Solve the equation using Guassian elimination method

X1+2x2+3x3=4

X1+3x2+5x3=1

X1+5x2-12x3=2

To solve the system of linear equations using Gaussian elimination, we first represent the system in augmented matrix form, then perform row operations to achieve a row-echelon form, and finally back-substitute to find the solution.

### Step 1: Write the augmented matrix

The given system of equations is:

\[
\begin{aligned}
1. & \quad x_1 + 2x_2 + 3x_3 = 4 \\
2. & \quad x_1 + 3x_2 + 5x_3 = 1 \\
3. & \quad x_1 + 5x_2 - 12x_3 = 2 \\
\end{aligned}
\]

This can be written as an augmented matrix:

\[
\begin{pmatrix}
1 & 2 & 3 & | & 4 \\
1 & 3 & 5 & | & 1 \\
1 & 5 & -12 & | & 2
\end{pmatrix}
\]

### Step 2: Perform row operations

**First, eliminate \(x_1\) from rows 2 and 3.**

- Subtract Row 1 from Row 2:
\[
R2 = R2 - R1 \Rightarrow (1 - 1) & \quad (3 - 2) & \quad (5 - 3) & \quad (1 - 4) \\
\Rightarrow \begin{pmatrix}
0 & 1 & 2 & | & -3
\end{pmatrix}
\]

- Subtract Row 1 from Row 3:
\[
R3 = R3 - R1 \Rightarrow (1 - 1) & \quad (5 - 2) & \quad (-12 - 3) & \quad (2 - 4) \\
\Rightarrow \begin{pmatrix}
0 & 3 & -15 & | & -2
\end{pmatrix}
\]

So, the augmented matrix becomes:

\[
\begin{pmatrix}
1 & 2 & 3 & | & 4 \\
0 & 1 & 2 & | & -3 \\
0 & 3 & -15 & | & -2
\end{pmatrix}
\]

**Next, eliminate the coefficient of \(x_2\) in Row 3.**

- Subtract 3 times Row 2 from Row 3:
\[
R3 = R3 - 3R2 \Rightarrow \begin{pmatrix}
0 & 3 & -15 & | & -2
\end{pmatrix} - 3\begin{pmatrix}
0 & 1 & 2 & | & -3
\end{pmatrix}
\Rightarrow \begin{pmatrix}
0 & 0 & -21 & | & 7
\end{pmatrix}
\]

The matrix now looks like this:

\[
\begin{pmatrix}
1 & 2 & 3 & | & 4 \\
0 & 1 & 2 & | & -3 \\
0 & 0 & -21 & | & 7
\end{pmatrix}
\]

### Step 3: Back-substitution

Now that the matrix is in row-echelon form, we can solve for the variables starting from the bottom:

1. From Row 3: \(-21x_3 = 7\)
   \[
   x_3 = \frac{7}{-21} = -\frac{1}{3}
   \]

2. From Row 2: \(x_2 + 2x_3 = -3\)
   \[
   x_2 + 2\left(-\frac{1}{3}\right) = -3 \Rightarrow x_2 - \frac{2}{3} = -3 \Rightarrow x_2 = -3 + \frac{2}{3} = -\frac{7}{3}
   \]

3. From Row 1: \(x_1 + 2x_2 + 3x_3 = 4\)
   \[
   x_1 + 2\left(-\frac{7}{3}\right) + 3\left(-\frac{1}{3}\right) = 4 \Rightarrow x_1 - \frac{14}{3} - 1 = 4 \Rightarrow x_1 = 4 + \frac{14}{3} + 1 = \frac{31}{3}
   \]

### Final Solution
The solution to the system of equations is:

\[
x_1 = \frac{31}{3}, \quad x_2 = -\frac{7}{3}, \quad x_3 = -\frac{1}{3}
\]

Would you like to go into more details or do you have any questions? Here are some related problems:

1. Solve a 3x3 system with a different set of coefficients using Gaussian elimination.
2. How would you handle a system with no solutions or infinite solutions using Gaussian elimination?
3. Compare Gaussian elimination with the Gauss-Jordan method.
4. Solve a 4x4 system using Gaussian elimination.
5. What is the significance of the determinant in solving systems of equations?

**Tip:** Always check the determinant of the coefficient matrix before applying Gaussian elimination to ensure the system has a unique solution.

Solve the equation using Gaussian elimination method

X1+2x2+3x3=4

X1+3x2+5x3=1

X1+5x2-12x3=2

Learn how to solve a 3x3 system of linear equations using the Gaussian elimination method. This problem involves transforming the system into an augmented matrix, performing row operations, and applying back-substitution to find the solution. Suitable for students in Grades 10-12.

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