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

Correct Answer: 5

Given matrix \( A = \begin{bmatrix} a & b \\ b & d \end{bmatrix} \).

The condition \( A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \) implies \( a + b = 3 \) and \( b + d = 7 \).

Additionally, the determinant of \( A \) is \( ad - b^2 = 1 \).

We solve the system of equations:

1. From \( a + b = 3 \), derive \( a = 3 - b \).
2. From \( b + d = 7 \), derive \( d = 7 - b \).
3. Substitute \( a \) and \( d \) into the determinant equation: \( (3 - b)(7 - b) - b^2 = 1 \).

Simplifying the equation \( 21 - 10b + b^2 - b^2 = 1 \) yields \( 21 - 10b = 1 \), so \( b = 2 \).

Substituting \( b = 2 \) gives \( a = 1 \) and \( d = 5 \).

Therefore, \( A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} \).

The inverse of \( A \) is calculated as:

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

We are given \( A^{-1} = \alpha A + \beta I \). Equating this with the calculated inverse yields:

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

This results in the system of equations:

• \( \alpha + \beta = 5 \)
• \( 2\alpha = -2 \implies \alpha = -1 \)
• \( 5\alpha + \beta = 1 \)

Solving these equations, we find \( \beta = 6 \).

The sum \( \alpha + \beta \) equals \( -1 + 6 = 5 \).

Answer: 5.