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

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.