Understanding the Problem

We are given a $2 \times 2$ matrix $A$ and a relationship involving its inverse, $A^{-1} = \alpha I + \beta A$, where $I$ is the $2 \times 2$ identity matrix and $\alpha, \beta$ are real numbers. Our goal is to find the value of the expression $4(\alpha - \beta)$. This problem tests our ability to compute matrix inverses, perform matrix algebra (scalar multiplication and addition), and solve systems of linear equations derived from matrix equality.

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Key Concepts and Formulas

1. Inverse of a $2 \times 2$ Matrix: For a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its inverse $A^{-1}$ exists if and only if its determinant $|A| = ad - bc \neq 0$. The formula for the inverse is: $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ The $\text{adj}(A)$ (adjoint of $A$) for a $2 \times 2$ matrix is found by swapping the diagonal elements and negating the off-diagonal elements.

2. Matrix Scalar Multiplication: If $k$ is a scalar and $A = [a_{ij}]$ is a matrix, then $kA = [ka_{ij}]$. Each element of the matrix is multiplied by the scalar.

3. Matrix Addition: If $A = [a_{ij}]$ and $B = [b_{ij}]$ are matrices of the same dimension, then $A+B = [a_{ij} + b_{ij}]$. Corresponding elements are added.

4. Matrix Equality: Two matrices $X$ and $Y$ are equal if and only if they have the same dimensions and their corresponding elements are equal. If $X = Y$, then $x_{ij} = y_{ij}$ for all $i, j$. This property allows us to form a system of linear equations.

5. Identity Matrix: For a $2 \times 2$ matrix, the identity matrix is $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. It acts like the number '1' in scalar multiplication, i.e., $AI = IA = A$.

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Step-by-Step Solution

Given the matrix $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and the relation $A^{-1} = \alpha I + \beta A$.

Step 1: Calculate the Inverse of Matrix A ($A^{-1}$)

To find $A^{-1}$, we first need its determinant and adjoint.

• Calculate the Determinant of A ($|A|$): The determinant tells us if the inverse exists and is a crucial part of the inverse formula. $|A| = (1)(4) - (2)(-1)$ $|A| = 4 - (-2)$ $|A| = 4 + 2$ $|A| = 6$ Since $|A| = 6 \neq 0$, the inverse $A^{-1}$ exists.

• Calculate the Adjoint of A ($\text{adj}(A)$): For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. For $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$, we have $a=1, b=2, c=-1, d=4$. $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -(-1) & 1 \end{bmatrix}$ $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$

• Find $A^{-1}$: Now, use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$: $A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$ Distribute the scalar $\frac{1}{6}$ to each element to get the explicit inverse matrix: $A^{-1} = \begin{bmatrix} \frac{4}{6} & \frac{-2}{6} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$ This is the left-hand side of our given matrix equation.

Step 2: Express the Right-Hand Side ($\alpha I + \beta A$) as a Single Matrix

We are given $A^{-1} = \alpha I + \beta A$. Let's construct the right-hand side using the given matrices and scalars $\alpha, \beta$. Recall $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$.

• Perform Scalar Multiplication: $\alpha I = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix}$ $\beta A = \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \beta & 2\beta \\ -\beta & 4\beta \end{bmatrix}$

• Perform Matrix Addition: Add the two resulting matrices: $\alpha I + \beta A = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} + \begin{bmatrix} \beta & 2\beta \\ -\beta & 4\beta \end{bmatrix}$ $\alpha I + \beta A = \begin{bmatrix} \alpha + \beta & 0 + 2\beta \\ 0 + (-\beta) & \alpha + 4\beta \end{bmatrix}$ $\alpha I + \beta A = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}$ This is the right-hand side of our matrix equation, expressed as a single matrix.

Step 3: Equate Corresponding Elements to Form a System of Equations

Now we equate the calculated $A^{-1}$ from Step 1 with the expression for $\alpha I + \beta A$ from Step 2: $\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}$ By the principle of matrix equality, corresponding elements in the same position must be equal. This gives us a system of four linear equations:

1. $\alpha + \beta = \frac{2}{3}$ (from the $(1,1)$ element)
2. $2\beta = -\frac{1}{3}$ (from the $(1,2)$ element)
3. $-\beta = \frac{1}{6}$ (from the $(2,1)$ element)
4. $\alpha + 4\beta = \frac{1}{6}$ (from the $(2,2)$ element)

Step 4: Solve the System of Equations for $\alpha$ and $\beta$

We can solve for $\beta$ using either equation (2) or (3), as they are simpler. From equation (2): $2\beta = -\frac{1}{3}$ $\beta = -\frac{1}{3} \times \frac{1}{2}$ $\beta = -\frac{1}{6}$

We can verify this with equation (3): $-\beta = \frac{1}{6}$ $\beta = -\frac{1}{6}$ Both equations consistently give $\beta = -\frac{1}{6}$.

