Let `{:A=[(1,0,0),(2,1,0),(3,2,1)]:}and U_1,U_2,U_3` be column matrices satisfying `{:AU_1=[(1),(0),(0)],AU_2=[(2),(3),(6)],AU_3=[(2),(3),(1)]:}`.If U is `3xx3` matrix whose columns are `U_1,U_2,U_3," then "absU=`

Options

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

Explanation

To solve the problem step by step, we will first define the matrices and then calculate the columns \( U_1, U_2, \) and \( U_3 \) based on the given equations. Finally, we will compute the determinant of the matrix \( U \).

### Step 1: Define the Matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \]

### Step 2: Set Up the Column Matrices
Let:
\[ U_1 = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad U_2 = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \quad U_3 = \begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} \]

### Step 3: Solve for \( U_1 \)
We know:
\[ AU_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]
This gives us the equation:
\[ A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]
Multiplying out gives:
\[ \begin{pmatrix} 1 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 \\ 2 \cdot x_1 + 1 \cdot x_2 + 0 \cdot x_3 \\ 3 \cdot x_1 + 2 \cdot x_2 + 1 \cdot x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]
From this, we get the equations:
1. \( x_1 = 1 \)
2. \( 2x_1 + x_2 = 0 \)
3. \( 3x_1 + 2x_2 + x_3 = 0 \)

Substituting \( x_1 = 1 \) into the second equation:
\[ 2(1) + x_2 = 0 \implies x_2 = -2 \]
Substituting \( x_1 = 1 \) and \( x_2 = -2 \) into the third equation:
\[ 3(1) + 2(-2) + x_3 = 0 \implies 3 - 4 + x_3 = 0 \implies x_3 = 1 \]
Thus, we have:
\[ U_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \]

### Step 4: Solve for \( U_2 \)
Given:
\[ AU_2 = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} \]
This gives:
\[ A \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} \]
Multiplying out gives:
\[ \begin{pmatrix} y_1 \\ 2y_1 + y_2 \\ 3y_1 + 2y_2 + y_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} \]
From this, we get:
1. \( y_1 = 2 \)
2. \( 2y_1 + y_2 = 3 \)
3. \( 3y_1 + 2y_2 + y_3 = 6 \)

Substituting \( y_1 = 2 \) into the second equation:
\[ 2(2) + y_2 = 3 \implies 4 + y_2 = 3 \implies y_2 = -1 \]
Substituting \( y_1 = 2 \) and \( y_2 = -1 \) into the third equation:
\[ 3(2) + 2(-1) + y_3 = 6 \implies 6 - 2 + y_3 = 6 \implies y_3 = 2 \]
Thus, we have:
\[ U_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \]

### Step 5: Solve for \( U_3 \)
Given:
\[ AU_3 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]
This gives:
\[ A \begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]
Multiplying out gives:
\[ \begin{pmatrix} z_1 \\ 2z_1 + z_2 \\ 3z_1 + 2z_2 + z_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]
From this, we get:
1. \( z_1 = 2 \)
2. \( 2z_1 + z_2 = 3 \)
3. \( 3z_1 + 2z_2 + z_3 = 1 \)

Substituting \( z_1 = 2 \) into the second equation:
\[ 2(2) + z_2 = 3 \implies 4 + z_2 = 3 \implies z_2 = -1 \]
Substituting \( z_1 = 2 \) and \( z_2 = -1 \) into the third equation:
\[ 3(2) + 2(-1) + z_3 = 1 \implies 6 - 2 + z_3 = 1 \implies z_3 = -3 \]
Thus, we have:
\[ U_3 = \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix} \]

### Step 6: Form the Matrix \( U \)
Now we can form the matrix \( U \):
\[ U = \begin{pmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & 2 & -3 \end{pmatrix} \]

### Step 7: Calculate the Determinant of \( U \)
To find the determinant of \( U \):
\[ \text{det}(U) = 1 \cdot \begin{vmatrix} -1 & -1 \\ 2 & -3 \end{vmatrix} - 2 \cdot \begin{vmatrix} -2 & -1 \\ 1 & -3 \end{vmatrix} + 2 \cdot \begin{vmatrix} -2 & -1 \\ 1 & 2 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -1 & -1 \\ 2 & -3 \end{vmatrix} = (-1)(-3) - (-1)(2) = 3 + 2 = 5 \)

2. \( \begin{vmatrix} -2 & -1 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (-1)(1) = 6 + 1 = 7 \)

3. \( \begin{vmatrix} -2 & -1 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (-1)(1) = -4 + 1 = -3 \)

Putting it all together:
\[ \text{det}(U) = 1 \cdot 5 - 2 \cdot 7 + 2 \cdot (-3) = 5 - 14 - 6 = -15 \]

### Final Result
Thus, the determinant of \( U \) is:
\[ \text{det}(U) = -15 \]