Wikipedia
row reduction : In linear algebra, Gaussian elimination is an efficient algorithm for solving systems of linear equations, finding the rank of a matrix, and calculating the inverse of an invertible square matrix. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss. Elementary row operations are used to reduce a matrix to row echelon form. An extension of this algorithm, Gauss–Jordan elimination, reduces the matrix further to reduced row echelon form. Gaussian elimination alone is sufficient for many applications. The method of Gaussian elimination appears in Chapter Eight, Rectangular Arrays, of the important Chinese mathematical text Jiuzhang suanshu or The Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen problems, with two to five equations. The first reference to the book by this title is dated to 179 CE, but parts of it were written as early as approximately 150 BCE. It was commented on by Liu Hui in the 3rd ce....   More from Wikipedia
row reduction : In commutative algebra and algebraic geometry, elimination theory is the classical name for algorithmic approaches to eliminating between polynomials of several variables. The linear case would now routinely be handled by Gaussian elimination, rather than the theoretical solution provided by..   More from Wikipedia
Linear Algebra
Types of linear systems Gauss-Jordan elimination, Gauss-Jordan method Vectors and vector addition Geometrical solution sets of systems of equations Rectangular matrices to row echelon form Matrix multiplication Inverse to a square matrix Determinants of 2 by 2 and 3 by 3 matrices..
  System of Equation Using Row Reduction Matrix
  Example of row reduction of an augmented matrix to reduced row echelon form
Question : Use row reduction to find the inverse of the matrix:
1 3 6
2 4 3
0 2 -1
Please can you help me do this question? I am more interested in finding out the method of doing this question than simply getting the answer, therefore it would be appreciated if you could work it out for me step by step with explanations?
Answer : ----- 1 3 6 2 4 3 0 2 -1 1 0 0 0 1 0 0 0 1 ----- 1 3 6 0 -2 -9 0 2 -1 1 0 0 -2 1 0 0 0 1 ----- 1 3 6 0 -2 -9 0 0 -10 1 0 0 -2 1 0 -2 1 1 ----- 1 0 -7.5 0 -2 -9 0 0 -10 -2 1.5 0 -2 1 0 -2 1 1 ----- Continue the process on your own. Too long.....   More from Yahoo Answers
Answer : ----- 1 3 6 2 4 3 0 2 -1 1 0 0 0 1 0 0 0 1 ----- 1 3 6 0 -2 -9 0 2 -1 1 0 0 -2 1 0 0 0 1 ----- 1 3 6 0 -2 -9 0 0 -10 1 0 0 -2 1 0 -2 1 1 ----- 1 0 -7.5 0 -2 -9 0 0 -10 -2 1.5 0 -2 1 0 -2 1 1 ----- Continue the process on your own. Too long.....   More from Yahoo Answers
Question : I understand the concept of the row reduction method, but I don't understand how one finds the right equation to use. for example, if its a 3x3 matrix, the equation might become R3 -> R3+2R2
but where does that come from? Do I just have to figure it out?
I do have a program on my calculator that can do it but I want to be able to do it manually.
Answer : Row reduction is actually reasonably systematic. Just follow these steps: 1) Swap rows so that the rows STARTING with the most zeros are on the bottom e.g. Make we change the following matrix: 0 1 -2 5 0 0 -3 -1 2 4 0 3 0 2 -4 -2 into this matrix: 2 4 0 3 0 2 -2 5 0 2 -4 -2 0 0 -3 -1 Notice that all the 0s are in the bottom left hand corner. 2) Add or subtract multiples of the first row to the others so that the first column has only one non-zero entry in it In the previous example, this was already done. However, had it been this: 2 4 0 3 4 2 -2 5 0 2 -4 -2 0 0 -3 -1 we would have subtracted 2 lots of the first row from the second, or, in other words, R2 -> R2 - 2*R1: 2 4 0 3 0 -6 -2 -1 0 2 -4 -2 0 0 -3 -1 3) Do this with the second row/second column, etc, until there are as many zeros in the bottom left corner as possible. In this example, you could either divide row 2 by 3 and add it to row 3, or we can do some simplifying and rearr....   More from Yahoo Answers
Answer : Row reduction is actually reasonably systematic. Just follow these steps: 1) Swap rows so that the rows STARTING with the most zeros are on the bottom e.g. Make we change the following matrix: 0 1 -2 5 0 0 -3 -1 2 4 0 3 0 2 -4 -2 into this matrix: 2 4 0 3 0 2 -2 5 0 2 -4 -2 0 0 -3 -1 Notice that all the 0s are in the bottom left hand corner. 2) Add or subtract multiples of the first row to the others so that the first column has only one non-zero entry in it In the previous example, this was already done. However, had it been this: 2 4 0 3 4 2 -2 5 0 2 -4 -2 0 0 -3 -1 we would have subtracted 2 lots of the first row from the second, or, in other words, R2 -> R2 - 2*R1: 2 4 0 3 0 -6 -2 -1 0 2 -4 -2 0 0 -3 -1 3) Do this with the second row/second column, etc, until there are as many zeros in the bottom left corner as possible. In this example, you could either divide row 2 by 3 and add it to row 3, or we can do some simplifying and rearr....   More from Yahoo Answers
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