Let `A=[(-4, -3, -3),(1, a, 1),(4, b, 3)] and A=A^(-1)`, then `a+2b` is equal to

Options

A. 0
B. 4
C. 8
D. 5

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the matrix \( A \) is equal to its inverse \( A^{-1} \). Given that \( A = A^{-1} \), we can conclude that \( A^2 = I \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Define the matrix \( A \)**:
\[ A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \]

2. **Set up the equation for \( A^2 \)**:
Since \( A = A^{-1} \), we have:
\[ A^2 = I \]
This means:
\[ A \cdot A = I \]

3. **Calculate \( A^2 \)**:
We need to multiply \( A \) by itself:
\[ A^2 = \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \cdot \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \]

4. **Perform the multiplication**:
Let's calculate each element of the resulting matrix \( A^2 \):

- First row, first column:
\[ (-4)(-4) + (-3)(1) + (-3)(4) = 16 - 3 - 12 = 1 \]

- First row, second column:
\[ (-4)(-3) + (-3)(a) + (-3)(b) = 12 - 3a - 3b \]

- First row, third column:
\[ (-4)(-3) + (-3)(1) + (-3)(3) = 12 - 3 - 9 = 0 \]

- Second row, first column:
\[ (1)(-4) + (a)(1) + (1)(4) = -4 + a + 4 = a \]

- Second row, second column:
\[ (1)(-3) + (a)(a) + (1)(b) = -3 + a^2 + b \]

- Second row, third column:
\[ (1)(-3) + (a)(1) + (1)(3) = -3 + a + 3 = a \]

- Third row, first column:
\[ (4)(-4) + (b)(1) + (3)(4) = -16 + b + 12 = b - 4 \]

- Third row, second column:
\[ (4)(-3) + (b)(a) + (3)(b) = -12 + ab + 3b \]

- Third row, third column:
\[ (4)(-3) + (b)(1) + (3)(3) = -12 + b + 9 = b - 3 \]

5. **Set up equations from \( A^2 = I \)**:
From the calculations, we have:
\[ A^2 = \begin{pmatrix} 1 & 12 - 3a - 3b & 0 \\ a & -3 + a^2 + b & a \\ b - 4 & -12 + ab + 3b & b - 3 \end{pmatrix} \]
Setting this equal to the identity matrix \( I \):
\[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:
- From (1,2): \( 12 - 3a - 3b = 0 \) (1)
- From (2,1): \( a = 0 \) (2)
- From (2,2): \( -3 + a^2 + b = 1 \) (3)
- From (3,1): \( b - 4 = 0 \) (4)
- From (3,3): \( b - 3 = 0 \) (5)

6. **Solve the equations**:
From equation (2), we find \( a = 0 \).
From equation (4), we find \( b = 4 \).
Now substituting \( a = 0 \) into equation (3):
\[ -3 + 0 + b = 1 \implies b = 4 \]

7. **Calculate \( a + 2b \)**:
Now we can calculate:
\[ a + 2b = 0 + 2(4) = 0 + 8 = 8 \]

### Final Answer:
\[ a + 2b = 8 \]