Let `X=[{:(x_(1)),(x_(2)),(x_(3)):}],A=[{:(1,-1,2),(2,0,1),(3,2,1):}]` and `B=[{:(3),(1),(4):}]`.If AX=B, then X is equal to

Options

A. `[{:(1),(2),(3):}]`
B. `[{:(-1),(-2),(3):}]`
C. `[{:(-1),(-2),(-3):}]`
D. `[{:(-1),(2),(3):}]`

Answer

To solve the equation \( AX = B \) for \( X \), we will follow these steps:

### Step 1: Write down the matrices
Given:
- \( A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{pmatrix} \)
- \( B = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \)

### Step 2: Find the inverse of matrix \( A \)
To find \( X \), we need to calculate \( A^{-1} \) and then multiply it by \( B \):
\[ X = A^{-1}B \]

### Step 3: Calculate the determinant of \( A \)
The determinant of \( A \) can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
\[ \text{det}(A) = 1 \cdot (0 \cdot 1 - 1 \cdot 2) - (-1) \cdot (2 \cdot 1 - 1 \cdot 3) + 2 \cdot (2 \cdot 2 - 0 \cdot 3) \]
Calculating each term:
\[ = 1 \cdot (0 - 2) + 1 \cdot (2 - 3) + 2 \cdot (4 - 0) \]
\[ = -2 + (-1) + 8 = 5 \]

### Step 4: Find the adjoint of \( A \)
To find the adjoint, we need to calculate the cofactor matrix and then take the transpose. The cofactor \( C_{ij} \) is calculated as:
\[ C_{ij} = (-1)^{i+j} M_{ij} \]
where \( M_{ij} \) is the minor of element \( a_{ij} \).

Calculating the minors and cofactors:
1. For \( a_{11} = 1 \): Minor = \( \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 2 = -2 \) → Cofactor = \( -2 \)
2. For \( a_{12} = -1 \): Minor = \( \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 3 = -1 \) → Cofactor = \( 1 \)
3. For \( a_{13} = 2 \): Minor = \( \begin{vmatrix} 2 & 0 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot 3 = 4 \) → Cofactor = \( 4 \)
4. For \( a_{21} = 2 \): Minor = \( \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} = -1 \cdot 1 - 2 \cdot 2 = -5 \) → Cofactor = \( 5 \)
5. For \( a_{22} = 0 \): Minor = \( \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 3 = -5 \) → Cofactor = \( 0 \)
6. For \( a_{23} = 1 \): Minor = \( \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 - (-1) \cdot 3 = 5 \) → Cofactor = \( 5 \)
7. For \( a_{31} = 3 \): Minor = \( \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -1 \cdot 1 - 2 \cdot 0 = -1 \) → Cofactor = \( -1 \)
8. For \( a_{32} = 2 \): Minor = \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 2 = -3 \) → Cofactor = \( 3 \)
9. For \( a_{33} = 1 \): Minor = \( \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} = 1 \cdot 0 - (-1) \cdot 2 = 2 \) → Cofactor = \( 2 \)

The cofactor matrix is:
\[ C = \begin{pmatrix} -2 & 1 & 4 \\ 5 & 0 & 5 \\ -1 & 3 & 2 \end{pmatrix} \]

Taking the transpose gives us the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} -2 & 5 & -1 \\ 1 & 0 & 3 \\ 4 & 5 & 2 \end{pmatrix} \]

### Step 5: Calculate the inverse of \( A \)
Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]
Thus:
\[ A^{-1} = \frac{1}{5} \begin{pmatrix} -2 & 5 & -1 \\ 1 & 0 & 3 \\ 4 & 5 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{2}{5} & 1 & -\frac{1}{5} \\ \frac{1}{5} & 0 & \frac{3}{5} \\ \frac{4}{5} & 1 & \frac{2}{5} \end{pmatrix} \]

### Step 6: Multiply \( A^{-1} \) by \( B \)
Now we multiply \( A^{-1} \) by \( B \):
\[ X = A^{-1}B = \begin{pmatrix} -\frac{2}{5} & 1 & -\frac{1}{5} \\ \frac{1}{5} & 0 & \frac{3}{5} \\ \frac{4}{5} & 1 & \frac{2}{5} \end{pmatrix} \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \]

Calculating each element:
1. First row:
\[ -\frac{2}{5} \cdot 3 + 1 \cdot 1 - \frac{1}{5} \cdot 4 = -\frac{6}{5} + 1 - \frac{4}{5} = -\frac{10}{5} + 1 = -2 + 1 = -1 \]
2. Second row:
\[ \frac{1}{5} \cdot 3 + 0 \cdot 1 + \frac{3}{5} \cdot 4 = \frac{3}{5} + 0 + \frac{12}{5} = \frac{15}{5} = 3 \]
3. Third row:
\[ \frac{4}{5} \cdot 3 + 1 \cdot 1 + \frac{2}{5} \cdot 4 = \frac{12}{5} + 1 + \frac{8}{5} = \frac{12}{5} + \frac{5}{5} + \frac{8}{5} = \frac{25}{5} = 5 \]

Thus, we have:
\[ X = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} \]

### Final Answer
The solution for \( X \) is:
\[ X = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} \]