1. The problem is to find the values of $\alpha$, $b$, $c$, and $d$ such that the matrices $$\begin{bmatrix} \alpha & 3 \\ -1 & \alpha + b \end{bmatrix} = \begin{bmatrix} 4 & d - 2c \\ d + 2c & -2 \end{bmatrix}$$ are equal. 2. Two matrices are equal if and only if their corresponding elements are equal. This means: $$\alpha = 4$$ $$3 = d - 2c$$ $$-1 = d + 2c$$ $$\alpha + b = -2$$ 3. From the first equation, we have: $$\alpha = 4$$ 4. Substitute $\alpha = 4$ into the fourth equation: $$4 + b = -2$$ Solve for $b$: $$b = -2 - 4 = -6$$ 5. Now solve the system of two equations for $c$ and $d$: $$3 = d - 2c$$ $$-1 = d + 2c$$ 6. Add the two equations to eliminate $c$: $$3 + (-1) = (d - 2c) + (d + 2c)$$ $$2 = 2d$$ $$d = 1$$ 7. Substitute $d = 1$ into the first equation: $$3 = 1 - 2c$$ Solve for $c$: $$-2c = 3 - 1 = 2$$ $$c = -1$$ 8. Final values are: $$\alpha = 4, b = -6, c = -1, d = 1$$