Mathematics 13: Lecture 12 Elementary Matrices and Finding Inverses Dan Sloughter Furman University January 30, 2008 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 1 / 24 I A is invertible. I The only solution to A~x = ~0 is the trivial solution ~x = ~0. I The reduced row-echelon form of A is In. ~ ~ n I A~x = b has a solution for b in R . I There exists an n × n matrix C for which AC = In. Result from previous lecture I The following statements are equivalent for an n × n matrix A: Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 I The only solution to A~x = ~0 is the trivial solution ~x = ~0. I The reduced row-echelon form of A is In. ~ ~ n I A~x = b has a solution for b in R . I There exists an n × n matrix C for which AC = In. Result from previous lecture I The following statements are equivalent for an n × n matrix A: I A is invertible. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 I The reduced row-echelon form of A is In. ~ ~ n I A~x = b has a solution for b in R . I There exists an n × n matrix C for which AC = In. Result from previous lecture I The following statements are equivalent for an n × n matrix A: I A is invertible. I The only solution to A~x = ~0 is the trivial solution ~x = ~0. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 ~ ~ n I A~x = b has a solution for b in R . I There exists an n × n matrix C for which AC = In. Result from previous lecture I The following statements are equivalent for an n × n matrix A: I A is invertible. I The only solution to A~x = ~0 is the trivial solution ~x = ~0. I The reduced row-echelon form of A is In. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 I There exists an n × n matrix C for which AC = In. Result from previous lecture I The following statements are equivalent for an n × n matrix A: I A is invertible. I The only solution to A~x = ~0 is the trivial solution ~x = ~0. I The reduced row-echelon form of A is In. ~ ~ n I A~x = b has a solution for b in R . Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 Result from previous lecture I The following statements are equivalent for an n × n matrix A: I A is invertible. I The only solution to A~x = ~0 is the trivial solution ~x = ~0. I The reduced row-echelon form of A is In. ~ ~ n I A~x = b has a solution for b in R . I There exists an n × n matrix C for which AC = In. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 2 / 24 I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. I Note: this requires solving n systems of n equations in n unknowns. I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. −1 I That is, the same row operations which take A to In take In to A . Computing an inverse I The proof of the previous theorem provides a method for finding inverses. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. I Note: this requires solving n systems of n equations in n unknowns. I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. −1 I That is, the same row operations which take A to In take In to A . Computing an inverse I The proof of the previous theorem provides a method for finding inverses. I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 I Note: this requires solving n systems of n equations in n unknowns. I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. −1 I That is, the same row operations which take A to In take In to A . Computing an inverse I The proof of the previous theorem provides a method for finding inverses. I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. −1 I That is, the same row operations which take A to In take In to A . Computing an inverse I The proof of the previous theorem provides a method for finding inverses. I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. I Note: this requires solving n systems of n equations in n unknowns. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 −1 I That is, the same row operations which take A to In take In to A . Computing an inverse I The proof of the previous theorem provides a method for finding inverses. I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. I Note: this requires solving n systems of n equations in n unknowns. I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 Computing an inverse I The proof of the previous theorem provides a method for finding inverses. I Namely: given an n × n matrix A, let ~c1, ~c2,..., ~cn be the solutions of A~x = ~ej , for j = 1; 2;:::; n. −1 I Then A is the matrix with columns ~c1, c~2,..., c~n. I Note: this requires solving n systems of n equations in n unknowns. I Note: we can do this at once by writing an augmented matrix consisting of A and In and reducing A to In. −1 I That is, the same row operations which take A to In take In to A . Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 3 / 24 I Using elementary row operations, we find 21 2 −1 1 0 03 21 2 −1 1 0 03 42 2 4 0 1 05 −! 40 −2 6 −2 1 05 1 3 −3 0 0 1 0 1 −2 −1 0 1 Example I We will find the inverse of 21 2 −13 A = 42 2 45 : 1 3 −3 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 4 / 24 Example I We will find the inverse of 21 2 −13 A = 42 2 45 : 1 3 −3 I Using elementary row operations, we find 21 2 −1 1 0 03 21 2 −1 1 0 03 42 2 4 0 1 05 −! 40 −2 6 −2 1 05 1 3 −3 0 0 1 0 1 −2 −1 0 1 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 4 / 24 I Hence 2 3 3 9 − 2 −5 −1 6 7 A = 4−5 1 35 : 1 −2 2 1 Example (cont'd) I Continuing: 21 2 −1 1 0 03 21 0 5 −1 1 03 −! 60 1 −3 1 − 1 07 −! 60 1 −3 1 − 1 07 4 2 5 4 2 5 1 0 1 −2 −1 0 1 0 0 1 −2 2 1 2 3 3 1 0 0 9 − 2 −5 6 7 −! 40 1 0 −5 1 35 : 1 0 0 1 −2 2 1 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 5 / 24 Example (cont'd) I Continuing: 21 2 −1 1 0 03 21 0 5 −1 1 03 −! 60 1 −3 1 − 1 07 −! 60 1 −3 1 − 1 07 4 2 5 4 2 5 1 0 1 −2 −1 0 1 0 0 1 −2 2 1 2 3 3 1 0 0 9 − 2 −5 6 7 −! 40 1 0 −5 1 35 : 1 0 0 1 −2 2 1 I Hence 2 3 3 9 − 2 −5 −1 6 7 A = 4−5 1 35 : 1 −2 2 1 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 5 / 24 I Using elementary row operations, we find that 2 2 1 −4 1 0 03 22 1 −4 1 0 03 4−4 −1 6 0 1 05 −! 40 1 −2 2 1 05 −2 2 −2 0 0 1 0 3 −6 1 0 1 Example I Suppose we wish to find an inverse for 2 2 1 −43 A = 4−4 −1 65 : −2 2 −2 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 6 / 24 Example I Suppose we wish to find an inverse for 2 2 1 −43 A = 4−4 −1 65 : −2 2 −2 I Using elementary row operations, we find that 2 2 1 −4 1 0 03 22 1 −4 1 0 03 4−4 −1 6 0 1 05 −! 40 1 −2 2 1 05 −2 2 −2 0 0 1 0 3 −6 1 0 1 Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 6 / 24 I Note: this shows that it is not possible to reduce A to I3 using elementary row operations.