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
1 2 −1 1 0 0 1 2 −1 1 0 0 2 2 4 0 1 0 −→ 0 −2 6 −2 1 0 1 3 −3 0 0 1 0 1 −2 −1 0 1
Example
I We will find the inverse of 1 2 −1 A = 2 2 4 . 1 3 −3
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 4 / 24 Example
I We will find the inverse of 1 2 −1 A = 2 2 4 . 1 3 −3
I Using elementary row operations, we find
1 2 −1 1 0 0 1 2 −1 1 0 0 2 2 4 0 1 0 −→ 0 −2 6 −2 1 0 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 3 9 − 2 −5 −1 A = −5 1 3 . 1 −2 2 1
Example (cont’d)
I Continuing:
1 2 −1 1 0 0 1 0 5 −1 1 0 −→ 0 1 −3 1 − 1 0 −→ 0 1 −3 1 − 1 0 2 2 1 0 1 −2 −1 0 1 0 0 1 −2 2 1 3 1 0 0 9 − 2 −5 −→ 0 1 0 −5 1 3 . 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:
1 2 −1 1 0 0 1 0 5 −1 1 0 −→ 0 1 −3 1 − 1 0 −→ 0 1 −3 1 − 1 0 2 2 1 0 1 −2 −1 0 1 0 0 1 −2 2 1 3 1 0 0 9 − 2 −5 −→ 0 1 0 −5 1 3 . 1 0 0 1 −2 2 1
I Hence 3 9 − 2 −5 −1 A = −5 1 3 . 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 1 −4 1 0 0 2 1 −4 1 0 0 −4 −1 6 0 1 0 −→ 0 1 −2 2 1 0 −2 2 −2 0 0 1 0 3 −6 1 0 1
Example
I Suppose we wish to find an inverse for
2 1 −4 A = −4 −1 6 . −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 1 −4 A = −4 −1 6 . −2 2 −2
I Using elementary row operations, we find that
2 1 −4 1 0 0 2 1 −4 1 0 0 −4 −1 6 0 1 0 −→ 0 1 −2 2 1 0 −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.
I It follows that A does not have an inverse.
Example (cont’d)
I Continuing, we have
2 1 −4 1 0 0 −→ 0 1 −2 2 1 0 . 0 0 0 −5 −3 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 7 / 24 I It follows that A does not have an inverse.
Example (cont’d)
I Continuing, we have
2 1 −4 1 0 0 −→ 0 1 −2 2 1 0 . 0 0 0 −5 −3 1
I Note: this shows that it is not possible to reduce A to I3 using elementary row operations.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 7 / 24 Example (cont’d)
I Continuing, we have
2 1 −4 1 0 0 −→ 0 1 −2 2 1 0 . 0 0 0 −5 −3 1
I Note: this shows that it is not possible to reduce A to I3 using elementary row operations.
I It follows that A does not have an inverse.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 7 / 24 I apply rref to the augmented matrix
1 2 3 1 0 0 −1 0 1 0 1 0 , 4 2 1 0 0 1
I or use inv(A).
Inverses in Octave
I To find the inverse of 1 2 3 A = −1 0 1 , 4 2 1
we can either
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 8 / 24 I or use inv(A).
Inverses in Octave
I To find the inverse of 1 2 3 A = −1 0 1 , 4 2 1
we can either
I apply rref to the augmented matrix
1 2 3 1 0 0 −1 0 1 0 1 0 , 4 2 1 0 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 8 / 24 Inverses in Octave
I To find the inverse of 1 2 3 A = −1 0 1 , 4 2 1
we can either
I apply rref to the augmented matrix
1 2 3 1 0 0 −1 0 1 0 1 0 , 4 2 1 0 0 1
I or use inv(A).
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 8 / 24 I Use the menus to find inverses in wxMaxima.
I In wxMaxima, A.B multiplies the matrices A and B.
I In wxMaxima, A^^3 raises A to the third power.
Inverses in Octave (cont’d)
I In either case, we find that −1.0000 2.0000 1.0000 −1 A = 2.5000 −5.5000 −2.0000 −1.0000 3.0000 1.0000
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 9 / 24 I In wxMaxima, A.B multiplies the matrices A and B.
I In wxMaxima, A^^3 raises A to the third power.
Inverses in Octave (cont’d)
I In either case, we find that −1.0000 2.0000 1.0000 −1 A = 2.5000 −5.5000 −2.0000 −1.0000 3.0000 1.0000
I Use the menus to find inverses in wxMaxima.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 9 / 24 I In wxMaxima, A^^3 raises A to the third power.
Inverses in Octave (cont’d)
I In either case, we find that −1.0000 2.0000 1.0000 −1 A = 2.5000 −5.5000 −2.0000 −1.0000 3.0000 1.0000
I Use the menus to find inverses in wxMaxima.
