This problem requires a strong understanding of matrix properties, including symmetric matrices, determinants, matrix multiplication, and finding the inverse. A crucial aspect is setting up the matrix based on given properties and then using algebraic manipulation to determine its elements and subsequently solve for the desired scalar values.

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1. Key Concepts and Formulas

• Symmetric Matrix: A square matrix $A$ is symmetric if it is equal to its transpose, i.e., $A^T = A$. For a $2 \times 2$ matrix, if $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$. For $A$ to be symmetric, we must have $b=c$. Thus, a general $2 \times 2$ symmetric matrix is of the form $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$.
• Determinant of a $2 \times 2$ Matrix: For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant is given by $|A| = ad - bc$.
• Inverse of a $2 \times 2$ Matrix: For an invertible matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ (i.e., $|A| \neq 0$), its inverse is $A^{-1} = \frac{1}{|A|} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
• Matrix Equality: Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal.
• Cayley-Hamilton Theorem (Pro Tip): Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, the characteristic equation is $\lambda^2 - (\text{Tr}(A))\lambda + |A| = 0$. By the Cayley-Hamilton Theorem, $A^2 - (\text{Tr}(A))A + |A|I = 0$. If $|A| \neq 0$, we can multiply this equation by $A^{-1}$ to get $A - (\text{Tr}(A))I + |A|A^{-1} = 0$. This can be rearranged to express $A^{-1}$ in terms of $A$ and $I$: $A^{-1} = \frac{1}{|A|} (\text{Tr}(A)I - A)$. This provides a direct way to find $\alpha$ and $\beta$ once $A$ is known.

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2. Step-by-Step Solution

Step 1: Represent the symmetric matrix $A$ and use the determinant condition.

We are given that $A$ is a $2 \times 2$ symmetric matrix.

• Why? By definition of a symmetric matrix, its off-diagonal elements must be equal. This reduces the number of unknowns from four to three. Let $A$ be represented as: $A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ We are also given that the determinant of $A$ is $1$, i.e., $|A|=1$.
• Why? This condition provides an algebraic relationship between the elements $a, b, c$. Using the determinant formula for a $2 \times 2$ matrix: $|A| = ac - b^2$ So, we have our first equation: $ac - b^2 = 1 \quad \text{... (1)}$

Step 2: Use the given matrix multiplication to form linear equations.

We are provided with the matrix equation $A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]$.

• Why? Performing this matrix multiplication will relate the unknown elements of $A$ to the known components of the resulting vector, generating more equations. Substitute the matrix $A$ into the given equation: $\begin{pmatrix} a & b \\ b & c \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ Perform the matrix-vector multiplication: $\begin{pmatrix} a \cdot 1 + b \cdot 1 \\ b \cdot 1 + c \cdot 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ $\begin{pmatrix} a+b \\ b+c \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$
• Why? By the principle of matrix equality, corresponding elements of equal matrices (or vectors) must be equal. This gives us two linear equations: $a+b = 3 \quad \text{... (2)}$ $b+c = 7 \quad \text{... (3)}$

Step 3: Solve the system of equations to find the elements of $A$.

We now have a system of three equations with three unknowns ($a, b, c$):

1. $ac - b^2 = 1$
2. $a+b = 3$
3. $b+c = 7$

• Why? Our goal is to determine the specific values of $a, b, c$ to fully define matrix $A$. We can solve this system by substitution. From equation (2), express $a$ in terms of $b$: $a = 3-b$ From equation (3), express $c$ in terms of $b$: $c = 7-b$ Now, substitute these expressions for $a$ and $c$ into the quadratic equation (1): $(3-b)(7-b) - b^2 = 1$ Expand the product: $(21 - 3b - 7b + b^2) - b^2 = 1$ $21 - 10b + b^2 - b^2 = 1$ Notice that the $b^2$ terms cancel out, simplifying the equation significantly: $21 - 10b = 1$ Solve for $b$: $10b = 21 - 1$ $10b = 20$ $b = 2$ Now, substitute $b=2$ back into the expressions for $a$ and $c$: $a = 3 - b = 3 - 2 = 1$ $c = 7 - b = 7 - 2 = 5$ Thus, the elements of matrix $A$ are $a=1, b=2, c=5$.
• Common Mistake: Algebraic errors during expansion or substitution are common. Double-check your calculations, especially when quadratic terms cancel out.

Step 4: Construct matrix $A$ and find its inverse $A^{-1}$.

Using the determined values $a=1, b=2, c=5$, the matrix $A$ is: $A = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}$ Now, we find $A^{-1}$.

• Why? The problem asks for a relationship involving $A^{-1}$, so calculating it is a necessary step. Since we know $|A|=1$, the formula for $A^{-1}$ simplifies: $A^{-1} = \frac{1}{|A|} \begin{pmatrix} c & -b \\ -b & a \end{pmatrix} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix}$ $A^{-1} = \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix}$

Step 5: Determine $\alpha$ and $\beta$ using the given matrix equation $A^{-1}=\alpha A+\beta I$.

We are given the matrix equation $A^{-1} = \alpha A + \beta I$.

• Why? This equation allows us to equate the known $A^{-1}$ with an expression involving $A$ and $I$ to find the scalar coefficients $\alpha$ and $\beta$. Substitute $A$, $A^{-1}$, and the $2 \times 2$ identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$: $\begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} = \alpha \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix} + \beta \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ Perform scalar multiplication and matrix addition on the right-hand side: $\begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} \alpha & 2\alpha \\ 2\alpha & 5\alpha \end{pmatrix} + \begin{pmatrix} \beta & 0 \\ 0 & \beta \end{pmatrix}$ $\begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} \alpha+\beta & 2\alpha \\ 2\alpha & 5\alpha+\beta \end{pmatrix}$
• Why? By equating corresponding elements of the two matrices, we can form a system of linear equations for $\alpha$ and $\beta$. From the $(1,2)$ (or $(2,1)$) element: $2\alpha = -2 \implies \alpha = -1$ From the $(1,1)$ element: $\alpha+\beta = 5$ Substitute $\alpha = -1$ into this equation: $-1+\beta = 5 \implies \beta = 6$
• Verification: Always check consistency with other elements. For the $(2,2)$ element: $5\alpha+\beta = 5(-1)+6 = -5+6=1$. This matches the $(2,2)$ element of $A^{-1}$, confirming our values for $\alpha$ and $\beta$ are correct.

Step 6: Calculate $\alpha+\beta$.

Finally, we need to find the value of $\alpha+\beta$. $\alpha+\beta = -1 + 6 = 5$

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3. Summary and Key Takeaway

The problem demonstrates a comprehensive application of matrix algebra concepts.

1. We started by defining the matrix $A$ using its symmetric property.
2. We then used the given determinant and matrix multiplication conditions to set up a system of equations for the elements of $A$.
3. Solving this system allowed us to fully determine $A$.
4. Next, we calculated $A^{-1}$ using the standard formula.
5. Finally, by equating $A^{-1}$ to the given linear combination $\alpha A + \beta I$, we solved for $\alpha$ and $\beta$.
6. The result $\alpha+\beta = 5$.

An important takeaway is the utility of the Cayley-Hamilton Theorem for $2 \times 2$ matrices. Had we used it earlier, we would directly identify $\alpha = -1$ and $\beta = \text{Tr}(A)$. Since $\text{Tr}(A) = a+c = 1+5=6$, we would get $\alpha=-1$ and $\beta=6$ directly, leading to $\alpha+\beta = -1+6=5$. This method often saves time in competitive exams.