Let`A= [[1,0,0],[2,1,0],[3,2,1]]` be a square matrix and ` C_(1), C_(2), C_(3)` be three
comumn matrices satisfying `AC_(1) = [[1],[0],[0]], AC_(2) = [[2],[3],[0]] and AC_(3)= [[2],[3],[1]]` of matrix B. If the matrix `C= 1/3 (AcdotB).`
The value of `det(B^(-1))`, is

Options

A. 2
B. `1/2`
C. 3
D. `1/3`

Explanation

To solve the problem, we need to find the value of \( \det(B^{-1}) \) given the matrix \( A \) and the conditions on matrices \( C_1, C_2, C_3 \).

### Step 1: Define the matrices
Let
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \]
Let \( C_1, C_2, C_3 \) be column matrices such that:
\[ AC_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad AC_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}, \quad AC_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \]

### Step 2: Find \( C_1 \)
Let \( C_1 = \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{bmatrix} \).
From \( AC_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \):
\[ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]
This gives us the equations:
1. \( \alpha_1 = 1 \)
2. \( 2\alpha_1 + \alpha_2 = 0 \) → \( \alpha_2 = -2 \)
3. \( 3\alpha_1 + 2\alpha_2 + \alpha_3 = 0 \) → \( 3 - 4 + \alpha_3 = 0 \) → \( \alpha_3 = 1 \)

Thus,
\[ C_1 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \]

### Step 3: Find \( C_2 \)
Let \( C_2 = \begin{bmatrix} \beta_1 \\ \beta_2 \\ \beta_3 \end{bmatrix} \).
From \( AC_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \):
\[ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} \beta_1 \\ \beta_2 \\ \beta_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \]
This gives us the equations:
1. \( \beta_1 = 2 \)
2. \( 2\beta_1 + \beta_2 = 3 \) → \( 4 + \beta_2 = 3 \) → \( \beta_2 = -1 \)
3. \( 3\beta_1 + 2\beta_2 + \beta_3 = 0 \) → \( 6 - 2 + \beta_3 = 0 \) → \( \beta_3 = -4 \)

Thus,
\[ C_2 = \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} \]

### Step 4: Find \( C_3 \)
Let \( C_3 = \begin{bmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{bmatrix} \).
From \( AC_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \):
\[ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \]
This gives us the equations:
1. \( \gamma_1 = 2 \)
2. \( 2\gamma_1 + \gamma_2 = 3 \) → \( 4 + \gamma_2 = 3 \) → \( \gamma_2 = -1 \)
3. \( 3\gamma_1 + 2\gamma_2 + \gamma_3 = 1 \) → \( 6 - 2 + \gamma_3 = 1 \) → \( \gamma_3 = -3 \)

Thus,
\[ C_3 = \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \]

### Step 5: Construct matrix \( B \)
Now, we can construct matrix \( B \):
\[ B = \begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix} \]

### Step 6: Calculate \( \det(B) \)
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \det(B) = 1 \cdot \begin{vmatrix} -1 & -1 \\ -4 & -3 \end{vmatrix} - 2 \cdot \begin{vmatrix} -2 & -1 \\ 1 & -3 \end{vmatrix} + 2 \cdot \begin{vmatrix} -2 & -1 \\ 1 & -4 \end{vmatrix} \]
Calculating the minors:
1. \( \begin{vmatrix} -1 & -1 \\ -4 & -3 \end{vmatrix} = (-1)(-3) - (-1)(-4) = 3 - 4 = -1 \)
2. \( \begin{vmatrix} -2 & -1 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (-1)(1) = 6 + 1 = 7 \)
3. \( \begin{vmatrix} -2 & -1 \\ 1 & -4 \end{vmatrix} = (-2)(-4) - (-1)(1) = 8 + 1 = 9 \)

Now substituting back:
\[ \det(B) = 1 \cdot (-1) - 2 \cdot 7 + 2 \cdot 9 = -1 - 14 + 18 = 3 \]

### Step 7: Find \( \det(B^{-1}) \)
Using the property of determinants:
\[ \det(B^{-1}) = \frac{1}{\det(B)} = \frac{1}{3} \]

### Final Answer:
\[ \det(B^{-1}) = \frac{1}{3} \]