I In wxMaxima, A.B multiplies the matrices A and B.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 9 / 24 Inverses in Octave (cont’d)
I In either case, we find that −1.0000 2.0000 1.0000 −1 A = 2.5000 −5.5000 −2.0000 −1.0000 3.0000 1.0000
I Use the menus to find inverses in wxMaxima.
I In wxMaxima, A.B multiplies the matrices A and B.
I In wxMaxima, A^^3 raises A to the third power.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 9 / 24 I Examples: Each of the following is an elementary matrix:
1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 1 0 , 0 0 1 , and . −5 0 1 0 0 0 1 0 1 0 0 0 0 1
Definition
I An elementary matrix is any matrix which may be obtained by applying an elementary row operation to an identity matrix.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 10 / 24 Definition
I An elementary matrix is any matrix which may be obtained by applying an elementary row operation to an identity matrix.
I Examples: Each of the following is an elementary matrix:
1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 1 0 , 0 0 1 , and . −5 0 1 0 0 0 1 0 1 0 0 0 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 10 / 24 I Note: E is an elementary matrix obtained by multiplying the second row of I3 by 3.
I Note: 1 0 0 1 2 3 1 1 2 3 1 EA = 0 3 0 4 5 6 2 = 12 15 18 6 , 0 0 1 7 8 9 1 7 8 9 1
which is what we would obtain applying the same elementary row operation to A.
Example
I Let 1 2 3 1 1 0 0 A = 4 5 6 2 and E = 0 3 0 . 7 8 9 1 0 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 11 / 24 I Note: 1 0 0 1 2 3 1 1 2 3 1 EA = 0 3 0 4 5 6 2 = 12 15 18 6 , 0 0 1 7 8 9 1 7 8 9 1
which is what we would obtain applying the same elementary row operation to A.
Example
I Let 1 2 3 1 1 0 0 A = 4 5 6 2 and E = 0 3 0 . 7 8 9 1 0 0 1
I Note: E is an elementary matrix obtained by multiplying the second row of I3 by 3.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 11 / 24 Example
I Let 1 2 3 1 1 0 0 A = 4 5 6 2 and E = 0 3 0 . 7 8 9 1 0 0 1
I Note: E is an elementary matrix obtained by multiplying the second row of I3 by 3.
I Note: 1 0 0 1 2 3 1 1 2 3 1 EA = 0 3 0 4 5 6 2 = 12 15 18 6 , 0 0 1 7 8 9 1 7 8 9 1
which is what we would obtain applying the same elementary row operation to A.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 11 / 24 I Now 1 0 0 1 2 3 1 1 2 3 1 EA = 3 1 0 4 5 6 2 = 7 11 15 5 , 0 0 1 7 8 9 1 7 8 9 1
which is what we would obtain from applying the same row operation to A.
Example (cont’d)
I Let 1 0 0 E = 3 1 0 , 0 0 1 the elementary 3 × 3 matrix obtained by adding 3 times the first row to the second row of I3.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 12 / 24 Example (cont’d)
I Let 1 0 0 E = 3 1 0 , 0 0 1 the elementary 3 × 3 matrix obtained by adding 3 times the first row to the second row of I3.
I Now 1 0 0 1 2 3 1 1 2 3 1 EA = 3 1 0 4 5 6 2 = 7 11 15 5 , 0 0 1 7 8 9 1 7 8 9 1
which is what we would obtain from applying the same row operation to A.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 12 / 24 I Then 1 0 0 1 2 3 1 1 2 3 1 EA = 0 0 1 4 5 6 2 = 7 8 9 1 , 0 1 0 7 8 9 1 4 5 6 2
which is what we would obtain from applying the same row operation to A.
Example (cont’d)
I Let 1 0 0 E = 0 0 1 , 0 1 0 the elementary 3 × 3 matrix obtained by interchanging the second and third rows of I3.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 13 / 24 Example (cont’d)
I Let 1 0 0 E = 0 0 1 , 0 1 0 the elementary 3 × 3 matrix obtained by interchanging the second and third rows of I3.
I Then 1 0 0 1 2 3 1 1 2 3 1 EA = 0 0 1 4 5 6 2 = 7 8 9 1 , 0 1 0 7 8 9 1 4 5 6 2
which is what we would obtain from applying the same row operation to A.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 13 / 24 I Let E be the elementary matrix obtained by performing the same row operation on Im.
I Then B = EA.
Row operations
I Suppose an elementary row operation is performed on an m × n matrix A to obtain the matrix B.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 14 / 24 I Then B = EA.
Row operations
I Suppose an elementary row operation is performed on an m × n matrix A to obtain the matrix B.
I Let E be the elementary matrix obtained by performing the same row operation on Im.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 14 / 24 Row operations
I Suppose an elementary row operation is performed on an m × n matrix A to obtain the matrix B.
