In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists. 3x + 4y + 2z = 3 4x - 2y - 8z = - 4 x + y - z = 3

Verified step by step guidance

1. Step 1: Write the system of equations as an augmented matrix: $$\left[\begin{array}{ccccc}3& 4& 2& |& 3\\ 4& -2& -8& |& -4\\ 1& 1& -1& |& 3\end{array}\right]$$
2. Step 2: Use row operations to get a leading 1 in the first row, first column. You can swap Row 1 and Row 3: $$\left[\begin{array}{ccccc}1& 1& -1& |& 3\\ 4& -2& -8& |& -4\\ 3& 4& 2& |& 3\end{array}\right]$$
3. Step 3: Eliminate the first column below the leading 1 by replacing Row 2 with Row 2 minus 4 times Row 1, and Row 3 with Row 3 minus 3 times Row 1.
4. Step 4: Simplify the matrix to get a leading 1 in the second row, second column. You may need to divide Row 2 by a constant.
5. Step 5: Continue using row operations to achieve a row-echelon form, then back-substitute to find the values of $x$, $y$, and $z$.

Key Concepts

Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations. It involves transforming the system's augmented matrix into row echelon form using a series of row operations, which include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting rows. This process simplifies the equations, making it easier to find the values of the variables.

Row Echelon Form

Row echelon form is a specific arrangement of a matrix where all non-zero rows are above any rows of all zeros, and the leading coefficient of each non-zero row (the first non-zero number from the left) is to the right of the leading coefficient of the previous row. This structure is crucial for back substitution, allowing for the systematic solving of the variables in a linear system.

Existence of Solutions

The existence of solutions in a system of equations refers to whether there are one, infinitely many, or no solutions at all. This can be determined through the analysis of the row echelon form of the augmented matrix. If a row leads to a contradiction (like 0 = 1), the system has no solution; if there are free variables, the system has infinitely many solutions.