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Problem 1

# Use Gauss-Jordan row reduction to solve the given systems of equation. We suggest doing some by hand and others using technology. $$x+y=4$$ $$x-y=2$$

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The given system of equations is: $$x+y=4 \\ x-y=2$$ Using Gauss-Jordan row reduction, we find the reduced row echelon form (RREF) of the augmented matrix: $$\begin{bmatrix} 1 & 0 & | & 3\\ 0 & 1 & | & 1 \end{bmatrix}$$ Converting the RREF matrix back into a system of equations, we get the solution $$x = 3$$ and $$y = 1$$.
See the step by step solution

## Step 1: Write the system of equations as an augmented matrix

We can represent the system of equations as an augmented matrix. This will allow us to perform row operations more easily. The given system of equations is: $$x+y=4\\ x-y=2$$ The corresponding augmented matrix will be: $$\begin{bmatrix} 1 & 1 & | & 4\\ 1 & -1 & | & 2 \end{bmatrix}$$

## Step 2: Perform row operations to transform the matrix into its RREF

Our goal is to perform row operations to transform the augmented matrix into the reduced row echelon form (RREF) which has the following structure: $$\begin{bmatrix} 1 & 0 & | & a\\ 0 & 1 & | & b \end{bmatrix}$$ where 'a' and 'b' are the solutions for 'x' and 'y' respectively. Let's perform row operations: 1. Swap Row 1 and Row 2 (for easier calculations): $$\begin{bmatrix} 1 & -1 & | & 2\\ 1 & 1 & | & 4 \end{bmatrix}$$ 2. Subtract Row 1 from Row 2, and replace Row 2 with the result: $$\begin{bmatrix} 1 & -1 & | & 2\\ 0 & 2 & | & 2 \end{bmatrix}$$ 3. Divide Row 2 by 2, and replace Row 2 with the result: $$\begin{bmatrix} 1 & -1 & | & 2\\ 0 & 1 & | & 1 \end{bmatrix}$$ 4. Add Row 2 to Row 1, and replace Row 1 with the result: $$\begin{bmatrix} 1 & 0 & | & 3\\ 0 & 1 & | & 1 \end{bmatrix}$$ Now, the matrix is in its reduced row echelon form (RREF).

## Step 3: Convert the RREF matrix back into a system of equations and solve for the variables

Now that we have the matrix in RREF, we can easily convert it back into a system of equations to find the solutions: $$\begin{cases} x &= 3\\ y &= 1 \end{cases}$$ The solution of this system of equations is $$x = 3$$ and $$y = 1$$.

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