I Let E be the elementary matrix obtained by performing the same row operation on Im.
I Then B = EA.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 14 / 24 −1 I If E multiplies a row by a nonzero scalar c, E multiplies that same 1 row by c . −1 I If E multiplies row i by scalar c and adds it to row j, E multiplies row i by −c and adds it to row j. −1 I If E interchanges rows i and j, E interchanges rows i and j (and, so, E = E −1).
Inverses of elementary matrices
−1 I Every elementary matrix E is invertible, and E is itself an elementary matrix:
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 15 / 24 −1 I If E multiplies row i by scalar c and adds it to row j, E multiplies row i by −c and adds it to row j. −1 I If E interchanges rows i and j, E interchanges rows i and j (and, so, E = E −1).
Inverses of elementary matrices
−1 I Every elementary matrix E is invertible, and E is itself an elementary matrix: −1 I If E multiplies a row by a nonzero scalar c, E multiplies that same 1 row by c .
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 15 / 24 −1 I If E interchanges rows i and j, E interchanges rows i and j (and, so, E = E −1).
Inverses of elementary matrices
−1 I Every elementary matrix E is invertible, and E is itself an elementary matrix: −1 I If E multiplies a row by a nonzero scalar c, E multiplies that same 1 row by c . −1 I If E multiplies row i by scalar c and adds it to row j, E multiplies row i by −c and adds it to row j.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 15 / 24 Inverses of elementary matrices
−1 I Every elementary matrix E is invertible, and E is itself an elementary matrix: −1 I If E multiplies a row by a nonzero scalar c, E multiplies that same 1 row by c . −1 I If E multiplies row i by scalar c and adds it to row j, E multiplies row i by −c and adds it to row j. −1 I If E interchanges rows i and j, E interchanges rows i and j (and, so, E = E −1).
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 15 / 24 I B = Ek Ek−1 ··· E1A, where E1, E2,..., Ek are elementary matrices. I B = UA for some invertible matrix U. I U is the matrix obtained from Im by performing the row operations on Im which reduce A to B.
Theorem
I Suppose an m × n matrix A is reducible to an m × n matrix B by a series of elementary row operations. Then:
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 16 / 24 I B = UA for some invertible matrix U. I U is the matrix obtained from Im by performing the row operations on Im which reduce A to B.
Theorem
I Suppose an m × n matrix A is reducible to an m × n matrix B by a series of elementary row operations. Then:
I B = Ek Ek−1 ··· E1A, where E1, E2,..., Ek are elementary matrices.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 16 / 24 I U is the matrix obtained from Im by performing the row operations on Im which reduce A to B.
Theorem
I Suppose an m × n matrix A is reducible to an m × n matrix B by a series of elementary row operations. Then:
I B = Ek Ek−1 ··· E1A, where E1, E2,..., Ek are elementary matrices. I B = UA for some invertible matrix U.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 16 / 24 Theorem
I Suppose an m × n matrix A is reducible to an m × n matrix B by a series of elementary row operations. Then:
I B = Ek Ek−1 ··· E1A, where E1, E2,..., Ek are elementary matrices. I B = UA for some invertible matrix U. I U is the matrix obtained from Im by performing the row operations on Im which reduce A to B.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 16 / 24 I Applying row reduction, we have
2 −1 4 6 1 0 0 1 2 1 2 0 1 0 1 2 1 2 0 1 0 −→ 2 −1 4 6 1 0 0 5 0 9 14 0 0 1 5 0 9 14 0 0 1
Example
I Let 2 −1 4 6 A = 1 2 1 2 . 5 0 9 14
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 17 / 24 Example
I Let 2 −1 4 6 A = 1 2 1 2 . 5 0 9 14
I Applying row reduction, we have
2 −1 4 6 1 0 0 1 2 1 2 0 1 0 1 2 1 2 0 1 0 −→ 2 −1 4 6 1 0 0 5 0 9 14 0 0 1 5 0 9 14 0 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 17 / 24 Example (cont’d)
I Continuing,
1 2 1 2 0 1 0 −→ 0 −5 2 2 1 −2 0 0 −10 4 4 0 −5 1 1 2 1 2 0 1 0 −→ 0 1 − 2 − 2 − 1 2 0 5 5 5 5 0 −10 4 4 0 −5 1 9 14 2 1 1 0 5 5 5 5 0 −→ 0 1 − 2 − 2 − 1 2 0 . 5 5 5 5 0 0 0 0 −2 −1 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 18 / 24 Example (cont’d)
I Hence if 9 14 1 0 5 5 B = 0 1 − 2 − 2 , 5 5 0 0 0 0 the reduced row-echelon form of A, and
2 1 5 5 0 U = − 1 2 0 , 5 5 −2 −1 1
then U is invertible and B = UA.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 19 / 24 I For example, using Octave to perform row reductions to reduced row-echelon form, we obtain
0.00000 0.00000 0.20000 V = 0.00000 0.50000 −0.10000 1.00000 0.50000 −0.50000
with the property that V is invertible and B = VA.
