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# 202 Class Note for MA 26500 at Purdue

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This 11 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 25 views.

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Date Created: 02/06/15
MA 265 Lectures 4 and 5 15 Echelon Form of 3 Matrix In this section we Will bring a matrix to a simpler form in order to be able to solve systems in a easier way De nition 1 A matrix A is said to be in row echelon form if 1 The leading entry the rst nonzero entry of each row is 1 if it exists it may happen that the Whole row is zero The leading entry of each row is to the right of the leading entry of the previous row forms a staire case All rows consisting entirely of zeros are at the bottom The matrix is in reduced row echelon form if it also satis es If a column contains the leading entry of a row all other entries in the column are 0 Example 1 This matrix is in row echelon form 4 1 1 l 1 I 0 J This matrix is in reduced row echelon form 3 O 1 1 O O O O O O 1 O O 5 1 2 O O O O Elementary Row Column Operations De nition 2 An elementary row column operation is any of the following TWO matrices are said to be r0u equivalent if one can be obtained from the other by performing a sequence of elementary row operations Theorem 3 Every matricc is raw equivalent to a matricc in the row eanchelon form and to a unique matricc in the reduce r0u eanchelon form Finding the RRE matrix Given a matrix A we follow these steps 1 Pivoting the matrix Find the pivot of the given matrix the rst nonzero entry in the rst column With nonzero entries Interchange rows so that the pivot is now in the top row Multiply the rst row by the inverse of the pivot Use the pivot Which now has value 1 to make zero all entries below it by adding multiples of the rst row to the other rows 2 Repeat the pivoting procedure to the submatrix obtained by disregarding the rst row This is repeated until there are no more pivots Then the matrix is in row echelon form To obtain the BBB matrix we have one extra step 3 Use the leading 1 s to eliminate the nonzero entries above them starting from the lowest row Solving Systems of Linear Equations We use the augmented matrix AlB associated to the system AX B Theorem 4 If AB and ClD are raw equivalent then the corresponding systems have euactly the same solutions METHOD 1 Transform AlB into a row equivalent row echelon matrix Then use back substitution This method is called GA USSIAN ELIMINATION METHOD II Transform AlB into a row equivalent reduced row echelon matrix This produces the solutions immediately This method is called GA USS JORDAN REDUCTION We Will mostly use the GaussJordan reduction Eazample 2 Solve the following system 96y223w13 x 2yzw8 3yz w1 Homogeneous systems Remember A homogeneous system always has at least one solution the trivial solution x 0 If it has another solution then it has in nitely many of them a Whole subspace of them because we can multyply one of the solution by any constant and we get a different solution Theorem 5 A homogeneous sytem ofm equations in n variables always has a nontrivial solution if m lt n In general homogeneous systems are easier to handle Observe the following if Xp is a solution of a system AX B With B 7E 0 and Xh is a solution for the system AX 0 then Xp Xh is also a solution of AX B In fact all solutions of AX B can be obtained in this fashion Homework Section 15

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