If `[(2,-1,3),(1,3,-1),(3,2,1)],[(x),(y),(z)]=[(9),(4),(10)]` ,then `[(x),(y),(z)]` is equal to

Options

A. `[(3),(2),(1)]`
B. `[(2),(3),(1)]`
C. `[(1),(2),(3)]`
D. `[(2),(1),(3)]`

Explanation

To solve the equation given by the matrix equation:

\[ \begin{pmatrix} 2 & -1 & 3 \\ 1 & 3 & -1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ 10 \end{pmatrix} \]

we will follow these steps:

### Step 1: Define the Matrices
Let \( A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 3 & -1 \\ 3 & 2 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 9 \\ 4 \\ 10 \end{pmatrix} \). We need to find the matrix \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \).

### Step 2: Find the Inverse of Matrix A
The solution can be found using the formula \( X = A^{-1}B \). First, we need to calculate the determinant of \( A \).

#### Determinant Calculation:
\[ \text{det}(A) = 2 \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = (3)(1) - (-1)(2) = 3 + 2 = 5 \)
2. \( \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = (1)(1) - (-1)(3) = 1 + 3 = 4 \)
3. \( \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = (1)(2) - (3)(3) = 2 - 9 = -7 \)

Now substituting back into the determinant formula:
\[ \text{det}(A) = 2(5) + 1(4) + 3(-7) = 10 + 4 - 21 = -7 \]

### Step 3: Calculate the Adjoint of Matrix A
Next, we need to find the adjoint of \( A \). The adjoint is the transpose of the cofactor matrix.

Calculating the cofactors:
1. For \( A_{11} \): \( C_{11} = \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 5 \)
2. For \( A_{12} \): \( C_{12} = -\begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = -4 \)
3. For \( A_{13} \): \( C_{13} = \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = -7 \)
4. For \( A_{21} \): \( C_{21} = -\begin{vmatrix} -1 & 3 \\ 2 & 1 \end{vmatrix} = -(-1 - 6) = 7 \)
5. For \( A_{22} \): \( C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = 2 - 9 = -7 \)
6. For \( A_{23} \): \( C_{23} = -\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = -4 + 3 = 1 \)
7. For \( A_{31} \): \( C_{31} = \begin{vmatrix} -1 & 3 \\ 3 & -1 \end{vmatrix} = 3 + 9 = 12 \)
8. For \( A_{32} \): \( C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -(-2 - 3) = 5 \)
9. For \( A_{33} \): \( C_{33} = \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = 6 + 1 = 7 \)

The cofactor matrix is:
\[ \begin{pmatrix} 5 & -4 & -7 \\ 7 & -7 & 1 \\ 12 & 5 & 7 \end{pmatrix} \]

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} 5 & 7 & 12 \\ -4 & -7 & 5 \\ -7 & 1 & 7 \end{pmatrix} \]

### Step 4: Calculate the Inverse of Matrix A
Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{-7} \begin{pmatrix} 5 & 7 & 12 \\ -4 & -7 & 5 \\ -7 & 1 & 7 \end{pmatrix} \]

### Step 5: Multiply \( A^{-1} \) by B
Now we compute \( X = A^{-1}B \):
\[ X = \frac{1}{-7} \begin{pmatrix} 5 & 7 & 12 \\ -4 & -7 & 5 \\ -7 & 1 & 7 \end{pmatrix} \begin{pmatrix} 9 \\ 4 \\ 10 \end{pmatrix} \]

Calculating the product:
1. First row: \( 5(9) + 7(4) + 12(10) = 45 + 28 + 120 = 193 \)
2. Second row: \( -4(9) + -7(4) + 5(10) = -36 - 28 + 50 = -14 \)
3. Third row: \( -7(9) + 1(4) + 7(10) = -63 + 4 + 70 = 11 \)

Thus,
\[ X = \frac{1}{-7} \begin{pmatrix} 193 \\ -14 \\ 11 \end{pmatrix} = \begin{pmatrix} -\frac{193}{7} \\ 2 \\ -\frac{11}{7} \end{pmatrix} \]

### Final Answer
The values of \( x, y, z \) are:
\[ x = -\frac{193}{7}, \quad y = 2, \quad z = -\frac{11}{7} \]