Let A be a `3xx3` real matrix such that `A((1),(1),(0))=((1),(1),(0)),A((1),(0),(1))=((-1),(0),(1))and ((0),(0),(1))=((1),(1),(2))`
If `X=(x_1, x_2, x_3)^T` and I is an identity matrix of order 3, then the system `(A-2I)X=((4),(1),(1))` has :

Options

A. no solution
B. infinitely many solutions
C. unique solution
D. exactly two solutions

Answer

To solve the problem, we need to analyze the given conditions and find the matrix \( A \) and then determine the nature of the solutions to the system \( (A - 2I)X = (4, 1, 1)^T \).

### Step 1: Formulate the matrix \( A \)

We have the following conditions:
1. \( A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \)
2. \( A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \)
3. \( A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)

Let \( A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \).

Using the first condition:
\[ A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a_{11} + a_{12} \\ a_{21} + a_{22} \\ a_{31} + a_{32} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \]
This gives us the equations:
1. \( a_{11} + a_{12} = 1 \)
2. \( a_{21} + a_{22} = 1 \)
3. \( a_{31} + a_{32} = 0 \)

Using the second condition:
\[ A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a_{11} + a_{13} \\ a_{21} + a_{23} \\ a_{31} + a_{33} \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \]
This gives us the equations:
4. \( a_{11} + a_{13} = -1 \)
5. \( a_{21} + a_{23} = 0 \)
6. \( a_{31} + a_{33} = 1 \)

Using the third condition:
\[ A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
This gives us:
7. \( a_{13} = 1 \)
8. \( a_{23} = 1 \)
9. \( a_{33} = 2 \)

### Step 2: Solve for the coefficients

From equation 7:
- \( a_{13} = 1 \)

Substituting \( a_{13} \) into equation 4:
- \( a_{11} + 1 = -1 \) ⇒ \( a_{11} = -2 \)

From equation 5:
- \( a_{21} + 1 = 0 \) ⇒ \( a_{21} = -1 \)

From equations 1 and 2:
- \( -2 + a_{12} = 1 \) ⇒ \( a_{12} = 3 \)
- \( -1 + a_{22} = 1 \) ⇒ \( a_{22} = 2 \)

From equations 3 and 6:
- \( a_{31} + a_{32} = 0 \) ⇒ \( a_{32} = -a_{31} \)
- \( a_{31} + 2 = 1 \) ⇒ \( a_{31} = -1 \) ⇒ \( a_{32} = 1 \)

### Step 3: Construct matrix \( A \)

Now we can construct the matrix \( A \):
\[ A = \begin{pmatrix} -2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2 \end{pmatrix} \]

### Step 4: Calculate \( A - 2I \)

Now, we calculate \( A - 2I \):
\[ A - 2I = \begin{pmatrix} -2-2 & 3 & 1 \\ -1 & 2-2 & 1 \\ -1 & 1 & 2-2 \end{pmatrix} = \begin{pmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix} \]

### Step 5: Solve the system \( (A - 2I)X = (4, 1, 1)^T \)

We need to solve:
\[ \begin{pmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} \]

### Step 6: Form the augmented matrix and row reduce

The augmented matrix is:
\[ \begin{pmatrix} -4 & 3 & 1 & | & 4 \\ -1 & 0 & 1 & | & 1 \\ -1 & 1 & 0 & | & 1 \end{pmatrix} \]

Perform row operations to reduce this matrix. After performing the necessary row operations, we will find that the system has infinitely many solutions.

### Conclusion

The system \( (A - 2I)X = (4, 1, 1)^T \) has infinitely many solutions.