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.

Answer 2

Given matrix \( A \) is symmetric, \( 2 \times 2 \), and satisfies:
\[ A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \] and \( \det(A) = 1 \).
Also, \[ A^{-1} = \alpha A + \beta I \] where \( \alpha, \beta \in \mathbb{R} \) and \( I \) is the identity matrix.
Step 1: Find entries of \( A \)
Let, \[ A = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \] From, \[ A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a + b \\ b + c \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \] So, \[ a + b = 3, \quad b + c = 7 \]
Step 2: Use determinant condition
\[ \det A = ac - b^2 = 1 \]
From the equations, \[ a = 3 - b, \quad c = 7 - b \] Substitute into determinant, \[ (3 - b)(7 - b) - b^2 = 1 \] \[ 21 - 3b - 7b + b^2 - b^2 = 1 \] \[ 21 - 10b = 1 \implies 10b = 20 \implies b = 2 \] Using \( b = 2 \): \[ a = 3 - 2 = 1, \quad c = 7 - 2 = 5 \] So, \[ A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} \]
Step 3: Find \( A^{-1} \)
\[ A^{-1} = \frac{1}{\det A} \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} \]
Step 4: Use \( A^{-1} = \alpha A + \beta I \)
\[ \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \alpha \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\alpha \\ 2\alpha & 5\alpha + \beta \end{bmatrix} \] Equate entries:
\[ 5 = \alpha + \beta, \quad -2 = 2\alpha, \quad -2 = 2\alpha, \quad 1 = 5\alpha + \beta \] From \( -2 = 2\alpha \),
\[ \alpha = -1 \] Substitute in \( 5 = \alpha + \beta \): \[ 5 = -1 + \beta \implies \beta = 6 \] Check with \( 1 = 5\alpha + \beta \): \[ 1 = 5(-1) + 6 = -5 + 6 = 1 \] True, consistent.

Step 5: Find \( \alpha + \beta \)
\[ \alpha + \beta = -1 + 6 = 5 \]
Final answer: \( \alpha + \beta = 5 \)