Matrix Equation Analysis

The problem provides a specific polynomial equation involving an $n \times n$ matrix $A$ and the $n \times n$ unit (or identity) matrix $I$. The given equation is:

\[A^3 - 3A^2 + 4A - 6I = 0\]

We are also given the crucial information that the inverse of matrix $A$, denoted as $A^{-1}$, exists. Our objective is to find an algebraic expression for this $A^{-1}$ using the provided matrix equation.

Deriving Matrix Inverse \(A^{-1}\)

Since the inverse matrix $A^{-1}$ is stated to exist, we can multiply every term in the given matrix equation by $A^{-1}$. When multiplying by the inverse, it's essential to recall fundamental properties of matrix multiplication:

• The product of a matrix and its inverse results in the identity matrix: $A^{-1}A = I$.
• Multiplying any matrix by the identity matrix yields the matrix itself: $IA = A$ and $AI = A$.
• The inverse matrix multiplied by the identity matrix gives the inverse matrix itself: $A^{-1}I = A^{-1}$.
• Multiplying any matrix by the zero matrix results in the zero matrix: $A^{-1}0 = 0$.

Let's proceed by multiplying the entire given equation by $A^{-1}$ from the left:

\[A^{-1}(A^3 - 3A^2 + 4A - 6I) = A^{-1}(0)\]

Now, distribute $A^{-1}$ to each term inside the parenthesis on the left side:

\[A^{-1}A^3 - A^{-1}(3A^2) + A^{-1}(4A) - A^{-1}(6I) = 0\]

Next, we simplify each term using the matrix properties mentioned above:

• The first term: $A^{-1}A^3 = (A^{-1}A)A^2 = IA^2 = A^2$.
• The second term: $-A^{-1}(3A^2) = -3(A^{-1}A)A = -3IA = -3A$.
• The third term: $A^{-1}(4A) = 4(A^{-1}A) = 4I$.
• The fourth term: $-A^{-1}(6I) = -6(A^{-1}I) = -6A^{-1}$.

Substitute these simplified terms back into the equation:

\[A^2 - 3A + 4I - 6A^{-1} = 0\]

Our objective is to find $A^{-1}$. To do this, we need to isolate the $A^{-1}$ term. Let's move the term $-6A^{-1}$ to the right side of the equation:

\[A^2 - 3A + 4I = 6A^{-1}\]

Finally, to express $A^{-1}$ explicitly, divide both sides of the equation by 6:

\[A^{-1} = \frac{1}{6}(A^2 - 3A + 4I)\]

Comparing with Options

Let's compare our derived expression for $A^{-1}$ with the given options to identify the correct match:

The expression we derived, $A^{-1} = \frac{1}{6}(A^2 - 3A + 4I)$, perfectly matches Option 4.