The solution of the system of equations is `x-y+2z=1,2y-3z=1` and `3x-2y+4y=2` is

Options

A. `x=1, y=5` and `z=3`
B. `x=0, y=5` and `z=3`
C. `x=0, y=5` and `z=-3`
D. None of the above

Explanation

To solve the system of equations given by:

1. \( x - y + 2z = 1 \) (Equation 1)
2. \( 2y - 3z = 1 \) (Equation 2)
3. \( 3x - 2y + 4z = 2 \) (Equation 3)

we can represent this system in matrix form \( AX = B \), where:

\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]

### Step 1: Find the Determinant of Matrix A

To find the inverse of matrix \( A \), we first need to calculate its determinant \( \text{det}(A) \).

\[ \text{det}(A) = 1 \cdot (2 \cdot 4 - (-3) \cdot (-2)) - (-1) \cdot (0 \cdot 4 - (-3) \cdot 3) + 2 \cdot (0 \cdot (-2) - 2 \cdot 3) \]

Calculating each term:

1. \( 1 \cdot (8 - 6) = 1 \cdot 2 = 2 \)
2. \( 1 \cdot (0 - 9) = -9 \) (since we have a negative sign from the equation)
3. \( 2 \cdot (0 - 6) = -12 \)

Now, summing these results:

\[ \text{det}(A) = 2 + 9 - 12 = -1 \]

### Step 2: Find the Adjoint of Matrix A

Next, we need to find the adjoint of matrix \( A \). The adjoint is found by calculating the cofactor matrix and then taking its transpose.

The cofactor matrix is calculated as follows:

1. For element \( a_{11} \): \( (-1)^{1+1} \cdot \text{det} \begin{pmatrix} 2 & -3 \\ -2 & 4 \end{pmatrix} = 1 \cdot (2 \cdot 4 - (-3)(-2)) = 8 - 6 = 2 \)
2. For element \( a_{12} \): \( (-1)^{1+2} \cdot \text{det} \begin{pmatrix} 0 & -3 \\ 3 & 4 \end{pmatrix} = -1 \cdot (0 \cdot 4 - (-3)(3)) = -(-9) = 9 \)
3. For element \( a_{13} \): \( (-1)^{1+3} \cdot \text{det} \begin{pmatrix} 0 & 2 \\ 3 & -2 \end{pmatrix} = 1 \cdot (0 \cdot (-2) - 2 \cdot 3) = -6 \)

Continuing this process for all elements, we find:

\[ \text{Cofactor matrix} = \begin{pmatrix} 2 & 9 & -6 \\ -6 & 4 & 3 \\ 6 & 3 & 2 \end{pmatrix} \]

Now, taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} 2 & -6 & 6 \\ 9 & 4 & 3 \\ -6 & 3 & 2 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A

The inverse of matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = -1 \cdot \begin{pmatrix} 2 & -6 & 6 \\ 9 & 4 & 3 \\ -6 & 3 & 2 \end{pmatrix} \]

Thus,

\[ A^{-1} = \begin{pmatrix} -2 & 6 & -6 \\ -9 & -4 & -3 \\ 6 & -3 & -2 \end{pmatrix} \]

### Step 4: Multiply \( A^{-1} \) by \( B \)

Now we can find \( X \) by multiplying \( A^{-1} \) with \( B \):

\[ X = A^{-1}B = \begin{pmatrix} -2 & 6 & -6 \\ -9 & -4 & -3 \\ 6 & -3 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]

Calculating the product:

1. First element: \( -2 \cdot 1 + 6 \cdot 1 - 6 \cdot 2 = -2 + 6 - 12 = -8 \)
2. Second element: \( -9 \cdot 1 - 4 \cdot 1 - 3 \cdot 2 = -9 - 4 - 6 = -19 \)
3. Third element: \( 6 \cdot 1 - 3 \cdot 1 - 2 \cdot 2 = 6 - 3 - 4 = -1 \)

Thus,

\[ X = \begin{pmatrix} -8 \\ -19 \\ -1 \end{pmatrix} \]

### Final Result

The solution to the system of equations is:

\[ x = -8, \quad y = -19, \quad z = -1 \]