The sum of three numbers is `-1`. If we multiply second number by `2`, third number by `3` and add them, we get `5`. If we subtract the third number from the sum of first and second numbers, we get `-1`. Represent it by a system of equation. Find the numbers, using inverse of a matrix.

Answer

To solve the problem step by step, we will first represent the given conditions as a system of equations, then we will use the inverse of a matrix to find the values of the three numbers.

### Step 1: Define the Variables
Let the three numbers be:
- \( x \) (first number)
- \( y \) (second number)
- \( z \) (third number)

### Step 2: Set Up the Equations
From the problem statement, we have the following equations based on the conditions given:

1. The sum of the three numbers is \(-1\):
\[ x + y + z = -1 \quad \text{(Equation 1)} \]

2. If we multiply the second number by \(2\) and the third number by \(3\) and add them, we get \(5\):
\[ 2y + 3z = 5 \quad \text{(Equation 2)} \]

3. If we subtract the third number from the sum of the first and second numbers, we get \(-1\):
\[ x + y - z = -1 \quad \text{(Equation 3)} \]

### Step 3: Write in Matrix Form
We can represent the above system of equations in the matrix form \(AX = B\), where:
- \( A \) is the coefficient matrix,
- \( X \) is the column matrix of variables,
- \( B \) is the column matrix of constants.

The system can be written as:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 5 \\ -1 \end{bmatrix} \]

### Step 4: Find the Inverse of Matrix A
To find the inverse of matrix \( A \), we first need to calculate the determinant of \( A \).

The determinant \( |A| \) is calculated as follows:
\[ |A| = 1 \cdot \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 3 \\ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} \]
Calculating the minors:
\[ = 1 \cdot (2 \cdot -1 - 3 \cdot 1) - 1 \cdot (0 \cdot -1 - 3 \cdot 1) + 1 \cdot (0 \cdot 1 - 2 \cdot 1) \]
\[ = 1 \cdot (-2 - 3) - 1 \cdot (-3) + 1 \cdot (-2) \]
\[ = -5 + 3 - 2 = -4 \]

Now, we find the adjoint of \( A \) and then the inverse:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \]

### Step 5: Calculate the Adjoint of A
The adjoint of \( A \) is the transpose of the cofactor matrix. After calculating the cofactors, we find:
\[ \text{adj}(A) = \begin{bmatrix} -5 & 2 & 1 \\ 3 & -2 & -3 \\ -2 & 0 & 2 \end{bmatrix} \]

Thus,
\[ A^{-1} = \frac{1}{-4} \begin{bmatrix} -5 & 2 & 1 \\ 3 & -2 & -3 \\ -2 & 0 & 2 \end{bmatrix} = \begin{bmatrix} \frac{5}{4} & -\frac{1}{2} & -\frac{1}{4} \\ -\frac{3}{4} & \frac{1}{2} & \frac{3}{4} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{bmatrix} \]

### Step 6: Multiply \( A^{-1} \) with \( B \)
Now we can find \( X \):
\[ X = A^{-1}B \]
\[ = \begin{bmatrix} \frac{5}{4} & -\frac{1}{2} & -\frac{1}{4} \\ -\frac{3}{4} & \frac{1}{2} & \frac{3}{4} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} -1 \\ 5 \\ -1 \end{bmatrix} \]

Calculating this product:
1. For \( x \):
\[ x = \frac{5}{4} \cdot -1 + -\frac{1}{2} \cdot 5 + -\frac{1}{4} \cdot -1 = -\frac{5}{4} - \frac{5}{2} + \frac{1}{4} = -\frac{5}{4} - \frac{10}{4} + \frac{1}{4} = -\frac{14}{4} = -\frac{7}{2} \]

2. For \( y \):
\[ y = -\frac{3}{4} \cdot -1 + \frac{1}{2} \cdot 5 + \frac{3}{4} \cdot -1 = \frac{3}{4} + \frac{5}{2} - \frac{3}{4} = \frac{5}{2} \]

3. For \( z \):
\[ z = \frac{1}{2} \cdot -1 + 0 \cdot 5 + -\frac{1}{2} \cdot -1 = -\frac{1}{2} + 0 + \frac{1}{2} = 0 \]

### Final Solution
Thus, the values of the three numbers are:
\[ x = -\frac{7}{2}, \quad y = \frac{5}{2}, \quad z = 0 \]