https://file.aiquickdraw.com/mathbot/file/1728922027167l1rsa3eu.jpg

The system of linear equations given in your image is:

\[
\begin{aligned}
4x + 1y + 1z &= 3 \\
8x - 2y + 2z &= 3 \\
12x - 1y + 3z &= 6
\end{aligned}
\]

You are asked to form an augmented matrix from this system and then reduce it to find the solution in terms of \( z \). Let's start by writing the augmented matrix.

### Step 1: Write the Augmented Matrix

From the equations, we can create the augmented matrix:

\[
\begin{pmatrix}
4 & 1 & 1 & | & 3 \\
8 & -2 & 2 & | & 3 \\
12 & -1 & 3 & | & 6
\end{pmatrix}
\]

This matrix is made up of the coefficients of the variables \( x \), \( y \), and \( z \), with the constants on the right side forming the last column.

### Step 2: Row Reduction to Reduced Row Echelon Form (RREF)

We aim to perform elementary row operations to reduce this matrix to a simpler form.

Starting matrix:
\[
\begin{pmatrix}
4 & 1 & 1 & | & 3 \\
8 & -2 & 2 & | & 3 \\
12 & -1 & 3 & | & 6
\end{pmatrix}
\]

**Row 1 (R1)**: Divide the first row by 4 to make the pivot element 1:
\[
R1 \to \frac{1}{4}R1 = \begin{pmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{3}{4} \end{pmatrix}
\]

Updated matrix:
\[
\begin{pmatrix}
1 & \frac{1}{4} & \frac{1}{4} & | & \frac{3}{4} \\
8 & -2 & 2 & | & 3 \\
12 & -1 & 3 & | & 6
\end{pmatrix}
\]

**Row 2 (R2)**: Subtract 8 times the first row from the second row to eliminate the first entry of the second row:
\[
R2 \to R2 - 8R1 = \begin{pmatrix} 0 & -4 & 0 & | & -3 \end{pmatrix}
\]

**Row 3 (R3)**: Subtract 12 times the first row from the third row:
\[
R3 \to R3 - 12R1 = \begin{pmatrix} 0 & -4 & 0 & | & -3 \end{pmatrix}
\]

The matrix now looks like this:
\[
\begin{pmatrix}
1 & \frac{1}{4} & \frac{1}{4} & | & \frac{3}{4} \\
0 & -4 & 0 & | & -3 \\
0 & -4 & 0 & | & -3
\end{pmatrix}
\]

Notice that rows 2 and 3 are the same, meaning this system has dependent equations, and thus, an infinite number of solutions.

**Row 2 (R2)**: Divide by -4 to simplify:
\[
R2 \to \frac{1}{-4}R2 = \begin{pmatrix} 0 & 1 & 0 & | & \frac{3}{4} \end{pmatrix}
\]

The matrix is now:
\[
\begin{pmatrix}
1 & \frac{1}{4} & \frac{1}{4} & | & \frac{3}{4} \\
0 & 1 & 0 & | & \frac{3}{4} \\
0 & -4 & 0 & | & -3
\end{pmatrix}
\]

Thus, the solution involves a free parameter \( z \), as this system is dependent.

Would you like further clarification on the steps or a more detailed breakdown of the solution? Here are five follow-up questions:

1. Would you like to see the final solution written in terms of \( z \)?
2. Should I explain why this system has infinite solutions?
3. Do you want more examples of solving systems with infinite solutions?
4. Would you like a more detailed explanation of row reduction?
5. Do you need help interpreting the meaning of dependent equations?

**Tip:** When dealing with systems of linear equations, if two rows in the augmented matrix are multiples of each other, it usually indicates that the system has infinitely many solutions or is dependent.

Form an augmented matrix for the system of equations, reduce it, and express the solution in terms of z: 
4x + y + z = 3
8x - 2y + 2z = 3
12x - y + 3z = 6.

This problem involves forming an augmented matrix from a system of linear equations and using row reduction to find the solution in terms of z. Explore concepts such as RREF, dependent equations, and infinite solutions.

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