1. Key Concepts and Formulas

• Cayley-Hamilton Theorem: This fundamental theorem states that every square matrix satisfies its own characteristic polynomial equation. If $P(\lambda) = 0$ is the characteristic equation of a matrix $A$, then $P(A) = 0$ (where constant terms are replaced by constant times the identity matrix).
• Characteristic Equation: For a square matrix $A$, its characteristic equation is found by solving $\det(A - \lambda I) = 0$, where $\lambda$ is a scalar variable and $I$ is the identity matrix of the same order as $A$. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic equation is $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$.
• Matrix Inverse Properties: Key properties used when manipulating equations involving $A^{-1}$ include $A^{-1}A = I$, $A^{-1}I = A^{-1}$, and for a scalar $k$ and matrix $X$, $A^{-1}(kX) = k(A^{-1}X)$.

2. Step-by-Step Solution

Step 1: Form the matrix $(A - \lambda I)$ We are given the matrix $A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}$ and the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. To find the characteristic equation, we first construct the matrix $(A - \lambda I)$. This operation involves subtracting $\lambda$ times the identity matrix from $A$, which effectively subtracts $\lambda$ from each element on the main diagonal of $A$. $A - \lambda I = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 - \lambda & 2 \\ 9 & 4 - \lambda \end{pmatrix}$

Step 2: Calculate the determinant of $(A - \lambda I)$ and set it to zero The characteristic equation is defined as $\det(A - \lambda I) = 0$. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant is $ad - bc$. Applying this to our matrix $(A - \lambda I)$: $\det(A - \lambda I) = (2 - \lambda)(4 - \lambda) - (2)(9) = 0$

Step 3: Expand and simplify the equation to find the characteristic polynomial Now, we expand the product and combine the terms to form a quadratic polynomial in $\lambda$. $(2 - \lambda)(4 - \lambda) - 18 = 0$ $(8 - 2\lambda - 4\lambda + \lambda^2) - 18 = 0$ $\lambda^2 - 6\lambda + 8 - 18 = 0$ $\lambda^2 - 6\lambda + 10 = 0$ This is the characteristic equation of matrix $A$.

Step 4: Apply the Cayley-Hamilton Theorem According to the Cayley-Hamilton Theorem, matrix $A$ must satisfy its own characteristic equation. This means we can substitute $A$ for $\lambda$, $A^2$ for $\lambda^2$, and the constant term $10$ must be replaced by $10I$ (where $I$ is the $2 \times 2$ identity matrix) to maintain dimensional consistency within the matrix equation. Substituting $A$ into the characteristic equation $\lambda^2 - 6\lambda + 10 = 0$: $A^2 - 6A + 10I = 0$ This is a fundamental matrix equation derived from the given matrix $A$.

Step 5: Solve for $10A^{-1}$ Our objective is to find an expression for $10A^{-1}$. We can achieve this by manipulating the matrix equation we just derived. First, we rearrange the equation to isolate the term containing the identity matrix: $10I = 6A - A^2$ Next, we multiply the entire equation by $A^{-1}$. Since we are looking for $A^{-1}$, we must ensure $A$ is invertible. The determinant of $A$ is $(2)(4) - (2)(9) = 8 - 18 = -10 \neq 0$, so $A^{-1}$ exists. We'll multiply each term by $A^{-1}$ from the left. $A^{-1}(10I) = A^{-1}(6A - A^2)$ Now, we apply the properties of matrix inverses and multiplication:

• $A^{-1}(10I) = 10A^{-1}$
• $A^{-1}(6A) = 6(A^{-1}A) = 6I$
• $A^{-1}(A^2) = A^{-1}(AA) = (A^{-1}A)A = IA = A$ Substituting these simplified terms back into the equation: $10A^{-1} = 6I - A$ We have successfully derived the expression for $10A^{-1}$.

3. Common Mistakes & Tips

• Arithmetic Precision: Be extremely careful with arithmetic, especially when calculating the determinant and simplifying the characteristic equation. A single sign error can lead to an incorrect final answer.
• Constant Term Conversion: Always remember to convert the constant term $k$ in the characteristic polynomial to $kI$ when forming the matrix equation using the Cayley-Hamilton Theorem. This maintains matrix dimensional consistency.
• Matrix Multiplication Direction: While multiplying by $A^{-1}$, ensure you apply it consistently to all terms and from the same side (e.g., always pre-multiply or always post-multiply). For this problem, multiplying from either side yields the same result because $A$ commutes with $I$ and $A^2$ commutes with $A$.
• Understanding the Power: The Cayley-Hamilton Theorem allows us to find powers of matrices or their inverses without direct, lengthy computations, making it a very efficient tool for such problems.

4. Summary The problem asked us to find an expression for $10A^{-1}$ using the given matrix $A$. We employed the Cayley-Hamilton Theorem, which states that every matrix satisfies its own characteristic equation. We first calculated the characteristic equation $\det(A - \lambda I) = 0$. Then, we substituted $A$ into this equation, converting the constant term to a multiple of the identity matrix. Finally, by multiplying the resulting matrix equation by $A^{-1}$ and simplifying, we successfully isolated $10A^{-1}$ to find it equals $6I - A$.

5. Final Answer The final answer is $\boxed{\text{6I - A}}$, which corresponds to option (A).