Let $$\alpha$$ and $$\beta$$ be real numbers. Consider a $$3 \times 3$$ matrix $$A$$ such that $$A^2 = 3A + \alpha I$$. If $$A^4 = 21A + \beta I$$, then

To find the values of $$\alpha$$and$$\beta$$, we can use the given equation for $$A^2$$to find an expression for$$A^4$$.

Given:

$$A^2 = 3A + \alpha I$$

$$A^4 = 21A + \beta I$$

Step 1: Express $$A^4$$in terms of $$A$$ and $$I$$ using the first equation.

Square both sides of the equation for $$A^2$$:

$$A^4 = (A^2)^2 = (3A + \alpha I)^2$$

$$A^4 = 9A^2 + 6\alpha A + \alpha^2 I$$

Step 2: Substitute $$A^2$$ back into the expanded equation.

$$A^4 = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I$$

$$A^4 = 27A + 9\alpha I + 6\alpha A + \alpha^2 I$$

$$A^4 = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$$

Step 3: Compare this result with the given equation for $$A^4$$.

We are given that:

$$A^4 = 21A + \beta I$$

By equating the coefficients of $$A$$ and $$I$$ from both expressions, we get :

$$27 + 6\alpha = 21$$

$$9\alpha + \alpha^2 = \beta$$

Step 4: Solve for $$\alpha$$ and $$\beta$$.

From the first equation:

$$6\alpha = 21 - 27$$

$$\alpha = -1$$

Substitute $$\alpha = -1$$ into the second equation:

$$\beta = 9(-1) + (-1)^2$$

$$\beta = -9 + 1$$

$$\beta = -8$$

Final Answer

The values are $$\alpha = -1$$ and $$\beta = -8$$.