Key Concepts and Formulas

1. Matrix Inverse Properties: For an invertible matrix $M$, $M M^{-1} = M^{-1} M = I$, where $I$ is the identity matrix. This is used extensively for simplifying matrix equations.
2. Similar Matrices: If a matrix $C$ can be expressed as $P^{-1}AP$ for some invertible matrix $P$, then $C$ is similar to $A$. A powerful property of similar matrices is that $C^n = P^{-1}A^n P$. This simplifies calculations of powers of $C$.
3. Cayley-Hamilton Theorem (for $2 \times 2$ matrices): Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $M$, its characteristic equation is $\det(M - \lambda I) = 0$, which simplifies to $\lambda^2 - \text{tr}(M)\lambda + \det(M) = 0$. By Cayley-Hamilton, $M^2 - \text{tr}(M)M + \det(M)I = O$, where $\text{tr}(M)$ is the trace (sum of diagonal elements) and $\det(M)$ is the determinant of $M$.

Step-by-Step Solution

Step 1: Simplify the relationship between $A$ and $B$.

We are given the equation $A B^{-1}=A^{-1}$. Our goal is to find a direct relationship between $A$ and $B$.

• Given: $A B^{-1}=A^{-1}$
• Action: To eliminate $A^{-1}$ from the right side, we multiply both sides by $A$ from the left. $A (A B^{-1}) = A (A^{-1})$
• Explanation: Using the associative property of matrix multiplication and the definition of an inverse ($A A^{-1} = I$), we simplify the equation. $A^2 B^{-1} = I$
• Action: To isolate $A^2$, we multiply both sides by $B$ from the right. $(A^2 B^{-1}) B = I B$
• Explanation: Again, using associativity and the inverse property ($B^{-1} B = I$), we get: $A^2 (B^{-1} B) = B$ $A^2 I = B$ $\boxed{A^2 = B}$ This fundamental relationship is crucial for subsequent steps.

Step 2: Express matrix $C$ in terms of $A$ and $B$.

We are given $B C B^{-1}=A$. We want to isolate $C$ to work with its powers.

• Given: $B C B^{-1}=A$
• Action: To isolate $C$, we need to remove $B$ from its left and $B^{-1}$ from its right. First, multiply both sides by $B^{-1}$ from the left. $B^{-1} (B C B^{-1}) = B^{-1} A$
• Explanation: Using $B^{-1} B = I$, the expression simplifies to: $(B^{-1} B) C B^{-1} = B^{-1} A$ $I C B^{-1} = B^{-1} A$ $C B^{-1} = B^{-1} A$
• Action: Next, multiply both sides by $B$ from the right. $(C B^{-1}) B = (B^{-1} A) B$
• Explanation: Using $B^{-1} B = I$ again, we get: $C (B^{-1} B) = B^{-1} A B$ $C I = B^{-1} A B$ $\boxed{C = B^{-1} A B}$ This shows that $C$ is similar to $A$, which is a key insight for efficiently calculating powers of $C$.

Step 3: Calculate $C^2$ and $C^4$ in terms of $B$.

Using the similar matrix form of $C$ and the relationship $A^2=B$, we can simplify $C^2$ and $C^4$.

• Calculate $C^2$: $C^2 = (B^{-1} A B)(B^{-1} A B)$

• Explanation: We regroup terms using associativity, noting that $B B^{-1} = I$. $C^2 = B^{-1} A (B B^{-1}) A B$ $C^2 = B^{-1} A I A B$ $C^2 = B^{-1} A^2 B$

• Action: Now, substitute $A^2=B$ (from Step 1). $C^2 = B^{-1} (B) B$

• Explanation: Using $B^{-1} B = I$, we get: $C^2 = (B^{-1} B) B$ $C^2 = I B$ $\boxed{C^2 = B}$

• Calculate $C^4$: We can calculate $C^4$ as $(C^2)^2$. $C^4 = (C^2)^2$

• Action: Substitute $C^2=B$. $C^4 = B^2$ $\boxed{C^4 = B^2}$

Step 4: Substitute $C^2$ and $C^4$ into the polynomial equation.

