Solved In Exercises 17-36, use the Gauss-Jordan elimination ... - Free Printable

To solve the given systems of linear equations using the Gauss-Jordan elimination method, we will follow these steps:

1. Write the augmented matrix for the system.
2. Perform row operations to transform the matrix into reduced row echelon form (RREF).
3. Interpret the RREF to find the solutions.

Let's solve one of the problems as an example. We'll choose Exercise 17:

Exercise 17:

\[
\begin{cases}
2x - 4y = 6 \\
-x + 2y = -3
\end{cases}
\]

#### Step 1: Write the augmented matrix
The augmented matrix for the system is:
\[
\begin{bmatrix}
2 & -4 & 6 \\
-1 & 2 & -3
\end{bmatrix}
\]

#### Step 2: Perform row operations to get RREF

1. Swap rows if necessary: The first row already has a non-zero leading coefficient, so no swap is needed.
2. Make the leading coefficient of the first row 1: Divide the first row by 2:
\[
\begin{bmatrix}
1 & -2 & 3 \\
-1 & 2 & -3
\end{bmatrix}
\]
3. Eliminate the first entry in the second row: Add the first row to the second row:
\[
\begin{bmatrix}
1 & -2 & 3 \\
0 & 0 & 0
\end{bmatrix}
\]

At this point, the matrix is in RREF. The second row represents the equation \(0x + 0y = 0\), which is always true and does not provide new information.

#### Step 3: Interpret the RREF
The RREF corresponds to the system:
\[
\begin{cases}
x - 2y = 3 \\
0 = 0
\end{cases}
\]
The second equation is redundant, so we only need to consider the first equation:
\[
x - 2y = 3
\]
This is a dependent system with infinitely many solutions. We can express \(x\) in terms of \(y\):
\[
x = 3 + 2y
\]
Let \(y = t\) (a free variable). Then the solution can be written as:
\[
(x, y) = (3 + 2t, t)
\]
where \(t\) is any real number.

Final Answer for Exercise 17:

\[
\boxed{(3 + 2t, t)}
\]

This process can be applied to each of the other exercises in a similar manner. If you need solutions for other specific exercises, please let me know!