By the method of matrix inversion, solve the system.
`[(1,1,1),(2,5,7),(2,1,-1)][(x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3))]=[(9,2),(52,15),(0,-1)]`

Explanation

To solve the system of equations using the method of matrix inversion, we start with the given matrix equation:

\[ A \cdot X = B \]

Where:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad B = \begin{pmatrix} 9 \\ 52 \\ 0 \end{pmatrix} \]

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

To find the inverse of matrix A, we first need to calculate its determinant.

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} = (5)(-1) - (7)(1) = -5 - 7 = -12\)
2. \(\begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (7)(2) = -2 - 14 = -16\)
3. \(\begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} = (2)(1) - (5)(2) = 2 - 10 = -8\)

Now substituting back:

\[ \text{det}(A) = 1 \cdot (-12) - 1 \cdot (-16) + 1 \cdot (-8) = -12 + 16 - 8 = -4 \]

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

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

Calculating the cofactors:

1. For \(C_{11}\): \(\begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} = -12\)
2. For \(C_{12}\): \(-\begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} = 16\)
3. For \(C_{13}\): \(\begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} = -8\)

4. For \(C_{21}\): \(-\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-2) = 2\)
5. For \(C_{22}\): \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -3\)
6. For \(C_{23}\): \(-\begin{vmatrix} 1 & 1 \\ 2 & 5 \end{vmatrix} = -3\)

7. For \(C_{31}\): \(\begin{vmatrix} 1 & 1 \\ 5 & 7 \end{vmatrix} = -2\)
8. For \(C_{32}\): \(-\begin{vmatrix} 1 & 1 \\ 2 & 7 \end{vmatrix} = -5\)
9. For \(C_{33}\): \(\begin{vmatrix} 1 & 1 \\ 2 & 5 \end{vmatrix} = -3\)

The cofactor matrix is:

\[ C = \begin{pmatrix} -12 & 16 & -8 \\ 2 & -3 & -3 \\ -2 & -5 & -3 \end{pmatrix} \]

Taking the transpose to find the adjoint:

\[ \text{adj}(A) = C^T = \begin{pmatrix} -12 & 2 & -2 \\ 16 & -3 & -5 \\ -8 & -3 & -3 \end{pmatrix} \]

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

Using the formula for the inverse:

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

Substituting the values:

\[ A^{-1} = \frac{1}{-4} \cdot \begin{pmatrix} -12 & 2 & -2 \\ 16 & -3 & -5 \\ -8 & -3 & -3 \end{pmatrix} = \begin{pmatrix} 3 & -0.5 & 0.5 \\ -4 & 0.75 & 1.25 \\ 2 & 0.75 & 0.75 \end{pmatrix} \]

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

Now we compute \(X = A^{-1} \cdot B\):

\[ B = \begin{pmatrix} 9 \\ 52 \\ 0 \end{pmatrix} \]

Calculating the product:

\[ X = \begin{pmatrix} 3 & -0.5 & 0.5 \\ -4 & 0.75 & 1.25 \\ 2 & 0.75 & 0.75 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 52 \\ 0 \end{pmatrix} \]

Calculating each component:

1. \(x_1 = 3 \cdot 9 + (-0.5) \cdot 52 + 0.5 \cdot 0 = 27 - 26 = 1\)
2. \(x_2 = -4 \cdot 9 + 0.75 \cdot 52 + 1.25 \cdot 0 = -36 + 39 = 3\)
3. \(x_3 = 2 \cdot 9 + 0.75 \cdot 52 + 0.75 \cdot 0 = 18 + 39 = 57\)

Thus, we have:

\[ X = \begin{pmatrix} 1 \\ 3 \\ 57 \end{pmatrix} \]

### Conclusion

The solution to the system of equations is:

\[ x_1 = 1, \quad x_2 = 3, \quad x_3 = 57 \]