Solve the following equations using inverse of a matrix.
`x+y=5`, `y+z=3`, `z+x=4`

Answer

To solve the system of equations using the inverse of a matrix, we follow these steps:

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

We can express this system in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the column matrix of variables,
- \( B \) is the column matrix of constants.

From the equations, we can identify:
\[ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 3 \\ 4 \end{bmatrix} \]

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

\[ |A| = 1 \cdot (1 \cdot 1 - 1 \cdot 0) - 1 \cdot (0 \cdot 1 - 1 \cdot 1) + 0 \cdot (0 \cdot 1 - 1 \cdot 1) \]
Calculating this step-by-step:
\[ = 1 \cdot 1 - 1 \cdot (-1) + 0 \]
\[ = 1 + 1 = 2 \]

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

The cofactor matrix \( C \) is calculated as follows:
\[ C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]
Where each \( C_{ij} \) is calculated by taking the determinant of the submatrix formed by deleting the \( i \)-th row and \( j \)-th column, multiplied by \( (-1)^{i+j} \).

Calculating each cofactor:
- \( C_{11} = | \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} | = 1 \)
- \( C_{12} = -| \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} | = -(-1) = 1 \)
- \( C_{13} = | \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} | = -1 \)

- \( C_{21} = -| \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} | = -1 \)
- \( C_{22} = | \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} | = 1 \)
- \( C_{23} = -| \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} | = -(-1) = 1 \)

- \( C_{31} = | \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} | = -1 \)
- \( C_{32} = -| \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} | = -1 \)
- \( C_{33} = | \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} | = 1 \)

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 1 \\ -1 & -1 & 1 \end{bmatrix} \]

Now, we take the transpose of this matrix to get the adjoint:
\[ \text{Adj}(A) = \begin{bmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix} \]

### Step 4: Calculate the inverse of matrix \( A \)
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A) \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix} \]

### Step 5: Multiply \( A^{-1} \) by \( B \) to find \( X \)
Now we can find \( X \) by multiplying \( A^{-1} \) with \( B \):
\[ X = A^{-1}B = \frac{1}{2} \begin{bmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \\ 4 \end{bmatrix} \]

Calculating the multiplication:
1. First row: \( \frac{1}{2} (1 \cdot 5 + (-1) \cdot 3 + (-1) \cdot 4) = \frac{1}{2} (5 - 3 - 4) = \frac{1}{2} (-2) = -1 \)
2. Second row: \( \frac{1}{2} (1 \cdot 5 + 1 \cdot 3 + (-1) \cdot 4) = \frac{1}{2} (5 + 3 - 4) = \frac{1}{2} (4) = 2 \)
3. Third row: \( \frac{1}{2} ((-1) \cdot 5 + 1 \cdot 3 + 1 \cdot 4) = \frac{1}{2} (-5 + 3 + 4) = \frac{1}{2} (2) = 1 \)

Thus, we have:
\[ X = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \]

### Step 6: Interpret the results
From the matrix \( X \), we find:
- \( x = -1 \)
- \( y = 2 \)
- \( z = 1 \)

### Final Answer
The solution to the system of equations is:
\[ x = 3, \quad y = 2, \quad z = 1 \]