Solving Systems of Equations

Solving Systems of Equations - Solving Systems of Equations...

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Solving Systems of Equations In this section we are going to take a look at using linear algebra techniques to solve a system of linear equations. Once we have a couple of definitions out of the way we’ll see that the process is a fairly simple one. Well, it’s fairly simple to write down the process anyway. Applying the process is fairly simple as well but for large systems it can take quite a few steps. So, let’s get the definitions out of the way. A matrix (any matrix, not just an augmented matrix) is said to be in reduced row- echelon form if it satisfies all four of the following conditions. 1. If there are any rows of all zeros then they are at the bottom of the matrix. 2. If a row does not consist of all zeros then its first non-zero entry ( i.e. the left most non- zero entry) is a 1. This 1 is called a leading 1 . 3. In any two successive rows, neither of which consists of all zeroes, the leading 1 of the lower row is to the right of the leading 1 of the higher row. 4. If a column contains a leading 1 then all the other entries of that column are zero. A matrix (again any matrix) is said to be in row-echelon form if it satisfies items 1 3 of the reduced row-echelon form definition. Notice from these definitions that a matrix that is in reduced row-echelon form is also in row-echelon form while a matrix in row-echelon form may or may not be in reduced row-echelon form. Example 1 The following matrices are all in row-echelon form.
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None of the matrices in the previous example are in reduced row-echelon form. The entries that are preventing these matrices from being in reduced row-echelon form are highlighted in red and underlined (for those without color printers. ..). In order for these matrices to be in reduced row-echelon form all of these highlighted entries would need to be zeroes. Notice that we didn’t highlight the entries above the 1 in the fifth column of the third matrix. Since this 1 is not a leading 1 ( i.e. the leftmost non-zero entry) we don’t need the numbers above it to be zero in order for the matrix to be in reduced row-echelon form. Example 2 The following matrices are all in reduced row-echelon form.
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In the second matrix on the first row we have all zeroes in the entries. This is perfectly acceptable and so don’t worry about it. This matrix is in reduced row- echelon form, the fact that it doesn’t have any non-zero entries does not change that fact since it satisfies the conditions. Also, in the second matrix of the second row notice that the last column does not have zeroes above the 1 in that column. That is perfectly acceptable since the 1 in that column is not a leading 1 for the fourth row. Notice from Examples 1 and 2 that the only real difference between row-echelon form and reduced row-echelon form is that a matrix in row-echelon form is only required to have zeroes below a leading 1 while a matrix in reduced row-echelon from must have zeroes both below and above a leading 1. Okay, let’s now start thinking about how to use linear algebra techniques to solve
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This note was uploaded on 11/06/2011 for the course MATH 211 taught by Professor Staff during the Spring '11 term at Rutgers.

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Solving Systems of Equations - Solving Systems of Equations...

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