Sure! Let's solve the system of equations using the augmented matrix method.
The system of equations is given by:
[tex]\[
\begin{cases}
x_1 + 2x_2 + x_3 = 2 \\
x_1 - x_2 - 2x_3 = -1
\end{cases}
\][/tex]
First, we write this system as an augmented matrix:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 2 \\
1 & -1 & -2 & -1
\end{array}\right]
\][/tex]
We want to convert this augmented matrix into its row echelon form (REF). Here are the steps:
1. Step 1: We already have a leading 1 in the first row, first column. So, the first row remains the same:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 2 \\
1 & -1 & -2 & -1
\end{array}\right]
\][/tex]
2. Step 2: We need to make the element below this leading 1 a zero. We can do this by subtracting the first row from the second row:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 2 \\
0 & -3 & -3 & -3
\end{array}\right]
\][/tex]
3. Step 3: We simplify the second row by dividing it by -3:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 2 \\
0 & 1 & 1 & 1
\end{array}\right]
\][/tex]
4. Step 4: We need to make the element above the leading 1 in the second row a zero. We can do this by subtracting 2 times the second row from the first row:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 0 & -1 & 0 \\
0 & 1 & 1 & 1
\end{array}\right]
\][/tex]
Now, we have the matrix in row echelon form:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 0 & -1 & 0 \\
0 & 1 & 1 & 1
\end{array}\right]
\][/tex]
From this, we can write the system of equations:
[tex]\[
\begin{cases}
x_1 - x_3 = 0 \\
x_2 + x_3 = 1
\end{cases}
\][/tex]
Solving this system, we get:
1. From the first equation:
[tex]\[
x_1 = x_3
\][/tex]
2. From the second equation:
[tex]\[
x_2 = 1 - x_3
\][/tex]
So, the solution set in terms of [tex]\( x_3 \)[/tex] (which is the free variable) is:
[tex]\[
(x_1, x_2, x_3) = (x_3, 1 - x_3, x_3)
\][/tex]
Hence, the solution to the system of equations can be written as:
[tex]\[
\{(x_1, x_2, x_3) \mid x_1 = x_3, x_2 = 1 - x_3, x_3 \in \mathbb{R}\}
\][/tex]
where [tex]\(x_3\)[/tex] is any real number.
