Let `A=[(1,2),(-2,-5)] , alpha & beta` belongs to real numbers such that `alpha A^2+beta A+2I`, where I is an identity matrix of order `2xx2`. Then the value of `alpha+beta` is equal to :

Options

A. -10
B. -6
C. 6
D. 10

Answer

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) such that the equation \( \alpha A^2 + \beta A + 2I = 0 \) holds, where \( A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \) and \( I \) is the identity matrix of order \( 2 \times 2 \).

### Step 1: Calculate \( A^2 \)

First, we need to compute \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- The element at (1,1):
\[ 1 \cdot 1 + 2 \cdot (-2) = 1 - 4 = -3 \]

- The element at (1,2):
\[ 1 \cdot 2 + 2 \cdot (-5) = 2 - 10 = -8 \]

- The element at (2,1):
\[ -2 \cdot 1 + (-5) \cdot (-2) = -2 + 10 = 8 \]

- The element at (2,2):
\[ -2 \cdot 2 + (-5) \cdot (-5) = -4 + 25 = 21 \]

Thus, we have:

\[ A^2 = \begin{pmatrix} -3 & -8 \\ 8 & 21 \end{pmatrix} \]

### Step 2: Substitute \( A \) and \( A^2 \) into the equation

Now we substitute \( A \), \( A^2 \), and \( I \) into the equation \( \alpha A^2 + \beta A + 2I = 0 \):

\[ \alpha \begin{pmatrix} -3 & -8 \\ 8 & 21 \end{pmatrix} + \beta \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 0 \]

This expands to:

\[ \begin{pmatrix} -3\alpha + \beta + 2 & -8\alpha + 2\beta \\ 8\alpha - 2\beta & 21\alpha - 5\beta + 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 3: Set up the system of equations

From the above matrix equation, we can set up the following system of equations:

1. \( -3\alpha + \beta + 2 = 0 \)
2. \( -8\alpha + 2\beta = 0 \)
3. \( 8\alpha - 2\beta = 0 \)
4. \( 21\alpha - 5\beta + 2 = 0 \)

### Step 4: Solve the equations

From equation (2) \( -8\alpha + 2\beta = 0 \):

\[ 2\beta = 8\alpha \implies \beta = 4\alpha \]

Substituting \( \beta = 4\alpha \) into equation (1):

\[ -3\alpha + 4\alpha + 2 = 0 \implies \alpha + 2 = 0 \implies \alpha = -2 \]

Now substituting \( \alpha = -2 \) back into \( \beta = 4\alpha \):

\[ \beta = 4(-2) = -8 \]

### Step 5: Find \( \alpha + \beta \)

Now we can find \( \alpha + \beta \):

\[ \alpha + \beta = -2 - 8 = -10 \]

### Final Answer

Thus, the value of \( \alpha + \beta \) is:

\[ \boxed{-10} \]