Now, substitute $\beta = -\frac{1}{6}$ into equation (1) to find $\alpha$: $\alpha + \left(-\frac{1}{6}\right) = \frac{2}{3}$ $\alpha - \frac{1}{6} = \frac{2}{3}$ $\alpha = \frac{2}{3} + \frac{1}{6}$ To add these fractions, find a common denominator (which is 6): $\alpha = \frac{2 \times 2}{3 \times 2} + \frac{1}{6}$ $\alpha = \frac{4}{6} + \frac{1}{6}$ $\alpha = \frac{5}{6}$

It's good practice to verify these values with the remaining equation (4): $\alpha + 4\beta = \frac{5}{6} + 4\left(-\frac{1}{6}\right)$ $= \frac{5}{6} - \frac{4}{6}$ $= \frac{1}{6}$ This matches the $(2,2)$ element of $A^{-1}$, confirming our values for $\alpha = \frac{5}{6}$ and $\beta = -\frac{1}{6}$ are correct.

Step 5: Calculate the Required Expression $4(\alpha - \beta)$

Finally, substitute the determined values of $\alpha$ and $\beta$ into the expression $4(\alpha - \beta)$: First, calculate $\alpha - \beta$: $\alpha - \beta = \frac{5}{6} - \left(-\frac{1}{6}\right)$ $\alpha - \beta = \frac{5}{6} + \frac{1}{6}$ $\alpha - \beta = \frac{6}{6}$ $\alpha - \beta = 1$

Now, multiply by 4: $4(\alpha - \beta) = 4(1)$ $4(\alpha - \beta) = 4$

The final answer is 4.

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Tips and Common Mistakes

• Sign Errors: Be extremely careful with negative signs, especially when calculating the determinant ($ad-bc$) and the adjoint (negating off-diagonal elements). A single sign error can propagate through the entire solution.
• Fraction Arithmetic: Ensure accuracy when adding, subtracting, or multiplying fractions. Find common denominators carefully.
• Consistency Check: After finding $\alpha$ and $\beta$ from a subset of the equations, always use the remaining equations to verify your values. This helps catch errors early.
• Alternative Method (Cayley-Hamilton Theorem): For problems of the form $A^{-1} = \alpha I + \beta A$, the Cayley-Hamilton theorem can often provide a much faster route for $2 \times 2$ matrices. The theorem states that every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, the characteristic equation is $\det(A - \lambda I) = 0$, which simplifies to $\lambda^2 - (\text{trace}(A))\lambda + \det(A) = 0$. So, $A^2 - (\text{trace}(A))A + \det(A)I = 0$. For $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$: Trace$(A) = 1+4 = 5$ $\det(A) = 6$ (calculated earlier) So, $A^2 - 5A + 6I = 0$. To find $A^{-1}$, multiply the equation by $A^{-1}$ (assuming $A^{-1}$ exists, which it does since $\det(A) \neq 0$): $A^{-1}(A^2 - 5A + 6I) = A^{-1}(0)$ $A - 5I + 6A^{-1} = 0$ $6A^{-1} = 5I - A$ $A^{-1} = \frac{5}{6}I - \frac{1}{6}A$ Comparing this with $A^{-1} = \alpha I + \beta A$, we directly get $\alpha = \frac{5}{6}$ and $\beta = -\frac{1}{6}$. This method bypasses the explicit calculation of $A^{-1}$ and the system of equations, making it very efficient for this type of problem.

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Summary and Key Takeaway

This problem demonstrates a standard procedure for working with matrix inverses and matrix equations. We first computed the inverse of the given matrix $A$. Then, we expressed the right-hand side of the given relation in terms of $\alpha$, $\beta$, and the matrices. By equating the corresponding elements of the resulting matrices, we formed a system of linear equations which we solved to find $\alpha$ and $\beta$. Finally, we substituted these values into the expression $4(\alpha - \beta)$ to get the answer. The Cayley-Hamilton theorem offers a powerful shortcut for similar problems, especially for $2 \times 2$ matrices.

The final answer is $\boxed{4}$.