I want to solve a matrix with parameters, basically I need to find when it has no solutions, when it has a single one and when it has an infinite amount of them, when it does I need to find the general solution. The matrix is 4 8 7 3c 3b, 1 2 2 c b, 2 4 2 (c-1) b

To solve the system of linear equations represented by the matrix with parameters, we'll analyze the augmented matrix and determine the conditions under which the system has no solution, a unique solution, or infinitely many solutions. Here is the given matrix:

\[
\begin{pmatrix}
4 & 8 & 7 & 3c & 3b \\
1 & 2 & 2 & c & b \\
2 & 4 & 2 & (c-1) & b 
\end{pmatrix}
\]

Let's write down the corresponding system of equations:
1. \( 4x + 8y + 7z = 3c \)
2. \( x + 2y + 2z = c \)
3. \( 2x + 4y + 2z = c - 1 \)

To find the solutions, we need to reduce the augmented matrix to row-echelon form (REF) or reduced row-echelon form (RREF). Let's start by representing the augmented matrix:

\[
\left[\begin{array}{ccc|c}
4 & 8 & 7 & 3c \\
1 & 2 & 2 & c \\
2 & 4 & 2 & c-1 
\end{array}\right]
\]

### Step 1: Row Operations
1. Divide the first row by 4:
   \[
   \left[\begin{array}{ccc|c}
   1 & 2 & 1.75 & 0.75c \\
   1 & 2 & 2 & c \\
   2 & 4 & 2 & c-1 
   \end{array}\right]
   \]

2. Subtract the first row from the second row:
   \[
   \left[\begin{array}{ccc|c}
   1 & 2 & 1.75 & 0.75c \\
   0 & 0 & 0.25 & 0.25c \\
   2 & 4 & 2 & c-1 
   \end{array}\right]
   \]

3. Subtract 2 times the first row from the third row:
   \[
   \left[\begin{array}{ccc|c}
   1 & 2 & 1.75 & 0.75c \\
   0 & 0 & 0.25 & 0.25c \\
   0 & 0 & -1.5 & -0.5c-1 
   \end{array}\right]
   \]

### Step 2: Analyze the Reduced Matrix
Let's focus on the second and third rows:
\[
\left[\begin{array}{ccc|c}
0 & 0 & 0.25 & 0.25c \\
0 & 0 & -1.5 & -0.5c-1 
\end{array}\right]
\]

Divide the second row by 0.25 and the third row by -1.5:
\[
\left[\begin{array}{ccc|c}
0 & 0 & 1 & c \\
0 & 0 & 1 & \frac{1 + 0.5c}{1.5} 
\end{array}\right]
\]

\[
\left[\begin{array}{ccc|c}
0 & 0 & 1 & c \\
0 & 0 & 1 & \frac{1 + 0.5c}{-1.5} 
\end{array}\right]
\]

Since the last two rows are inconsistent for any value of \( c \), the system has no solution.

### General Solution Analysis
Since the last two rows in the augmented matrix are inconsistent, it indicates that there is no value of \( c \) or \( b \) that can make the system consistent. Hence, the system has no solution under the given circumstances.

Would you like more details or have any questions? Here are 5 numbered questions you might want to ask:

1. Can you explain how to find the row-echelon form (REF) of a matrix?
2. How do I interpret the row-echelon form to determine the number of solutions?
3. What are the next steps if the system had a solution?
4. How do I find the general solution for a system with infinitely many solutions?
5. Can you demonstrate a specific example where the system has a unique solution?

**Tip:** When dealing with systems of linear equations, always check the consistency of the system by looking for rows in the augmented matrix that suggest contradictions, such as having a row with all zeros on the left and a non-zero entry on the right.

Explore how to solve systems of linear equations with parameters using an augmented matrix approach. Understand the conditions for no solution, unique solutions, and infinitely many solutions, with a detailed step-by-step solution provided. Suitable for advanced high school students and beyond.

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