Question

Let
A = \(\begin{pmatrix}1 & 2\\ -2 & -5 \end{pmatrix}\)
Let \( \alpha, \beta \in \mathbb{R} \) be such that \( \alpha A^2 + \beta A = 2I \). Then \( \alpha + \beta \) is equal to 

• A. –10
• B. –6
• C. 6
• D. 10

Answer 1

The Correct Option is D

 To find the value of \( \alpha + \beta \), let's first understand the given equation:

We have the matrix \( A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \) and are given:

\( \alpha A^2 + \beta A = 2I \),

where \( I \) is the identity matrix \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

Step 1: Compute \( A^2 \).

We have:

\( A^2 = A \times A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} \)

\( = \begin{pmatrix} (1)(1) + (2)(-2) & (1)(2) + (2)(-5) \\ (-2)(1) + (-5)(-2) & (-2)(2) + (-5)(-5) \end{pmatrix} \)

\( = \begin{pmatrix} 1 - 4 & 2 - 10 \\ -2 + 10 & -4 - 25 \end{pmatrix} \)

\( = \begin{pmatrix} -3 & -8 \\ 8 & -29 \end{pmatrix} \)

Step 2: Substitute \( A^2 \) and \( A \) into the equation \( \alpha A^2 + \beta A = 2I \).

We have:

\( \alpha \begin{pmatrix} -3 & -8 \\ 8 & -29 \end{pmatrix} + \beta \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix} = 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)

Expanding this equation:

\( \begin{pmatrix} -3\alpha + \beta & -8\alpha + 2\beta \\ 8\alpha - 2\beta & -29\alpha - 5\beta \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \)

This results in a system of equations:

• \(-3\alpha + \beta = 2\)
• \(-8\alpha + 2\beta = 0\)

Step 3: Solve these equations.

From the second equation:

\(-8\alpha + 2\beta = 0 \Rightarrow \alpha = \frac{\beta}{4}\)

Substitute \(\alpha = \frac{\beta}{4}\) into the first equation:

\(-3\left(\frac{\beta}{4}\right) + \beta = 2\)

\(\Rightarrow -\frac{3\beta}{4} + \beta = 2\)

\(\Rightarrow \frac{\beta}{4} = 2\)

\(\Rightarrow \beta = 8\)

Substitute \(\beta = 8\) back to find \(\alpha\):

\(\alpha = \frac{\beta}{4} = \frac{8}{4} = 2\)

Therefore, \(\alpha + \beta = 2 + 8 = 10\).

Hence, the correct option is 10.