We are tasked with finding the value of $\alpha$ for the given matrix equation:

$A^2 - \alpha A + I = 0$

where $A = \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}$ and $I$ is the $2 \times 2$ identity matrix.

Step 1: Compute $A^2$

First, calculate $A^2 = A \cdot A$.
Using matrix multiplication:

$A^2 = \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} \cdot \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}.$

Perform the multiplication:

• Element (1,1): $(-1)(-1) + (2)(-4) = 1 - 8 = -7$,
• Element (1,2): $(-1)(2) + (2)(7) = -2 + 14 = 12$,
• Element (2,1): $(-4)(-1) + (7)(-4) = 4 - 28 = -24$,
• Element (2,2): $(-4)(2) + (7)(7) = -8 + 49 = 41$.

Thus:

$A^2 = \begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix}.$

Step 2: Plug $A^2$ and $A$ into the equation

The equation is:

$A^2 - \alpha A + I = 0.$

Substitute $A^2$, $A$, and $I$:

$\begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0.$

Combine the terms:

$\begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -7 + 1 & 12 + 0 \\ -24 + 0 & 41 + 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}.$

Simplify:

$\begin{bmatrix} -6 & 12 \\ -24 & 42 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} = 0.$

Step 3: Solve for $\alpha$

Expand $-\alpha A$:

$-\alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} = \begin{bmatrix} \alpha & -2\alpha \\ 4\alpha & -7\alpha \end{bmatrix}.$

Substitute back:

$\begin{bmatrix} -6 & 12 \\ -24 & 42 \end{bmatrix} + \begin{bmatrix} \alpha & -2\alpha \\ 4\alpha & -7\alpha \end{bmatrix} = 0.$

Combine matrices:

$\begin{bmatrix} -6 + \alpha & 12 - 2\alpha \\ -24 + 4\alpha & 42 - 7\alpha \end{bmatrix} = 0.$

This gives the system of equations by equating each entry to 0:

1. $-6 + \alpha = 0$,
2. $12 - 2\alpha = 0$,
3. $-24 + 4\alpha = 0$,
4. $42 - 7\alpha = 0$.

Step 4: Solve the system

From the first equation:

$\alpha = 6.$

Verify this value in the other equations:

• Second equation: $12 - 2(6) = 12 - 12 = 0$ (True),
• Third equation: $-24 + 4(6) = -24 + 24 = 0$ (True),
• Fourth equation: $42 - 7(6) = 42 - 42 = 0$ (True).

Thus, $\alpha = 6$ satisfies all equations.

Final Answer:

$\boxed{\alpha = 6}$

Would you like me to expand on any part of this solution?
Here are some related questions you might consider exploring:

1. What is the determinant of $A^2$?
2. How do eigenvalues relate to this matrix equation?
3. Can the characteristic equation of $A$ verify this result?
4. What happens if we change $A$ to a different matrix but keep the same equation?
5. How is the Cayley-Hamilton theorem relevant to this problem?

Tip: When solving matrix equations, always verify your solutions by substituting back into the original equation to ensure consistency.