ nor $\det(A - I)$ can be zero. Therefore, $\det(A) \neq 0$, which confirms that the inverse of matrix $A$, denoted as $A^{-1}$, exists.

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Step-by-Step Solution: Finding the Inverse of A

We are given the matrix equation $A^2 - A + I = 0$, and our goal is to find an expression for $A^{-1}$.

Step 1: Start with the given matrix equation. The foundation of our solution is the provided equation: $A^2 - A + I = 0$ Why this step? This is the starting point from which we will derive the expression for $A^{-1}$.

Step 2: Multiply the entire equation by $A^{-1}$. Since we've established that $A^{-1}$ exists, we can multiply both sides of the equation by $A^{-1}$. In matrix algebra, it's important to be consistent with the side of multiplication (left or right). For polynomial equations like this, where $A$ commutes with $A^{-1}$ and $I$, multiplying from either side yields the same result. Let's multiply from the left: $A^{-1}(A^2 - A + I) = A^{-1}(0)$ Why this step? Our objective is to find $A^{-1}$. Multiplying by $A^{-1}$ introduces it into the equation and allows us to use the definition $A^{-1}A = I$ to simplify terms.

Step 3: Distribute $A^{-1}$ and simplify the right side. Now, distribute $A^{-1}$ to each term inside the parenthesis on the left side. The product of any matrix with the zero matrix is the zero matrix itself. $A^{-1}A^2 - A^{-1}A + A^{-1}I = 0$ Why this step? This is an application of the distributive property of matrix multiplication, which allows us to break down the equation into simpler terms.

Step 4: Simplify each term using the definition of inverse and identity matrices. Let's simplify each term on the left-hand side using the properties $A^{-1}A = I$ and $AI = IA = A$:

• Term 1: $A^{-1}A^2$ We can write $A^2$ as $A \cdot A$. So, $A^{-1}A^2 = A^{-1}(A \cdot A)$. Using the associative property of matrix multiplication, this becomes $(A^{-1}A)A$. By the definition of an inverse matrix, $A^{-1}A = I$. Therefore, $(A^{-1}A)A = I \cdot A$. Any matrix multiplied by the identity matrix remains itself: $I \cdot A = A$. So, $A^{-1}A^2 = A$.

• Term 2: $A^{-1}A$ By the definition of an inverse matrix, $A^{-1}A = I$.

• Term 3: $A^{-1}I$ Any matrix multiplied by the identity matrix remains itself: $A^{-1}I = A^{-1}$.

Substitute these simplified terms back into the equation from Step 3: $A - I + A^{-1} = 0$ Why this step? This is the core simplification step where we apply the fundamental definitions of matrix inverse and identity to transform the equation into a form where $A^{-1}$ is present and can be isolated.

Step 5: Isolate $A^{-1}$. Finally, rearrange the equation to solve for $A^{-1}$: $A^{-1} = I - A$ Why this step? This is the final algebraic manipulation to express $A^{-1}$ explicitly, thereby providing the answer to the problem.

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Important Tips and Common Mistakes:

• Existence of Inverse: Always confirm that $A^{-1}$ exists before multiplying by it. While often implicitly assumed in multiple-choice questions, it's a critical prerequisite for the operation.
• Matrix vs. Scalar: Be extremely careful not to confuse the identity matrix $I$ with the scalar $1$. While $I$ acts like $1$ in matrix multiplication ($AI=IA=A$), it is a matrix, and operations like $A-I$ are matrix subtractions, not scalar subtractions.
• Order of Multiplication: In general matrix algebra, the order of multiplication matters ($AB \neq BA$). While $A^{-1}$ commutes with $A$ and $I$ in this context, it's good practice to be mindful of left vs. right multiplication for other matrix equations.
• Zero Matrix: Remember that $A^{-1}(0) = 0$, where $0$ represents the zero matrix of appropriate dimensions.

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Summary and Key Takeaway:

By systematically applying the definition of a matrix inverse ($A^{-1}A = I$) and the properties of the identity matrix ($IA = A$, $A^{-1}I = A^{-1}$), we can manipulate a given polynomial matrix equation to solve for the inverse.

For the given equation $A^2 - A + I = 0$, the inverse of $A$ is found to be: $A^{-1} = I - A$