The solutiion of `(x,y,z)` the equation `[(-1,0,1),(-1,1,0),(0,-1,1)][(x),(y),(z)]=[(1),(1),(2)]` is `(x,y,z)`

Options

A. `(1,1,1)`
B. `(0,-1,2)`
C. `(-1,2,2)`
D. `(-1,0,2)`

Explanation

To solve the equation represented by the matrix equation \( A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \), where

\[ A = \begin{pmatrix} -1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}, \]

we will follow these steps:

### Step 1: Write the equation in matrix form
We can express the equation as \( AX = B \) where

\[ X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}. \]

### Step 2: Find the inverse of matrix \( A \)
To find \( X \), we need to calculate \( A^{-1} \) using the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A). \]

### Step 3: Calculate the determinant of \( A \)
The determinant of a 3x3 matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg). \]

For our matrix \( A \):

\[ \text{det}(A) = -1 \cdot (1 \cdot 1 - 0 \cdot (-1)) - 0 \cdot (-1 \cdot 1 - 0 \cdot 0) + 1 \cdot (-1 \cdot (-1) - 1 \cdot 0) = -1 \cdot 1 + 0 + 1 = 0. \]

### Step 4: Calculate the adjoint of \( A \)
The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactors for each element of \( A \).

1. For \( a_{11} = -1 \):
\[ \text{Cofactor} = \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1. \]
2. For \( a_{12} = 0 \):
\[ \text{Cofactor} = -\begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = 0. \]
3. For \( a_{13} = 1 \):
\[ \text{Cofactor} = \begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix} = 1. \]
4. For \( a_{21} = -1 \):
\[ \text{Cofactor} = -\begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} = 1. \]
5. For \( a_{22} = 1 \):
\[ \text{Cofactor} = \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} = -1. \]
6. For \( a_{23} = 0 \):
\[ \text{Cofactor} = -\begin{vmatrix} -1 & 0 \\ 0 & -1 \end{vmatrix} = -1. \]
7. For \( a_{31} = 0 \):
\[ \text{Cofactor} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1. \]
8. For \( a_{32} = -1 \):
\[ \text{Cofactor} = -\begin{vmatrix} -1 & 1 \\ -1 & 0 \end{vmatrix} = -1. \]
9. For \( a_{33} = 1 \):
\[ \text{Cofactor} = \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} = -1. \]

Thus, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} 1 & 0 & 1 \\ 1 & -1 & -1 \\ -1 & -1 & -1 \end{pmatrix}. \]

The adjoint of \( A \) is the transpose of the cofactor matrix:

\[ \text{adj}(A) = \begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix}. \]

### Step 5: Calculate \( A^{-1} \)
Since we found that \( \text{det}(A) = 0 \), the matrix \( A \) is singular, and thus, \( A^{-1} \) does not exist.

### Step 6: Solve for \( X \)
Since \( A^{-1} \) does not exist, we cannot find a unique solution for \( X \). However, we can still analyze the system of equations represented by \( AX = B \).

### Conclusion
The system of equations does not have a unique solution due to the singular nature of matrix \( A \).