Reduce the matrix to its row-reduced form.

9	3	−6		−36	

6	8	6		−52
0	2	6		−28


Let's reduce the given matrix to its row-reduced echelon form (RREF). The matrix is:

\[
\begin{bmatrix}
9 & 3 & -6 & -36 \\
6 & 8 & 6 & -52 \\
0 & 2 & 6 & -28
\end{bmatrix}
\]

### Step 1: Form the augmented matrix:
\[
\left[ \begin{array}{ccc|c}
9 & 3 & -6 & -36 \\
6 & 8 & 6 & -52 \\
0 & 2 & 6 & -28
\end{array} \right]
\]

### Step 2: Make the first pivot entry a 1 by dividing the first row by 9:

\[
R_1 \to \frac{R_1}{9}
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
6 & 8 & 6 & -52 \\
0 & 2 & 6 & -28
\end{array} \right]
\]

### Step 3: Eliminate the first entry of the second row using row operations:

\[
R_2 \to R_2 - 6R_1
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
0 & 6 & 10 & -28 \\
0 & 2 & 6 & -28
\end{array} \right]
\]

### Step 4: Make the second pivot a 1 by dividing the second row by 6:

\[
R_2 \to \frac{R_2}{6}
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
0 & 1 & \frac{5}{3} & -\frac{14}{3} \\
0 & 2 & 6 & -28
\end{array} \right]
\]

### Step 5: Eliminate the second entry of the third row using row operations:

\[
R_3 \to R_3 - 2R_2
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
0 & 1 & \frac{5}{3} & -\frac{14}{3} \\
0 & 0 & \frac{4}{3} & -\frac{10}{3}
\end{array} \right]
\]

### Step 6: Make the third pivot a 1 by multiplying the third row by \(\frac{3}{4}\):

\[
R_3 \to \frac{3}{4}R_3
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
0 & 1 & \frac{5}{3} & -\frac{14}{3} \\
0 & 0 & 1 & -\frac{5}{2}
\end{array} \right]
\]

### Step 7: Eliminate the third entry in the second row and the third entry in the first row:

\[
R_2 \to R_2 - \frac{5}{3}R_3
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & -\frac{2}{3} & -4 \\
0 & 1 & 0 & \frac{4}{3} \\
0 & 0 & 1 & -\frac{5}{2}
\end{array} \right]
\]

\[
R_1 \to R_1 + \frac{2}{3}R_3
\]
\[
\left[ \begin{array}{ccc|c}
1 & \frac{1}{3} & 0 & -\frac{10}{3} \\
0 & 1 & 0 & \frac{4}{3} \\
0 & 0 & 1 & -\frac{5}{2}
\end{array} \right]
\]

### Step 8: Eliminate the second entry of the first row:

\[
R_1 \to R_1 - \frac{1}{3}R_2
\]
\[
\left[ \begin{array}{ccc|c}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & \frac{4}{3} \\
0 & 0 & 1 & -\frac{5}{2}
\end{array} \right]
\]

This is the row-reduced echelon form (RREF) of the matrix. The solution to the system of equations represented by this augmented matrix is:

\[
x_1 = -2, \quad x_2 = \frac{4}{3}, \quad x_3 = -\frac{5}{2}
\]

Would you like any further clarifications?

### Related Questions:
1. How do you perform row-reduction for a larger matrix?
2. What is the significance of row-reduced echelon form (RREF)?
3. How can row-reduction help solve systems of equations?
4. What happens if a matrix has no row-reduced form?
5. Can row-reduction be used for non-square matrices?

**Tip:** Always ensure the pivot elements (leading 1s) are in the correct columns to reach RREF successfully!

Reduce the matrix to its row-reduced form.

9  3  −6  −36  
6  8  6  −52
0  2  6  −28

Learn how to reduce the 3x4 matrix to its row-reduced echelon form using Gauss-Jordan elimination. This step-by-step solution covers essential linear algebra concepts, including row operations and solving systems of equations. Suitable for high school students (Grades 10-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation