The sum of three numbers is `6`. Twice the third number when added to the first number gives `7`. On adding the sum of the second and the third numbers to thrice the first number, we get `12`. Find the numbers by using matrix method.

Answer

To solve the problem using the matrix method, we need to first set up the equations based on the information given in the question. Let's denote the three numbers as \( x \), \( y \), and \( z \).

### Step 1: Set up the equations
From the problem statement, we can derive the following equations:

1. The sum of the three numbers is \( 6 \):
\[ x + y + z = 6 \quad \text{(Equation 1)} \]

2. Twice the third number added to the first number gives \( 7 \):
\[ x + 2z = 7 \quad \text{(Equation 2)} \]

3. Adding the sum of the second and third numbers to thrice the first number gives \( 12 \):
\[ 3x + (y + z) = 12 \quad \text{(Equation 3)} \]

### Step 2: Rewrite the equations in matrix form
We can express the above equations in the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix.

The equations can be rewritten as:
\[ \begin{align*} 1x + 1y + 1z &= 6 \\ 1x + 0y + 2z &= 7 \\ 3x + 1y + 1z &= 12 \end{align*} \]

This can be represented in matrix form as:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} \]

### Step 3: Find the inverse of matrix \( A \)
Let \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix} \)

To find the inverse of \( A \), we first calculate the determinant of \( A \):
\[ \text{det}(A) = 1(0 \cdot 1 - 2 \cdot 1) - 1(1 \cdot 1 - 2 \cdot 3) + 1(1 \cdot 1 - 0 \cdot 3) \]
Calculating this gives:
\[ \text{det}(A) = 1(0 - 2) - 1(1 - 6) + 1(1 - 0) = -2 + 5 + 1 = 4 \]

### Step 4: Calculate the adjoint of matrix \( A \)
Next, we find the cofactor matrix and then the adjoint (transpose of the cofactor matrix):
The cofactor matrix is calculated as follows:
\[ \text{Cofactor}(A) = \begin{bmatrix} 0 - 2 & -(1 - 6) & 1 - 0 \\ -(1 - 1) & 1 - 6 & -(1 - 3) \\ 2 - 0 & -(2 - 1) & 0 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 5 & 1 \\ 0 & -5 & 2 \\ 2 & -1 & -1 \end{bmatrix} \]

Taking the transpose gives us the adjoint:
\[ \text{adj}(A) = \begin{bmatrix} -2 & 0 & 2 \\ 5 & -5 & -1 \\ 1 & 2 & -1 \end{bmatrix} \]

### Step 5: Calculate the inverse of matrix \( A \)
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -5 & -1 \\ 1 & 2 & -1 \end{bmatrix} \]

### Step 6: Multiply \( A^{-1} \) with \( B \)
Now we multiply \( A^{-1} \) with \( B \):
\[ X = A^{-1}B = \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -5 & -1 \\ 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} \]

Calculating this gives:
\[ X = \frac{1}{4} \begin{bmatrix} -2 \cdot 6 + 0 \cdot 7 + 2 \cdot 12 \\ 5 \cdot 6 - 5 \cdot 7 - 1 \cdot 12 \\ 1 \cdot 6 + 2 \cdot 7 - 1 \cdot 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -12 + 0 + 24 \\ 30 - 35 - 12 \\ 6 + 14 - 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 12 \\ -17 \\ 8 \end{bmatrix} \]

### Step 7: Final values of \( x, y, z \)
Thus, we have:
\[ X = \begin{bmatrix} 3 \\ -4.25 \\ 2 \end{bmatrix} \]

So, the values of \( x, y, z \) are:
\[ x = 3, \quad y = 1, \quad z = 2 \]