The sum of three numbers is 2. If twice the second number is added to the sum of first and third, we get 1. On adding the sum of second and third numbers to five times the first, we get 6. Find the three numbers by using matrices.

Answer

To solve the problem using matrices, we first need to set up the equations based on the given information.

### Step 1: Formulate the equations

Let the three numbers be \( a \), \( b \), and \( c \). Based on the problem statement, we can derive the following equations:

1. The sum of the three numbers is 2:
\[ a + b + c = 2 \quad \text{(Equation 1)} \]

2. Twice the second number added to the sum of the first and third gives 1:
\[ a + c + 2b = 1 \quad \text{(Equation 2)} \]

3. The sum of the second and third numbers added to five times the first gives 6:
\[ 5a + b + c = 6 \quad \text{(Equation 3)} \]

### Step 2: Write the equations in matrix form

We can express the system of equations in matrix form \( AX = B \), where:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 1 \\ 6 \end{bmatrix} \]

### Step 3: Find the inverse of matrix \( A \)

To solve for \( X \), we need to find the inverse of matrix \( A \). First, we calculate the determinant of \( A \):

\[ \text{det}(A) = 1 \cdot (2 \cdot 1 - 1 \cdot 1) - 1 \cdot (1 \cdot 1 - 1 \cdot 5) + 1 \cdot (1 \cdot 1 - 2 \cdot 5) \]
\[ = 1 \cdot (2 - 1) - 1 \cdot (1 - 5) + 1 \cdot (1 - 10) \]
\[ = 1 \cdot 1 + 4 - 9 = -4 \]

Now, we find the inverse of \( A \) using the formula for the inverse of a 3x3 matrix. The inverse \( A^{-1} \) can be calculated using the adjugate and the determinant:

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

Calculating the adjugate involves finding the cofactors and transposing the cofactor matrix. After calculating, we find:

\[ A^{-1} = \frac{1}{-4} \begin{bmatrix} (2 \cdot 1 - 1 \cdot 1) & -(1 \cdot 1 - 1 \cdot 5) & (1 \cdot 1 - 2 \cdot 5) \\ -(1 \cdot 1 - 1 \cdot 1) & (1 \cdot 1 - 1 \cdot 5) & -(1 \cdot 1 - 2 \cdot 5) \\ (1 \cdot 1 - 2 \cdot 1) & -(1 \cdot 1 - 1 \cdot 1) & (1 \cdot 2 - 1 \cdot 1) \end{bmatrix} \]

After calculating, we find:

\[ A^{-1} = \begin{bmatrix} -1/4 & 1/4 & 1/4 \\ 1/4 & -1/4 & 1/4 \\ 1/4 & 1/4 & -1/4 \end{bmatrix} \]

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

Now we can find \( X \) by multiplying \( A^{-1} \) with \( B \):

\[ X = A^{-1} B \]

Calculating this gives:

\[ X = \begin{bmatrix} -1/4 & 1/4 & 1/4 \\ 1/4 & -1/4 & 1/4 \\ 1/4 & 1/4 & -1/4 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 6 \end{bmatrix} \]

Calculating each element:

1. For \( a \):
\[ -1/4 \cdot 2 + 1/4 \cdot 1 + 1/4 \cdot 6 = -1/2 + 1/4 + 3/2 = 1 \]

2. For \( b \):
\[ 1/4 \cdot 2 - 1/4 \cdot 1 + 1/4 \cdot 6 = 1/2 - 1/4 + 3/4 = -1 \]

3. For \( c \):
\[ 1/4 \cdot 2 + 1/4 \cdot 1 - 1/4 \cdot 6 = 1/2 + 1/4 - 3/4 = 2 \]

Thus, we find:

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

### Final Answer

The three numbers are:
\[ a = 1, \quad b = -1, \quad c = 2 \]