Linear Algebra/Gauss-Jordan Reduction/Solutions

Solutions Problem 1 Use Gauss-Jordan reduction to solve each system. <li> <li> <li> <li> Answer These answers show only the Gauss-Jordan reduction. With it, describing the solution set is easy. <li> <li> <li> <li> A row swap here makes the arithmetic easier. Problem 2 Find the reduced echelon form of each matrix. <li> <li> <li> <li> Answer Use Gauss-Jordan reduction. <li> <li> <li> <li> Problem 3 Find each solution set by using Gauss-Jordan reduction, then reading off the parametrization. <li> <li> <li> <li> Answer For the "Gauss" halves, see the answers to Problem I.2.5. <li> The "Jordan" half goes this way. The solution set is this <li> The second half is so the solution is this. <li> This Jordan half gives (of course, the zero vector could be omitted from the description). <li> The "Jordan" half ends with this solution set. Problem 4 Give two distinct echelon form versions of this matrix. Answer Routine Gauss' method gives one: and any cosmetic change, like multiplying the bottom row by , gives another. Problem 5exer:PossRedEchFrms List the reduced echelon forms possible for each size. <li> <li> <li> <li> Answer In the cases listed below, we take . Thus, some canonical forms listed below actually include infinitely many cases. In particular, they includes the cases and . <li> , , , <li> , , , , , , <li> , , , <li> , , , , , , , Problem 6 What results from applying Gauss-Jordan reduction to a nonsingular matrix? Answer A nonsingular homogeneous linear system has a unique solution. So a nonsingular matrix must reduce to a (square) matrix that is all 's except for 's down the upper-left to lower-right diagonal, e.g., Problem 7exer:INotJMakesRowOpsRev The proof of Lemma 4 contains a reference to the condition on the row pivoting operation. <li> The definition of row operations has an condition on the swap operation . Show that in this condition is not needed. <li> Write down a matrix with nonzero entries, and show that the operation is not reversed by . <li> Expand the proof of that lemma to make explicit exactly where the condition on pivoting is used. Answer <li> The operation does not change . <li> For instance, leaves the matrix changed. <li> If then does indeed give back. (Of course, if then the third matrix would have entries of the form .)