Question

Let $A$ be a $2 \times 2$ symmetric matrix such that $A \begin{bmatrix} 1 \\\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\\ 7 \end{bmatrix}$ and the determinant of $A$ be 1. If $A^{-1} = \alpha A + \beta I$, where $I$ is the identity matrix of order $2 \times 2$, then $\alpha + \beta$ equals $\dots$.

Answer 1

Let

$A = \begin{bmatrix} a & b \\\ b & d \end{bmatrix}$.

From the given condition, we have:

$A \begin{bmatrix} 1 \\\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\\ 7 \end{bmatrix},$

which gives:

$a + b = 3, \quad b + d = 7$.

Also, the determinant of $A$ is given by $ad - b^2 = 1$.

Using $a + b = 3$ and $b + d = 7$, we can solve these equations. Let's set up the system:

1. From $a + b = 3$, we get $a = 3 - b$.
2. Substitute into $b + d = 7$ to find $d = 7 - b$.
3. Substitute $a = 3 - b$ and $d = 7 - b$ into $ad - b^2 = 1$: $(3 - b)(7 - b) - b^2 = 1.$

Expanding and simplifying, we get:

$21 - 10b + b^2 - b^2 = 1 \implies 21 - 10b = 1 \implies b = 2$.

Then, $a = 1$ and $d = 5$.

Thus, we have: $A = \begin{bmatrix} 1 & 2 \\\ 2 & 5 \end{bmatrix}.$

Now, we find $A^{-1}$:

$A^{-1} = \frac{1}{1(5) - (2)(2)} \begin{bmatrix} 5 & -2 \\\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\\ -2 & 1 \end{bmatrix}.$

Since $A^{-1} = \alpha A + \beta I$, we equate:

$\begin{bmatrix} 5 & -2 \\\ -2 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\alpha \\\ 2\alpha & 5\alpha + \beta \end{bmatrix}.$

This gives the system:

• $\alpha + \beta = 5$,
• $2\alpha = -2 \implies \alpha = -1$,
• $5\alpha + \beta = 1$.

Solving, we find $\beta = 6$.

Thus, $\alpha + \beta = 5$.

Answer: 5.