Let `alpha` and `beta` be real numbers.Consider a `3xx3` matrix `A` such that `A^(2)=3A+alpha I`.If `A^(4)=21A+beta I`,then

Options

A. `beta=-8`
B. `beta=8`
C. `alpha=4`
D. `alpha=1`

Answer

To solve the problem, we start with the given equations for the matrix \( A \):

1. \( A^2 = 3A + \alpha I \)
2. \( A^4 = 21A + \beta I \)

We need to find the values of \( \alpha \) and \( \beta \).

### Step 1: Express \( A^3 \) in terms of \( A \) and \( I \)

From the first equation, we can multiply both sides by \( A \) to find \( A^3 \):

\[ A^3 = A \cdot A^2 = A(3A + \alpha I) = 3A^2 + \alpha A \]

Now, substitute \( A^2 \) from the first equation:

\[ A^3 = 3(3A + \alpha I) + \alpha A = 9A + 3\alpha I + \alpha A = (9 + \alpha)A + 3\alpha I \]

### Step 2: Express \( A^4 \) in terms of \( A \) and \( I \)

Now we can find \( A^4 \) by multiplying \( A^3 \) by \( A \):

\[ A^4 = A \cdot A^3 = A((9 + \alpha)A + 3\alpha I) = (9 + \alpha)A^2 + 3\alpha A \]

Substituting \( A^2 \) again:

\[ A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A = (27 + 12\alpha)A + (9 + \alpha^2)I \]

### Step 3: Set \( A^4 \) equal to the second equation

We know from the problem statement that:

\[ A^4 = 21A + \beta I \]

Setting the two expressions for \( A^4 \) equal gives:

\[ (27 + 12\alpha)A + (9 + \alpha^2)I = 21A + \beta I \]

### Step 4: Equate coefficients

From the equation above, we can equate the coefficients of \( A \) and \( I \):

1. For \( A \):
\[ 27 + 12\alpha = 21 \]
Solving this gives:
\[ 12\alpha = 21 - 27 \implies 12\alpha = -6 \implies \alpha = -\frac{1}{2} \]

2. For \( I \):
\[ 9 + \alpha^2 = \beta \]
Substituting \( \alpha = -\frac{1}{2} \):
\[ 9 + \left(-\frac{1}{2}\right)^2 = \beta \implies 9 + \frac{1}{4} = \beta \implies \beta = 9.25 \]

### Final Answer

Thus, the values of \( \alpha \) and \( \beta \) are:

\[ \alpha = -\frac{1}{2}, \quad \beta = 9.25 \]