Suppose the vectors `x_(1), x_(2) and x_(3)` are the solutions of the system of linear equations, `Ax=b` when the vector b on the right side is equal to `b_(1), b_(2) and b_(3)` respectively. If
`x_(1)=[(1),(1),(1)], x_(2)=[(0),(2),(1)], x_(3)=[(0),(0),(1)], b_(1)=[(1),(0),(0)], b_(2)=[(0),(2),(0)] and b_(3)=[(0),(0),(2)]`, then the determinant of A is equal to :

Options

A. `(1)/(2)`
B. 4
C. 2
D. `(3)/(2)`

Answer

To find the determinant of the matrix \( A \), we will first set up the equations based on the given vectors \( x_1, x_2, x_3 \) and the corresponding vectors \( b_1, b_2, b_3 \).

### Step 1: Set up the equations
We have the following equations based on the problem statement:

1. \( Ax_1 = b_1 \)
2. \( Ax_2 = b_2 \)
3. \( Ax_3 = b_3 \)

Where:
- \( x_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \), \( b_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \)
- \( x_2 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \), \( b_2 = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \)
- \( x_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \), \( b_3 = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} \)

### Step 2: Define the matrix \( A \)
Assume \( A \) is a \( 3 \times 3 \) matrix represented as:

\[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ p & q & r \end{bmatrix} \]

### Step 3: Write the equations based on \( Ax = b \)

From \( Ax_1 = b_1 \):
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ p & q & r \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]

This gives us the equations:
1. \( a + b + c = 1 \) (1)
2. \( d + e + f = 0 \) (2)
3. \( p + q + r = 0 \) (3)

From \( Ax_2 = b_2 \):
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ p & q & r \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \]

This gives us the equations:
1. \( 2b + c = 0 \) (4)
2. \( 2e + f = 2 \) (5)
3. \( 2q + r = 0 \) (6)

From \( Ax_3 = b_3 \):
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ p & q & r \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} \]

This gives us the equations:
1. \( c = 0 \) (7)
2. \( f = 2 \) (8)
3. \( r = 2 \) (9)

### Step 4: Substitute and solve the equations

From equation (7), we have \( c = 0 \).

Substituting \( c = 0 \) into equation (4):
\[ 2b + 0 = 0 \implies b = 0 \]

Substituting \( b = 0 \) and \( c = 0 \) into equation (1):
\[ a + 0 + 0 = 1 \implies a = 1 \]

Now substituting \( f = 2 \) into equation (5):
\[ 2e + 2 = 2 \implies 2e = 0 \implies e = 0 \]

Now substituting \( r = 2 \) into equation (6):
\[ 2q + 2 = 0 \implies 2q = -2 \implies q = -1 \]

### Step 5: Substitute back to find \( d \) and \( p \)

From equation (2):
\[ d + 0 + 2 = 0 \implies d = -2 \]

From equation (3):
\[ p - 1 + 2 = 0 \implies p + 1 = 0 \implies p = -1 \]

### Step 6: Form the matrix \( A \)
Now we have:
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 2 \\ -1 & -1 & 2 \end{bmatrix} \]

### Step 7: Calculate the determinant of \( A \)

To find the determinant of \( A \):
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 0 & 2 \\ -1 & 2 \end{vmatrix} - 0 + 0 \]
Calculating the \( 2 \times 2 \) determinant:
\[ \begin{vmatrix} 0 & 2 \\ -1 & 2 \end{vmatrix} = (0 \cdot 2) - (2 \cdot -1) = 2 \]
Thus,
\[ \text{det}(A) = 1 \cdot 2 = 2 \]

### Final Answer
The determinant of \( A \) is \( 2 \).