Solve the following system of equations by using matrix inversion method.
`x + y = 1, y + z = (5)/(3) and z + x = (4)/(3)`

Answer

To solve the system of equations using the matrix inversion method, we follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y = 1 \)
2. \( y + z = \frac{5}{3} \)
3. \( z + x = \frac{4}{3} \)

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

\[ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ \frac{5}{3} \\ \frac{4}{3} \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix A
To find the inverse of matrix \( A \), we first need to calculate its determinant.

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

### Step 3: Find the inverse of matrix A
Since \( \text{det}(A) \neq 0 \), the inverse exists. The formula for the inverse of a matrix is:

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

To find the adjugate of \( A \), we first calculate the cofactors and then take the transpose.

#### Cofactors of A:
- \( C_{11} = 1 \)
- \( C_{12} = -1 \)
- \( C_{13} = 1 \)
- \( C_{21} = -1 \)
- \( C_{22} = 1 \)
- \( C_{23} = -1 \)
- \( C_{31} = 1 \)
- \( C_{32} = -1 \)
- \( C_{33} = 1 \)

Thus, the cofactor matrix is:

\[ \text{Cof}(A) = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

Taking the transpose gives us the adjugate:

\[ \text{adj}(A) = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

Now, we can find the inverse:

\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \]

### Step 4: Multiply \( A^{-1} \) by \( B \)
Now we can find \( X \):

\[ X = A^{-1}B = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 \\ \frac{5}{3} \\ \frac{4}{3} \end{bmatrix} \]

Calculating the multiplication:

1. First row:
\[ \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{5}{3} + \frac{1}{2} \cdot \frac{4}{3} = \frac{1}{2} - \frac{5}{6} + \frac{2}{3} = \frac{1}{2} - \frac{5}{6} + \frac{4}{6} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} = \frac{1}{3} \]

2. Second row:
\[ -\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{5}{3} - \frac{1}{2} \cdot \frac{4}{3} = -\frac{1}{2} + \frac{5}{6} - \frac{2}{3} = -\frac{1}{2} + \frac{5}{6} - \frac{4}{6} = -\frac{1}{2} + \frac{1}{6} = -\frac{3}{6} + \frac{1}{6} = -\frac{2}{6} = -\frac{1}{3} \]

3. Third row:
\[ \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{5}{3} + \frac{1}{2} \cdot \frac{4}{3} = \frac{1}{2} - \frac{5}{6} + \frac{2}{3} = \frac{1}{2} - \frac{5}{6} + \frac{4}{6} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} = \frac{1}{3} \]

Thus, we have:

\[ X = \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ 1 \end{bmatrix} \]

### Final Result
The solution to the system of equations is:
- \( x = \frac{1}{3} \)
- \( y = \frac{2}{3} \)
- \( z = 1 \)