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 where we have the matrix \( A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \) and we know that \( A = A^{-1} \), we will follow these steps:

### Step 1: Set up the equation
Since \( A = A^{-1} \), we can multiply both sides by \( A \):
\[ A \cdot A = I \]
where \( I \) is the identity matrix. The identity matrix for a 3x3 matrix is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Multiply the matrix \( A \) by itself
We need to compute \( A \cdot A \):
\[ A \cdot A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \cdot \begin{pmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{pmatrix} \]

### Step 3: Calculate the elements of the resulting matrix
We will calculate each element of the resulting matrix:

- 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:
\[ (4)(-3) + (b)(1) + (3)(3) = -12 + b + 9 = b - 3 \]

### Step 4: Set the resulting matrix equal to the identity matrix
Now we set the resulting matrix equal to the identity matrix:
\[ \begin{pmatrix} 1 & 12 - 3a - 3b & 0 \\ a & -3 + a^2 + b & a \\ b - 4 & -12 + ab + 3b & b - 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Solve the equations
From the matrix comparison, we can derive the following equations:
1. \( 12 - 3a - 3b = 0 \)
2. \( a = 0 \)
3. \( -3 + a^2 + b = 1 \)
4. \( b - 4 = 0 \)
5. \( -12 + ab + 3b = 0 \)
6. \( b - 3 = 0 \)

From equation 2, we find \( a = 0 \).

Substituting \( a = 0 \) into equation 1:
\[ 12 - 3(0) - 3b = 0 \implies 12 - 3b = 0 \implies 3b = 12 \implies b = 4 \]

### Step 6: Calculate \( a + 2b \)
Now we can find \( a + 2b \):
\[ a + 2b = 0 + 2(4) = 8 \]

### Final Answer
Thus, \( a + 2b = 8 \).