Question:

To find the values of $\alpha$ and $\beta$ such that $A^{-1} = \alpha A + \beta I$, we first need to determine the matrix $A$. Step 1: Let $A = \begin{bmatrix}a & b \\ b & d\end{bmatrix}$ since $A$ is a $2 \times 2$ symmetric matrix. Given that $A\begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}3 \\ 7\end{bmatrix}$, we can write the equation as: $$\begin{bmatrix}a & b \\ b & d\end{bmatrix}\begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}3 \\ 7\end{bmatrix}$$ This simplifies to: $$\begin{bmatrix}a + b \\ b + d\end{bmatrix} = \begin{bmatrix}3 \\ 7\end{bmatrix}$$ So, we have two equations: $$a + b = 3$$ $$b + d = 7$$ Step 2: The determinant of $A$ is given as 1. Therefore, we have: $$ad - b^2 = 1$$ Step 3: From the equations $a + b = 3$ and $b + d = 7$, we can solve for $a$ and $d$ in terms of $b$: $$a = 3 - b$$ $$d = 7 - b$$ Substituting these into the determinant equation gives: $$(3 - b)(7 - b) - b^2 = 1$$ Expanding this, we get: $$21 - 3b - 7b + b^2 - b^2 = 1$$ Simplifying: $$21 - 10b = 1$$ $$10b = 20$$ $$b = 2$$ Step 4: With $b = 2$, we can find $a$ and $d$: $$a = 3 - b = 3 - 2 = 1$$ $$d = 7 - b = 7 - 2 = 5$$ So, the matrix $A$ is: $$A = \begin{bmatrix}1 & 2 \\ 2 & 5\end{bmatrix}$$ Step 5: To find $A^{-1}$, we first find the determinant of $A$, which we already know is 1. The inverse of $A$ is given by: $$A^{-1} = \frac{1}{\text{det}(A)}\begin{bmatrix}d & -b \\ -b & a\end{bmatrix} = \begin{bmatrix}5 & -2 \\ -2 & 1\end{bmatrix}$$ Step 6: Given $A^{-1} = \alpha A + \beta I$, we substitute $A$ and $A^{-1}$: $$\begin{bmatrix}5 & -2 \\ -2 & 1\end{bmatrix} = \alpha\begin{bmatrix}1 & 2 \\ 2 & 5\end{bmatrix} + \beta\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ This gives us the system of equations: $$\alpha + \beta = 5$$ $$2\alpha = -2$$ $$2\alpha = -2$$ $$5\alpha + \beta = 1$$ Step 7: Solving the equation $2\alpha = -2$ gives: $$\alpha = -1$$ Substituting $\alpha = -1$ into $\alpha + \beta = 5$: $$-1 + \beta = 5$$ $$\beta = 6$$ Therefore: $\alpha + \beta = -1 + 6 = 5$ <div class="key-concept"><strong>Key Concept:</strong> Let $A=\begin{bmatrix}a&b\\b&d\end{bmatrix}$. From $A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$: $a+b=3$, $b+d=7$. $\det A=ad-b^2=1$. Solve for $a,b,d$.</div> <div class="trap-box"><strong>Trap:</strong> $(3-b)(7-b)-b^2=1\Rightarrow21-10b=1\Rightarrow b=2$, $a=1$, $d=5$. $A^{-1}=\begin{bmatrix}5&-2\\-2&1\end{bmatrix}=\alpha A+\beta I\Rightarrow\alpha=-1$, $\beta=6$. $\alpha+\beta=5$.</div>