The following system of equations `5x-7y+3z=3, 5x+y+3z=7 and 5x+3y+2z=5` is

Options

A. Consistent with trivial solution
B. Consistent with a unique non trivial solution
C. Consistent with infinite solutions
D. Inconsistent with no solution

Answer

To solve the system of equations:

1. **Equations**:
- \( 5x - 7y + 3z = 3 \) (Equation 1)
- \( 5x + y + 3z = 7 \) (Equation 2)
- \( 5x + 3y + 2z = 5 \) (Equation 3)

2. **Form the Coefficient Matrix (A)**:
The coefficient matrix \( A \) is formed from the coefficients of \( x, y, z \) in the equations:
\[ A = \begin{bmatrix} 5 & -7 & 3 \\ 5 & 1 & 3 \\ 5 & 3 & 2 \end{bmatrix} \]

3. **Form the Augmented Matrix (A|B)**:
The augmented matrix combines the coefficient matrix with the constants from the right-hand side of the equations:
\[ A|B = \begin{bmatrix} 5 & -7 & 3 & | & 3 \\ 5 & 1 & 3 & | & 7 \\ 5 & 3 & 2 & | & 5 \end{bmatrix} \]

4. **Finding the Rank of Matrix A**:
To determine the rank, we will perform row operations to bring the matrix to row echelon form.

- Subtract the first row from the second and third rows:
\[ R_2 = R_2 - R_1 \Rightarrow \begin{bmatrix} 5 & -7 & 3 \\ 0 & 8 & 0 \\ 0 & 10 & -1 \end{bmatrix} \]
\[ R_3 = R_3 - R_1 \Rightarrow \begin{bmatrix} 5 & -7 & 3 \\ 0 & 8 & 0 \\ 0 & 10 & -1 \end{bmatrix} \]

- Now, we can simplify further by dividing the second row by 8:
\[ R_2 = \frac{1}{8}R_2 \Rightarrow \begin{bmatrix} 5 & -7 & 3 \\ 0 & 1 & 0 \\ 0 & 10 & -1 \end{bmatrix} \]

- Next, we can eliminate the second element of the third row:
\[ R_3 = R_3 - 10R_2 \Rightarrow \begin{bmatrix} 5 & -7 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \]

5. **Determining the Rank**:
The rank of matrix \( A \) is the number of non-zero rows in the row echelon form. Here, we have 3 non-zero rows, so:
\[ \text{Rank}(A) = 3 \]

6. **Conclusion**:
Since the rank of the coefficient matrix \( A \) is equal to the number of variables (3), the system of equations has a unique solution.

7. **Final Answer**:
The system of equations is consistent with a unique solution.

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