Solve the equations by matrix method : (i) x+2y+z=7 x+2y+z=7 x+3z=11

Answer

To solve the given equations using the matrix method, we will follow these steps:

### Given Equations:
1. \( x + 2y + z = 7 \) (Equation 1)
2. \( x + 2y + z = 7 \) (Equation 2)
3. \( x + 3z = 11 \) (Equation 3)

### Step 1: Write the equations in matrix form
We can express the equations in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

From the given equations, we can identify:
- Coefficient matrix \( A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix} \)
- Variable matrix \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)
- Constant matrix \( B = \begin{bmatrix} 7 \\ 7 \\ 11 \end{bmatrix} \)

So, we have:
\[ AX = B \]

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

\[ |A| = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{vmatrix} \]

Calculating the determinant using the first row:
\[ |A| = 1 \cdot \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} = (2)(3) - (1)(0) = 6 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(1) = 2 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} = (1)(0) - (2)(1) = -2 \)

Substituting back into the determinant calculation:
\[ |A| = 1 \cdot 6 - 2 \cdot 2 + 1 \cdot (-2) = 6 - 4 - 2 = 0 \]

### Step 3: Analyze the determinant
Since the determinant \( |A| = 0 \), this indicates that the matrix \( A \) is singular and does not have an inverse.

### Step 4: Conclusion about the system of equations
When the determinant of the coefficient matrix is zero, it implies that the system of equations is either inconsistent or has infinitely many solutions. In this case, since the first two equations are identical and the third equation is different, the system is inconsistent.

### Final Answer:
The solution to the given system of equations does not exist.

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