We are given the matrix polynomial equation $C^4+\alpha C^2+\beta I=O$.

• Given: $C^4+\alpha C^2+\beta I=O$
• Action: Substitute the simplified expressions $C^4=B^2$ and $C^2=B$. $B^2 + \alpha B + \beta I = O$
• Explanation: This transforms the equation into one involving only matrix $B$ and the scalar coefficients $\alpha, \beta$. Note that $\beta$ must be multiplied by the identity matrix $I$ for dimensional consistency when adding to other matrices, and $O$ represents the zero matrix.

Step 5: Use the Cayley-Hamilton Theorem to find $\alpha$ and $\beta$.

We have the matrix $B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$. The Cayley-Hamilton theorem states that every $2 \times 2$ matrix $M$ satisfies $M^2 - \text{tr}(M)M + \det(M)I = O$. We will apply this to matrix $B$.

• Calculate Trace of $B$ ($\text{tr}(B)$): The sum of the diagonal elements. $\text{tr}(B) = 1 + 5 = 6$
• Calculate Determinant of $B$ ($\det(B)$): For a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, $\det = ad-bc$. $\det(B) = (1)(5) - (3)(1) = 5 - 3 = 2$
• Apply Cayley-Hamilton Theorem for $B$: $B^2 - \text{tr}(B)B + \det(B)I = O$ $B^2 - 6B + 2I = O$
• Compare with the derived polynomial equation: We have $B^2 + \alpha B + \beta I = O$. Comparing this with $B^2 - 6B + 2I = O$, we can directly identify the coefficients: $\boxed{\alpha = -6}$ $\boxed{\beta = 2}$

Step 6: Calculate $2\beta-\alpha$.

Finally, substitute the values of $\alpha$ and $\beta$ into the expression $2\beta-\alpha$.

• Calculation: $2\beta-\alpha = 2(2) - (-6)$ $2\beta-\alpha = 4 + 6$ $\boxed{2\beta-\alpha = 10}$

Common Mistakes & Tips

• Order of Matrix Multiplication: Remember that matrix multiplication is not commutative ($AB \neq BA$ in general). The order of multiplication (left vs. right) is critical when manipulating matrix equations.
• Identity Matrix for Scalars: When adding or equating matrix expressions, scalar terms like $\beta$ must always be multiplied by the identity matrix $I$ (e.g., $\beta I$) to maintain dimensional consistency.
• Leverage Similar Matrices: Recognizing $C=P^{-1}AP$ simplifies powers of $C$ significantly to $C^n=P^{-1}A^n P$, avoiding tedious direct calculation of $C$.
• Cayley-Hamilton Theorem: For $2 \times 2$ matrices, this theorem is an incredibly powerful shortcut for finding polynomial relationships (like $B^2 + \alpha B + \beta I = O$) without calculating $B^2$ explicitly and solving a system of linear equations from its elements.

Summary

The problem started by establishing key relationships between matrices $A$, $B$, and $C$. We first simplified $A B^{-1}=A^{-1}$ to obtain $A^2=B$. Then, we expressed $C$ in terms of $A$ and $B$ as $C=B^{-1}AB$, showing that $C$ is similar to $A$. This allowed us to efficiently calculate $C^2 = B^{-1}A^2B = B^{-1}BB = B$ and subsequently $C^4 = (C^2)^2 = B^2$. Substituting these into the given polynomial equation $C^4+\alpha C^2+\beta I=O$ yielded $B^2+\alpha B+\beta I=O$. Finally, by applying the Cayley-Hamilton theorem to matrix $B$, we found its characteristic polynomial $B^2 - 6B + 2I = O$, from which we directly deduced $\alpha=-6$ and $\beta=2$. The desired expression $2\beta-\alpha$ was then calculated as $2(2) - (-6) = 10$.

The final answer is $\boxed{10}$, which corresponds to option (A).