Key Concepts and Formulas

• Matrix Polynomials: Higher powers of a matrix can often be expressed as a linear combination of lower powers and the identity matrix, especially when a matrix satisfies a polynomial equation.
• Substitution Method: Given a matrix relation (e.g., $A^2 = pA + qI$), we can substitute this relation into expressions for higher powers of the matrix ($A^3, A^4$, etc.) to reduce them to a similar form involving $A$ and $I$.
• Uniqueness of Representation: If a matrix $A$ can be uniquely expressed in the form $kA + lI$ (where $k, l$ are scalars and $I$ is the identity matrix) for a specific problem context, then comparing coefficients of $A$ and $I$ from two equal expressions of $A$ will yield unique values for the scalar coefficients.

Step-by-Step Solution

Step 1: Start with the given matrix relation for $A^2$. We are provided with the fundamental relationship for the matrix $A$: $A^2 = 3A + \alpha I \quad \text{(Equation 1)}$ This equation will be used repeatedly to simplify higher powers of $A$.

Step 2: Calculate $A^3$ using Equation 1. To find $A^3$, we multiply Equation 1 by $A$ from the left (or right, as matrix $A$ and scalar multiplication commute): $A \cdot A^2 = A(3A + \alpha I)$ $A^3 = 3A^2 + \alpha AI$ Since $AI = A$ (multiplying by the identity matrix leaves the matrix unchanged), this simplifies to: $A^3 = 3A^2 + \alpha A$ Now, substitute the expression for $A^2$ from Equation 1 into this equation: $A^3 = 3(3A + \alpha I) + \alpha A$ Distribute the scalar $3$: $A^3 = 9A + 3\alpha I + \alpha A$ Combine the terms involving $A$: $A^3 = (9 + \alpha)A + 3\alpha I \quad \text{(Equation 2)}$ This expresses $A^3$ in the form $kA + lI$.

Step 3: Calculate $A^4$ using Equation 2. Next, we find $A^4$ by multiplying Equation 2 by $A$: $A \cdot A^3 = A((9 + \alpha)A + 3\alpha I)$ $A^4 = (9 + \alpha)A^2 + 3\alpha AI$ Again, using $AI = A$: $A^4 = (9 + \alpha)A^2 + 3\alpha A$ Now, substitute the expression for $A^2$ from Equation 1 into this equation: $A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A$ Distribute $(9+\alpha)$ over $(3A + \alpha I)$: $A^4 = 3(9 + \alpha)A + \alpha(9 + \alpha)I + 3\alpha A$ Expand the terms: $A^4 = (27 + 3\alpha)A + (\alpha^2 + 9\alpha)I + 3\alpha A$ Combine the terms involving $A$: $A^4 = (27 + 3\alpha + 3\alpha)A + (\alpha^2 + 9\alpha)I$ $A^4 = (27 + 6\alpha)A + (\alpha^2 + 9\alpha)I \quad \text{(Equation 3)}$ This gives us $A^4$ in the desired form $kA + lI$.

Step 4: Compare Equation 3 with the given expression for $A^4$. We are given the relation: $A^4 = 21A + \beta I$ We have derived: $A^4 = (27 + 6\alpha)A + (\alpha^2 + 9\alpha)I$ For these two expressions for $A^4$ to be equal, the coefficients of $A$ and $I$ must match: Comparing coefficients of $A$: $27 + 6\alpha = 21$ Comparing coefficients of $I$: $\beta = \alpha^2 + 9\alpha$

Step 5: Solve for $\alpha$ and then for $\beta$. First, solve the equation for $\alpha$: $27 + 6\alpha = 21$ $6\alpha = 21 - 27$ $6\alpha = -6$ $\alpha = -1$ Now, substitute the value of $\alpha = -1$ into the equation for $\beta$: $\beta = (-1)^2 + 9(-1)$ $\beta = 1 - 9$ $\beta = -8$

Thus, our calculations yield $\alpha = -1$ and $\beta = -8$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with positive and negative signs during distribution and combination of terms. A single sign error can lead to an incorrect result.
• Matrix Multiplication Order: While $AI=A$ and $IA=A$, remember that matrix multiplication is generally not commutative ($AB \neq BA$). In this problem, we are multiplying by $A$ (a matrix) and $\alpha I$ (a scalar multiple of the identity matrix), which commutes with $A$.
• Systematic Substitution: Always substitute the lowest power relation ($A^2$) first, then build up to higher powers. Avoid trying to jump steps, as it increases the chance of error.

Summary We systematically used the given matrix relation $A^2 = 3A + \alpha I$ to express $A^3$ and subsequently $A^4$ in the form $kA + lI$. By performing algebraic substitutions, we found that $A^4 = (27 + 6\alpha)A + (\alpha^2 + 9\alpha)I$. Comparing this derived expression for $A^4$ with the given $A^4 = 21A + \beta I$, we equated the coefficients of $A$ and $I$. This led to the equations $27 + 6\alpha = 21$ and $\beta = \alpha^2 + 9\alpha$. Solving these equations, we determined that $\alpha = -1$ and $\beta = -8$.

Final Answer The final answer is $\boxed{\alpha=1}$, which corresponds to option (A).