The values of x, y, z for the equations `5x-y+4z=5, 2x+3y+5z=2, 5x-2y+6z=1` are

Options

A. `x=3, y=3, z=-2`
B. `x=1, y=2, z=1`
C. `x=1, y=3, z=5`
D. `x=3, y=2, z=-1`

Answer

To solve the system of equations given by:

1. \( 5x - y + 4z = 5 \)
2. \( 2x + 3y + 5z = 2 \)
3. \( 5x - 2y + 6z = 1 \)

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

\[ A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix} \]

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

To find \( X \), we first need to calculate the determinant of matrix \( A \).

\[ \text{det}(A) = 5 \cdot (3 \cdot 6 - 5 \cdot (-2)) - (-1) \cdot (2 \cdot 6 - 5 \cdot 5) + 4 \cdot (2 \cdot (-2) - 3 \cdot 5) \]

Calculating each term:

1. \( 5 \cdot (18 + 10) = 5 \cdot 28 = 140 \)
2. \( -(-1) \cdot (12 - 25) = 1 \cdot (-13) = -13 \)
3. \( 4 \cdot (-4 - 15) = 4 \cdot (-19) = -76 \)

Putting it all together:

\[ \text{det}(A) = 140 - 13 - 76 = 51 \]

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

Next, we need to find the cofactor matrix of \( A \) and then take its transpose to get the adjoint.

Calculating the cofactors:

1. The cofactor for \( a_{11} \) (5): \( C_{11} = 3 \cdot 6 - 5 \cdot (-2) = 28 \)
2. The cofactor for \( a_{12} \) (-1): \( C_{12} = -(2 \cdot 6 - 5 \cdot 5) = 13 \)
3. The cofactor for \( a_{13} \) (4): \( C_{13} = 2 \cdot (-2) - 3 \cdot 5 = -19 \)
4. The cofactor for \( a_{21} \) (2): \( C_{21} = -(5 \cdot 6 - 4 \cdot (-2)) = -38 \)
5. The cofactor for \( a_{22} \) (3): \( C_{22} = 5 \cdot 6 - 4 \cdot 5 = 10 \)
6. The cofactor for \( a_{23} \) (5): \( C_{23} = 5 \cdot (-2) - 5 \cdot 5 = -35 \)
7. The cofactor for \( a_{31} \) (5): \( C_{31} = 2 \cdot 3 - 5 \cdot (-1) = 11 \)
8. The cofactor for \( a_{32} \) (-2): \( C_{32} = 5 \cdot 4 - 2 \cdot 5 = 10 \)
9. The cofactor for \( a_{33} \) (6): \( C_{33} = 5 \cdot (-1) - 2 \cdot 3 = -11 \)

The cofactor matrix is:

\[ \text{Cof}(A) = \begin{bmatrix} 28 & -13 & -19 \\ -38 & 10 & -35 \\ 11 & 10 & -11 \end{bmatrix} \]

Taking the transpose to get the adjoint:

\[ \text{adj}(A) = \begin{bmatrix} 28 & -38 & 11 \\ -13 & 10 & 10 \\ -19 & -35 & -11 \end{bmatrix} \]

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

Now, we can find the inverse of matrix \( A \):

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{51} \cdot \begin{bmatrix} 28 & -38 & 11 \\ -13 & 10 & 10 \\ -19 & -35 & -11 \end{bmatrix} \]

### Step 4: Solve for X

Now we can find \( X \):

\[ X = A^{-1}B = \frac{1}{51} \begin{bmatrix} 28 & -38 & 11 \\ -13 & 10 & 10 \\ -19 & -35 & -11 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix} \]

Calculating the product:

1. First row: \( 28 \cdot 5 - 38 \cdot 2 + 11 \cdot 1 = 140 - 76 + 11 = 75 \)
2. Second row: \( -13 \cdot 5 + 10 \cdot 2 + 10 \cdot 1 = -65 + 20 + 10 = -35 \)
3. Third row: \( -19 \cdot 5 - 35 \cdot 2 - 11 \cdot 1 = -95 - 70 - 11 = -176 \)

Thus,

\[ X = \frac{1}{51} \begin{bmatrix} 75 \\ -35 \\ -176 \end{bmatrix} = \begin{bmatrix} \frac{75}{51} \\ \frac{-35}{51} \\ \frac{-176}{51} \end{bmatrix} \]

### Final Values

So the values of \( x, y, z \) are:

\[ x = \frac{75}{51}, \quad y = \frac{-35}{51}, \quad z = \frac{-176}{51} \]