Row-Reducing a MatrixDate: 04/10/98 at 00:49:41 From: audrey Subject: Row-reduced matrix form Dear Dr. Math, Could you please teach me how to row-reduce the matrix below? 1 1 1 4 2 1 2 0 -1 Thanks, Audrey Date: 04/10/98 at 10:17:05 From: Doctor Rob Subject: Re: Row-reduced matrix form Work with the first column first, then move progressively to the right. If there is no nonzero entry in the column in rows not already used, go on to the next column. Pick a nonzero entry in the column in a row not already used. That entry is called the "pivot." Call that row the "pivot row." Divide the pivot row by the pivot. Subtract constant multiples of the pivot row from the other rows to create zeroes in the column in all other rows. Go on to the next column. When you are through, the matrix will have the following properties. The first nonzero entry in each row will be equal to 1, and each column that contains the leading nonzero entry of some row will have all its other entries zero. In your case, we start with column 1, and pick the pivot in row 1, which is 1 and hence nonzero. I divide row 1 by the pivot 1 (this doesn't change anything). Next I subtract 4 times row 1 from row 2, and 2 times row 1 from row 3. That gives me: [1 1 1] [0 -2 -3] [0 -2 -3] Now I move to column 2, consider all rows but row 1 (already used), and pick the pivot to be in row 2, which is -2 and hence nonzero. I divide the pivot row, row 2, by the pivot -2. Next I subtract 1 times the row 2 from row 1, and -2 times the row 2 from row 3. That gives me: [1 0 -1/2] [0 1 3/2] [0 0 0 ] Now I move to column 3, and find that there is no nonzero entry in any row not yet used (only the row 3 has not yet been used, and it has a 0 in column 3. Since we have no more columns, we are through. This matrix satisfies the two properties. The first nonzero entry in any row is 1 (see rows 1 and 2), and each column which contains the leading nonzero entry of some row (i.e. columns 1 and 2) has all its other entries zero. Thus it is row-reduced. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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