Linear Algebra I Lecture 10
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1 Linear Algebra I Lecture 10 Xi Chen 1 1 University of Alberta January 30, 2019
2 Outline 1
3 Gauss-Jordan Algorithm ] Let A = [a ij m n be an m n matrix. To reduce A to a reduced row echelon form using elementary row operation, we do: STEP 0. Set i = j = 1. STEP 1. If i > m or j > n, we are done and stop. STEP 2. If a ij = a i+1,j =... = a mj = 0, increase j by 1 and go to STEP 1. STEP 3. Find a kj 0 for some i k m and exchange Row i and k if i k. STEP 4. Divide Row i by a ij (pivot). STEP 5. For each l i, substract a lj Row i from Row l (pivoting). STEP 6. Increase both i and j by 1 and go to STEP 1.
4 Examples of Gauss-Jordan Elimination For example, let us solve x 1 x 2 + x 3 = 1 2x 1 3x 2 + x 3 = 2 4x 1 5x 2 + 3x 3 = R2 2 R R3 4 R R2 ( 1) R1+R2 R3+R
5 Examples of Gauss-Jordan Elimination (CONT) We have reduced the augmented matrix to the RREF This corresponds to the system { x1 + 2x 3 = 1 x 2 + x 3 = 0 where x 1 and x 2 are the dependent variables and x 3 is the free variable. Setting x 3 = t, we obtain the solutions x 1 1 2t 1 2 x 2 = t = 0 + t 1 t 0 1 x 3
6 Examples of Gauss-Jordan Elimination For example, let us solve y + z = 1 2w 4x 2y + z = 1 w + 2x + y = R1 R R2+2 R1 R1 R R2 R R1+R R2 R
7 Examples of Gauss-Jordan Elimination (CONT) This corresponds to the system w + 2x = 4 y = 2 z = 3 where w, y and z are the dependent variables and x is the free variable. Setting x = t, we obtain the solutions w 4 2t 4 2 x y = t 2 = t 1 0 z 3 3 0
8 Solutions of SLE When applying Gauss-Jordan to solve a SLE, 1 whenever the resulting matrix contains a row [ ] b for some b 0, we can stop and the system is inconsistent; 2 if the system is homogeneous, we just need to apply the algorithm to the coefficient matrix, because the constant terms remain zero throughout.
9 Solutions of SLE Suppose that we have reduced the augmented matrix to a REF: Elementary Row Operations A b [A b ] m (n+1) 1 If [ A b ] contains a row [ ] b for some b 0, the system is inconsistent. 2 Every leading nonzero entry corresponds to a dependent variable: # of depend variables = # of leading nonzero entries = # of nonzero rows if the system is consistent.
10 Solutions of SLE (CONT) If the system is consistent, k = # of free variables = n # of dependent variables and the solution is = n # of nonzero rows in REF n m x = u + t 1 v 1 + t 2 v t k v k. A system of m linear equations in n variables has either no solutions or infinitely many solutions if n > m. If it is homogeneous, it has always infinitely many solutions if n > m.
11 Intersection of Two Lines in R 2 Let L 1 and L 2 be two lines in R 2 : L 1 = {a 11 x + a 12 y = b 1 } L 2 = {a 21 x + a 22 y = b 2 } Consider L 1 L 2, i.e., the solutions of { a11 x + a 12 y = b 1 a 21 x + a 22 y = b 2 1 L 1 L 2 = {p} the system has a unique solution. 2 L 1 L 2 = L 1 = L 2 the system has infinitely many solutions. 3 L 1 L 2 = the system has no solution.
12 Intersection of Two Lines in R 2 (CONT) Suppose that we have reduced the augmented matrix to a RREF: a11 a 12 b 1 Elementary Row Operations a 11 a 12 b 1 a 21 a 22 b 2 a 21 a 22 b 2 A list of all possible RREF: 1 0 c1 L 0 1 c 1 L 2 = {(c 1, c 2 )} 2 1 c1 0 L L 2 = L L 2 =
13 Intersection of Two Lines in R 2 (CONT) 1 c1 c 2 L L 2 = L 1 = L 2 = {x + c 1 y = c 2 } 0 1 c1 L L 2 = L 1 = L 2 = {y = c 2 } Since (a 11, a 12 ) (0, 0), a 11 a 12 a 21 a
14 Intersection of Three planes in R 3 Let P 1, P 2 and P 3 be three planes in R 3 given by P 1 = {a 11 x + a 12 y + a 13 z = b 1 } P 2 = {a 21 x + a 22 y + a 23 z = b 2 } P 3 = {a 31 x + a 32 y + a 33 z = b 3 } Consider P 1 P 2 P 3, i.e., the solutions of a 11 x + a 12 y + a 13 z = b 1 a 21 x + a 22 y + a 23 z = b 2 a 31 x + a 32 y + a 33 z = b 3 1 P 1 P 2 P 3 is a plane. 2 P 1 P 2 P 3 is a line. 3 P 1 P 2 P 3 is a point. 4 P 1 P 2 P 3 is empty.
15 Intersection of Three planes in R 3 (CONT) The augmented matrix can be reduced to one of the following RREF: c c c (a) c 2 (b) 0 1 c 2 0 (c) c c 1 c 2 1 c 1 0 c 2 (d) (e) 0 1 c 3 c 4 (f ) c c 1 c c c 1 0 (g) (h) c 2 (i)
16 Intersection of Three planes in R 3 (CONT) (j) (k) c 1 (m) c 1 c 2 c (l) c 1 c (b), (c), (d), (g), (i), (j): P 1 P 2 P 3 =. (a): P 1 P 2 P 3 is a point. (e), (f), (h): P 1 P 2 P 3 is a line. (k), (l), (m): P 1 P 2 P 3 = P 1 = P 2 = P 3 is a plane.
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