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To solve this system using Gaussian elimination, we aim to put it into an upper triangular form. Let's go through each step.

### Step 1: Set up the augmented matrix
The system of equations is:
\[
\begin{cases}
3x_1 + 4x_2 + 2x_3 = 10 \\
x_1 + x_3 = 3 \\
12x_2 - 4x_3 = 16
\end{cases}
\]

The augmented matrix is:
\[
\begin{pmatrix}
3 & 4 & 2 & | & 10 \\
1 & 0 & 1 & | & 3 \\
0 & 12 & -4 & | & 16
\end{pmatrix}
\]

### Step 2: Make the first entry of the first column into a leading 1
Divide the first row by 3:
\[
\begin{pmatrix}
1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\
1 & 0 & 1 & | & 3 \\
0 & 12 & -4 & | & 16
\end{pmatrix}
\]

### Step 3: Eliminate \( x_1 \) from the second row
Subtract the first row from the second row to make the second row’s first column entry zero:
\[
\begin{pmatrix}
1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\
0 & -\frac{4}{3} & \frac{1}{3} & | & \frac{-1}{3} \\
0 & 12 & -4 & | & 16
\end{pmatrix}
\]

### Step 4: Scale the second row to make the leading coefficient 1
Multiply the second row by \(-\frac{3}{4}\):
\[
\begin{pmatrix}
1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\
0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\
0 & 12 & -4 & | & 16
\end{pmatrix}
\]

### Step 5: Eliminate \( x_2 \) from the third row
Subtract 12 times the second row from the third row:
\[
\begin{pmatrix}
1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\
0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\
0 & 0 & -1 & | & 13
\end{pmatrix}
\]

### Step 6: Scale the third row to make the leading coefficient 1
Multiply the third row by \(-1\):
\[
\begin{pmatrix}
1 & \frac{4}{3} & \frac{2}{3} & | & \frac{10}{3} \\
0 & 1 & -\frac{1}{4} & | & \frac{1}{4} \\
0 & 0 & 1 & | & -13
\end{pmatrix}
\]

### Step 7: Back-substitute to find the solutions
1. From the third row: \( x_3 = -13 \)
2. Substitute \( x_3 = -13 \) into the second row: \( x_2 - \frac{1}{4}(-13) = \frac{1}{4} \Rightarrow x_2 = 3 \)
3. Substitute \( x_2 = 3 \) and \( x_3 = -13 \) into the first row: \( x_1 + \frac{4}{3}(3) + \frac{2}{3}(-13) = \frac{10}{3} \Rightarrow x_1 = 6 \)

### Solution
The unique solution is:
\[
(x_1, x_2, x_3) = (6, 3, -13)
\]

Would you like further details or clarification on any step?

Here are some related questions:
1. What is Gaussian elimination and how does it work in solving systems?
2. How does back-substitution work in a triangular system?
3. Can Gaussian elimination be used for non-square systems?
4. How do different row operations affect the solution of the system?
5. What are some common mistakes to avoid in Gaussian elimination?

**Tip:** Always verify your solution by substituting it back into the original equations.

Use Gaussian Elimination to solve the linear system by finding an equivalent system in triangular form.

3x1 + 4x2 + 2x3 = 10
x1 + x3 = 3
12x2 - 4x3 = 16

The unique solution to the system is (x1, x2, x3) = [ , , ].

Learn how to solve a linear system using Gaussian Elimination by transforming it into an equivalent system in triangular form. This example provides a step-by-step solution for the system 3x1 + 4x2 + 2x3 = 10, x1 + x3 = 3, and 12x2 - 4x3 = 16, suitable for students in Grades 10-12.

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