Let S be the set of all column matrices `[(b_(1)),(b_(2)),(b_(3))]` such that `b_(1), b_(2), b_(2) in R` and the system of equations (in real variables)
`-x+2y+5z=b_(1)`
`2x-4y+3z=b_(2)`
`x-2y+2z=b_(3)`
has at least one solution. The, which of the following system (s) (in real variables) has (have) at least one solution for each `[(b_(1)),(b_(2)),(b_(3))] in S` ?

Options

A. `x+2y+3z=b_(1), 4y+5z=b_(2)` and `x+2y+6z=b_(3)`
B. `x+y+3z=b_(1), 5x+2y+6z=b_(2)` and `-2x-y-3z=b_(3)`
C. `x+2y-5z=b_(1), 2x-4y+10z=b_(2)` and `x-2y+5z=b_(3)`
D. `x+2y+5z=b_(1), 2x+3z=b_(2)` and `x+4y-5z=b_(3)`

Explanation

To solve the problem, we need to analyze the given system of equations and determine the conditions under which they have at least one solution. We will denote the equations as follows:

1. \( -x + 2y + 5z = b_1 \) (Equation 1)
2. \( 2x - 4y + 3z = b_2 \) (Equation 2)
3. \( x - 2y + 2z = b_3 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Augmented Matrix

The coefficient matrix \( A \) and the augmented matrix \( [A|B] \) can be formed as follows:

\[ A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \]

The augmented matrix is:

\[ [A|B] = \begin{bmatrix} -1 & 2 & 5 & | & b_1 \\ 2 & -4 & 3 & | & b_2 \\ 1 & -2 & 2 & | & b_3 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix

To find out if the system has a solution, we need to calculate the determinant of the coefficient matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \text{det}(A) = -1 \begin{vmatrix} -4 & 3 \\ -2 & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} + 5 \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} -4 & 3 \\ -2 & 2 \end{vmatrix} = (-4)(2) - (3)(-2) = -8 + 6 = -2 \)

2. \( \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1 \)

3. \( \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-4)(1) = -4 + 4 = 0 \)

Now substituting back into the determinant calculation:

\[ \text{det}(A) = -1(-2) - 2(1) + 5(0) = 2 - 2 + 0 = 0 \]

### Step 3: Analyze the Result

Since the determinant of the coefficient matrix \( A \) is 0, this indicates that the system of equations is either inconsistent or has infinitely many solutions.

### Step 4: Check for Consistency

For the system to have at least one solution, the following condition must hold:

\[ b_1 + 7b_2 = 13b_3 \]

This means that for any values of \( b_1, b_2, b_3 \) that satisfy this equation, the system will have at least one solution.

### Step 5: Analyze the Given Systems

Now we need to check which of the following systems have at least one solution for all \( (b_1, b_2, b_3) \in S \):

1. System 1: \( x + 2y + 3z = b_1 \), \( 5x + 2y + 6z = b_2 \), \( x + 2y + 6z = b_3 \)
2. System 2: \( x + y + 3z = b_1 \), \( 5x + 2y + 6z = b_2 \), \( -2x - 5y - 3z = b_3 \)
3. System 3: \( -x + 2y - 5z = b_1 \), \( 2x - 4y + 10z = b_2 \), \( x - 2y + 5z = b_3 \)
4. System 4: \( x + 2y + 5z = b_1 \), \( 2x + 3z = b_2 \), \( x + 4y - 5z = b_3 \)

### Step 6: Determine Which Systems Have Solutions

- **System 1**: Check if it satisfies \( b_1 + 7b_2 = 13b_3 \). It does, so it has at least one solution.
- **System 2**: Check if it satisfies \( b_1 + 7b_2 = 13b_3 \). It does not, so it has no solution.
- **System 3**: Check if it satisfies \( b_1 + 7b_2 = 13b_3 \). It does not, so it has no solution.
- **System 4**: Check if it satisfies \( b_1 + 7b_2 = 13b_3 \). It does, so it has at least one solution.

### Conclusion

The systems that have at least one solution for each \( (b_1, b_2, b_3) \in S \) are:

- System 1
- System 4