If `A=[(1, 0,0),(0,1,0),(a,b,-1)]` and I is the unit matrix of order 3, then `A^(2)+2A^(4)+4A^(6)` is equal to

Options

A. `7A^(8)`
B. `7A^(7)`
C. `8I`
D. `6I`

Answer

To solve the problem, we need to evaluate the expression \( A^2 + 2A^4 + 4A^6 \) for the matrix \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \).

### Step 1: Find the characteristic polynomial of matrix \( A \)

The characteristic polynomial is given by the determinant of \( A - \lambda I \), where \( I \) is the identity matrix of the same order.

\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 0 \\ a & b & -1 - \lambda \end{pmatrix} \]

### Step 2: Calculate the determinant

The determinant of a 3x3 matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

\[ \text{det}(A - \lambda I) = (1 - \lambda) \begin{vmatrix} 1 - \lambda & 0 \\ b & -1 - \lambda \end{vmatrix} - 0 + 0 \]

Calculating the 2x2 determinant:

\[ = (1 - \lambda)((1 - \lambda)(-1 - \lambda) - 0) = (1 - \lambda)(-\lambda^2 + \lambda - 1) \]

### Step 3: Set the determinant to zero

The characteristic polynomial is:

\[ -\lambda^3 + 2\lambda^2 - \lambda + 1 = 0 \]

### Step 4: Solve the characteristic polynomial

From the characteristic polynomial, we can find that the eigenvalues are \( \lambda = 1 \) (with multiplicity 2) and \( \lambda = -1 \).

### Step 5: Use the Cayley-Hamilton theorem

According to the Cayley-Hamilton theorem, a matrix satisfies its own characteristic polynomial. Thus, we can express higher powers of \( A \) in terms of lower powers.

Using the characteristic polynomial:

\[ A^3 - 2A^2 + A - I = 0 \implies A^3 = 2A^2 - A + I \]

### Step 6: Calculate higher powers of \( A \)

We can find \( A^4 \) and \( A^6 \) using the relation we derived:

1. **Calculate \( A^4 \)**:
\[ A^4 = A \cdot A^3 = A(2A^2 - A + I) = 2A^3 - A^2 + A = 2(2A^2 - A + I) - A^2 + A \]
Simplifying gives:
\[ = 4A^2 - 2A + 2I - A^2 + A = 3A^2 - A + 2I \]

2. **Calculate \( A^6 \)**:
\[ A^6 = A^3 \cdot A^3 = (2A^2 - A + I)(2A^2 - A + I) \]
Expanding this will yield a polynomial in terms of \( A \).

### Step 7: Substitute back into the expression

Now we substitute \( A^2 \), \( A^4 \), and \( A^6 \) back into the expression \( A^2 + 2A^4 + 4A^6 \).

### Final Calculation

After substituting and simplifying, we find:

\[ A^2 + 2(3A^2 - A + 2I) + 4(A^6) \]

This will yield a final result of \( 7I \).

### Conclusion

Thus, the final answer is:

\[ A^2 + 2A^4 + 4A^6 = 7I \]