Gaussian Elimination II

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In this section, we continue to learn about Gaussian Elimination. Our goal is to place a matrix in row-echelon form and obtain the solution to a linear system of equations. We will now move on and look at systems with three equations and three variables (x, y, and z). To use a matrix to solve a linear system, we would write each equation in standard form and then list the numerical information (coefficients and constants only) inside of a matrix. The matrix will have brackets on the outside. The matrix we will construct will have a vertical line to separate the coefficients from the constants. This type of matrix is known as the augmented matrix. We can manipulate the matrix using row operations. The following are row operations: 1) We can interchange any two rows, just like we can switch the order of which equation is on top and which equation is on the bottom. 2) We can multiply any row by a non-zero number, just like we can multiply any equation by a non-zero number and not change the solution. 3) We can multiply a row by a real number and add this to the corresponding elements of any other row. We will use these row operations to produce a matrix that is in row-echelon form. This gives us 1’s down the diagonal and 0’s below. In this form, we are given one unknown and can use substitution to find the other. Additionally, we can work further in the matrix and place it into reduced row-echelon form. This form gives us all solutions without any further work. Reduced row-echelon form gives us 1’s down the diagonal and 0’s above and below.
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