Ubahlah matriks di bawah ini menjadi matriks eselon baris tereduksi melalui serangkaian operasi bariselementer!𝐴 = [1 −12 12 −1−2 −1−1 23 0−4 10 −3−1−21−3

1. Break Down the Problem

We need to convert the given matrix $A$ into reduced row echelon form (RREF) using elementary row operations.

2. Relevant Concepts

To achieve RREF, we will use the following elementary row operations:

1. Swap two rows.
2. Multiply a row by a non-zero scalar.
3. Add or subtract a scalar multiple of one row to another row.

The goal of RREF is:

• The leading entry of each non-zero row is 1.
• Each leading 1 is the only non-zero entry in its column.
• The leading 1s move to the right as we move down the rows.

3. Analysis and Detail

Given matrix: $A = \begin{bmatrix} 1 & -1 & 2 & 1 \\ -2 & -1 & -1 & 2 \\ 3 & 0 & -4 & 1 \\ -1 & -2 & 1 & -3 \end{bmatrix}$

Step 1: Make the leading coefficient of the first row a 1 (already is) and eliminate the first column.

Using $R_1$ to eliminate the first column below:

• $R_2 \leftarrow R_2 + 2R_1$
• $R_3 \leftarrow R_3 - 3R_1$
• $R_4 \leftarrow R_4 + R_1$

This gives: $\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & -3 & 3 & 4 \\ 0 & 3 & -10 & -2 \\ 0 & -3 & 3 & -2 \end{bmatrix}$

Step 2: Simplify $R_2$ and reduce the third and fourth rows.

Multiplying $R_2$ by $-\frac{1}{3}$: $R_2 \leftarrow -\frac{1}{3} R_2$ Gives: $\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1 & -\frac{4}{3} \\ 0 & 3 & -10 & -2 \\ 0 & -3 & 3 & -2 \end{bmatrix}$

Now eliminate the second column entries:

• $R_3 \leftarrow R_3 - 3R_2$
• $R_4 \leftarrow R_4 + 3R_2$

Results in: $\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1 & -\frac{4}{3} \\ 0 & 0 & -7 & 2 \\ 0 & 0 & 0 & -4 \end{bmatrix}$

Step 3: Simplify $R_3$ and $R_4$.

• For $R_3$, multiply by $-\frac{1}{7}$: $R_3 \leftarrow -\frac{1}{7}R_3$ Gives: $\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1 & -\frac{4}{3} \\ 0 & 0 & 1 & -\frac{2}{7} \\ 0 & 0 & 0 & -4 \end{bmatrix}$

• For $R_4$, multiplying by $-\frac{1}{4}$: $R_4 \leftarrow -\frac{1}{4}R_4$ Results in: $\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1 & -\frac{4}{3} \\ 0 & 0 & 1 & -\frac{2}{7} \\ 0 & 0 & 0 & 1 \end{bmatrix}$

Final Step: Eliminate non-zero entries above each leading 1.

Use $R_4$ to eliminate the last column in $R_1, R_2,$ and $R_3$:

• $R_1 \leftarrow R_1 - R_4$
• $R_2 \leftarrow R_2 + \frac{4}{3}R_4$
• $R_3 \leftarrow R_3 + \frac{2}{7}R_4$

Final RREF matrix obtained: $\begin{bmatrix} 1 & 0 & 0 & \frac{37}{21} \\ 0 & 1 & 0 & \frac{5}{3} \\ 0 & 0 & 1 & -\frac{2}{7} \\ 0 & 0 & 0 & 1 \end{bmatrix}$

4. Verify and Summarize

The operations performed lead to a reduced row echelon form. Each step confirmed the goals of RREF were achieved correctly.

Final Answer

The reduced row echelon form of the matrix $A$ is: $\begin{bmatrix} 1 & 0 & 0 & \frac{37}{21} \\ 0 & 1 & 0 & \frac{5}{3} \\ 0 & 0 & 1 & -\frac{2}{7} \\ 0 & 0 & 0 & 1 \end{bmatrix}$