2a + 3b + 1c = 7
a  -   b + 2c = 4
3a + 2b + 4c = 10
- Represent it in the form of an augmented matrix [A | b]
- Apply EOR to the matrix [A | b] to obtain a matrix in the form of RREF
- How is the solution to the SLE?

To solve the system of linear equations given by:

1. \( 2a + 3b + 1c = 7 \)
2. \(  a - b + 2c = 4 \)
3. \( 3a + 2b + 4c = 10 \)

we will follow these steps:

### 1. Represent it in the form of an augmented matrix \([A | b]\)

The augmented matrix \([A | b]\) is formed by writing the coefficients of the variables \(a\), \(b\), and \(c\) in the first three columns and the constants on the right-hand side of the equations in the last column.

\[
[A | b] = \begin{bmatrix}
2 & 3 & 1 & | & 7 \\
1 & -1 & 2 & | & 4 \\
3 & 2 & 4 & | & 10
\end{bmatrix}
\]

### 2. Apply Elementary Row Operations (EOR) to obtain the matrix in Row-Reduced Echelon Form (RREF)

To put the matrix into RREF, we perform a series of EORs:

**Step 1: Make the pivot of the first column 1**

Divide the first row by 2:

\[
\begin{bmatrix}
1 & 1.5 & 0.5 & | & 3.5 \\
1 & -1 & 2 & | & 4 \\
3 & 2 & 4 & | & 10
\end{bmatrix}
\]

**Step 2: Eliminate the first entry in the second and third rows**

Subtract the first row from the second row:

\[
R_2 = R_2 - R_1 \Rightarrow \begin{bmatrix}
1 & 1.5 & 0.5 & | & 3.5 \\
0 & -2.5 & 1.5 & | & 0.5 \\
3 & 2 & 4 & | & 10
\end{bmatrix}
\]

Subtract 3 times the first row from the third row:

\[
R_3 = R_3 - 3R_1 \Rightarrow \begin{bmatrix}
1 & 1.5 & 0.5 & | & 3.5 \\
0 & -2.5 & 1.5 & | & 0.5 \\
0 & -2.5 & 2.5 & | & -0.5
\end{bmatrix}
\]

**Step 3: Make the pivot of the second column in the second row -1**

Divide the second row by \(-2.5\):

\[
\begin{bmatrix}
1 & 1.5 & 0.5 & | & 3.5 \\
0 & 1 & -0.6 & | & -0.2 \\
0 & -2.5 & 2.5 & | & -0.5
\end{bmatrix}
\]

**Step 4: Eliminate the second entry in the first and third rows**

Subtract 1.5 times the second row from the first row:

\[
R_1 = R_1 - 1.5R_2 \Rightarrow \begin{bmatrix}
1 & 0 & 1.4 & | & 3.8 \\
0 & 1 & -0.6 & | & -0.2 \\
0 & -2.5 & 2.5 & | & -0.5
\end{bmatrix}
\]

Add 2.5 times the second row to the third row:

\[
R_3 = R_3 + 2.5R_2 \Rightarrow \begin{bmatrix}
1 & 0 & 1.4 & | & 3.8 \\
0 & 1 & -0.6 & | & -0.2 \\
0 & 0 & 1 & | & 0
\end{bmatrix}
\]

**Step 5: Make the pivot of the third column in the third row 1**

Since it's already 1, we skip this step.

**Step 6: Eliminate the third entry in the first and second rows**

Subtract 1.4 times the third row from the first row:

\[
R_1 = R_1 - 1.4R_3 \Rightarrow \begin{bmatrix}
1 & 0 & 0 & | & 3.8 \\
0 & 1 & -0.6 & | & -0.2 \\
0 & 0 & 1 & | & 0
\end{bmatrix}
\]

Add 0.6 times the third row to the second row:

\[
R_2 = R_2 + 0.6R_3 \Rightarrow \begin{bmatrix}
1 & 0 & 0 & | & 3.8 \\
0 & 1 & 0 & | & 0.1 \\
0 & 0 & 1 & | & 0
\end{bmatrix}
\]

### 3. How is the solution to the system of linear equations?

From the final RREF matrix, we can see the solution directly:

- \( a = 3.8 \)
- \( b = 0.1 \)
- \( c = 0 \)

Thus, the solution to the system of linear equations is \( a = 3.8 \), \( b = 0.1 \), and \( c = 0 \).

Would you like more details or have any questions?

### Relative Questions

1. How do you solve a system of equations using the substitution method?
2. What are some common mistakes to avoid when using elementary row operations?
3. Can all systems of linear equations be solved using row reduction?
4. How does the solution change if one of the equations is dependent on the others?
5. What is the geometric interpretation of a system of three linear equations?

### Tip

When performing row operations, always double-check your calculations to avoid arithmetic errors that could lead to incorrect results.

Learn how to solve the system of linear equations given by 2a + 3b + 1c = 7, a - b + 2c = 4, 3a + 2b + 4c = 10 using row reduction and matrix operations. This problem involves key concepts in linear algebra and matrix arithmetic. Suitable for college-level students studying linear equations and matrix algebra.

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