Let `A` is `3 xx 3` matrix such that `Ax_1=B_1,Ax_2=B_2,Ax_3=B_3` where
`x_1=[[1],[1],[1]],x_2=[[0],[2],[1]],x_3=[[0],[0],[1]]` ,br> `B_1=[[1],[0],[0]],B_2=[[0],[2],[0]],B_3=[[0],[0],[2]]` then find `|A|`.

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the determinant of the matrix \( A \) given the equations \( A x_1 = B_1 \), \( A x_2 = B_2 \), and \( A x_3 = B_3 \), we will follow these steps:

### Step 1: Define the Matrix \( A \)
Let \( A \) be represented as:
\[ A = \begin{pmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix} \]

### Step 2: Use the First Equation \( A x_1 = B_1 \)
From the first equation \( A x_1 = B_1 \) where \( x_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) and \( B_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \):
\[ A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]
This leads to the equations:
1. \( a_1 + a_4 + a_7 = 1 \)
2. \( a_2 + a_5 + a_8 = 0 \)
3. \( a_3 + a_6 + a_9 = 0 \)

### Step 3: Use the Second Equation \( A x_2 = B_2 \)
From the second equation \( A x_2 = B_2 \) where \( x_2 = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} \) and \( B_2 = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \):
\[ A \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \]
This leads to the equations:
1. \( 2a_4 + a_7 = 0 \)
2. \( 2a_5 + a_8 = 2 \)
3. \( 2a_6 + a_9 = 0 \)

### Step 4: Use the Third Equation \( A x_3 = B_3 \)
From the third equation \( A x_3 = B_3 \) where \( x_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \) and \( B_3 = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \):
\[ A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \]
This leads to the equations:
1. \( a_7 = 0 \)
2. \( a_8 = 0 \)
3. \( a_9 = 2 \)

### Step 5: Substitute Known Values
From the equations derived from \( A x_3 = B_3 \):
- \( a_7 = 0 \)
- \( a_8 = 0 \)
- \( a_9 = 2 \)

Now substitute these values into the equations from \( A x_1 = B_1 \):
1. \( a_1 + a_4 + 0 = 1 \) → \( a_1 + a_4 = 1 \)
2. \( a_2 + a_5 + 0 = 0 \) → \( a_2 + a_5 = 0 \)
3. \( a_3 + a_6 + 2 = 0 \) → \( a_3 + a_6 = -2 \)

Now substitute into the equations from \( A x_2 = B_2 \):
1. \( 2a_4 + 0 = 0 \) → \( a_4 = 0 \)
2. \( 2a_5 + 0 = 2 \) → \( a_5 = 1 \)
3. \( 2a_6 + 2 = 0 \) → \( a_6 = -1 \)

### Step 6: Solve for Remaining Variables
Substituting \( a_4 = 0 \) into \( a_1 + a_4 = 1 \):
- \( a_1 + 0 = 1 \) → \( a_1 = 1 \)

Substituting \( a_5 = 1 \) into \( a_2 + a_5 = 0 \):
- \( a_2 + 1 = 0 \) → \( a_2 = -1 \)

Substituting \( a_6 = -1 \) into \( a_3 + a_6 = -2 \):
- \( a_3 - 1 = -2 \) → \( a_3 = -1 \)

### Step 7: Construct Matrix \( A \)
Now we have:
\[ A = \begin{pmatrix} 1 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{pmatrix} \]

### Step 8: Calculate the Determinant of \( A \)
The determinant of \( A \) can be calculated as follows:
\[ |A| = 1 \cdot (1 \cdot 2 - (-1) \cdot 0) - (-1) \cdot (0 \cdot 2 - (-1) \cdot 0) + (-1) \cdot (0 \cdot (-1) - 1 \cdot 0) \]
Calculating:
\[ |A| = 1 \cdot 2 - 0 + 0 = 2 \]

### Final Answer
Thus, the determinant of matrix \( A \) is:
\[ \boxed{2} \]