I In particular, this shows that we may have UA = VA and yet U 6= V .
Example (cont’d)
I Note: U is not unique.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 20 / 24 I In particular, this shows that we may have UA = VA and yet U 6= V .
Example (cont’d)
I Note: U is not unique.
I For example, using Octave to perform row reductions to reduced row-echelon form, we obtain
0.00000 0.00000 0.20000 V = 0.00000 0.50000 −0.10000 1.00000 0.50000 −0.50000
with the property that V is invertible and B = VA.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 20 / 24 Example (cont’d)
I Note: U is not unique.
I For example, using Octave to perform row reductions to reduced row-echelon form, we obtain
0.00000 0.00000 0.20000 V = 0.00000 0.50000 −0.10000 1.00000 0.50000 −0.50000
with the property that V is invertible and B = VA.
I In particular, this shows that we may have UA = VA and yet U 6= V .
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 20 / 24 I Reason: Since A is reducible to I , there exist elementary matrices D1, D2,..., Dk such that
I = Dk Dk−1 ··· A.
I Hence −1 −1 −1 A = D1 D2 ··· Dk . −1 I So let Ei = Di , i = 1, 2,..., k.
Theorem
I If A is an invertible matrix, then there exist elementary matrices E1, E2,... Ek such that A = E1E2 ··· Ek .
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 21 / 24 I Hence −1 −1 −1 A = D1 D2 ··· Dk . −1 I So let Ei = Di , i = 1, 2,..., k.
Theorem
I If A is an invertible matrix, then there exist elementary matrices E1, E2,... Ek such that A = E1E2 ··· Ek .
I Reason: Since A is reducible to I , there exist elementary matrices D1, D2,..., Dk such that
I = Dk Dk−1 ··· A.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 21 / 24 −1 I So let Ei = Di , i = 1, 2,..., k.
Theorem
I If A is an invertible matrix, then there exist elementary matrices E1, E2,... Ek such that A = E1E2 ··· Ek .
I Reason: Since A is reducible to I , there exist elementary matrices D1, D2,..., Dk such that
I = Dk Dk−1 ··· A.
I Hence −1 −1 −1 A = D1 D2 ··· Dk .
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 21 / 24 Theorem
I If A is an invertible matrix, then there exist elementary matrices E1, E2,... Ek such that A = E1E2 ··· Ek .
I Reason: Since A is reducible to I , there exist elementary matrices D1, D2,..., Dk such that
I = Dk Dk−1 ··· A.
I Hence −1 −1 −1 A = D1 D2 ··· Dk . −1 I So let Ei = Di , i = 1, 2,..., k.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 21 / 24 Example
I If 3 2 A = , 1 −1
then I2 = D4D3D2D1A, where
0 1 1 0 1 0 1 1 D1 = , D2 = , D3 = 1 , and D4 = . 1 0 −3 1 0 5 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 22 / 24 Example (cont’d)
I It follows that A = E1E2E3E4, where
0 1 1 0 1 0 1 −1 E = , E = , E = , and E = . 1 1 0 2 3 1 3 0 5 4 0 1
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 23 / 24 I There exist invertible matrices P and Q such that R = PA and S = QA. −1 −1 −1 I Hence A = Q S and A = P R, so S = QP R. I That is, S = UR for some invertible matrix U.
I We will not prove this, but the proof is based on the following observations:
Theorem
I If R and S are both reduced row-echelon forms of a matrix A, then R = S.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 24 / 24 I There exist invertible matrices P and Q such that R = PA and S = QA. −1 −1 −1 I Hence A = Q S and A = P R, so S = QP R. I That is, S = UR for some invertible matrix U.
Theorem
I If R and S are both reduced row-echelon forms of a matrix A, then R = S.
I We will not prove this, but the proof is based on the following observations:
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 24 / 24 −1 −1 −1 I Hence A = Q S and A = P R, so S = QP R. I That is, S = UR for some invertible matrix U.
Theorem
I If R and S are both reduced row-echelon forms of a matrix A, then R = S.
I We will not prove this, but the proof is based on the following observations:
I There exist invertible matrices P and Q such that R = PA and S = QA.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 24 / 24 I That is, S = UR for some invertible matrix U.
Theorem
I If R and S are both reduced row-echelon forms of a matrix A, then R = S.
I We will not prove this, but the proof is based on the following observations:
I There exist invertible matrices P and Q such that R = PA and S = QA. −1 −1 −1 I Hence A = Q S and A = P R, so S = QP R.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 24 / 24 Theorem
I If R and S are both reduced row-echelon forms of a matrix A, then R = S.
I We will not prove this, but the proof is based on the following observations:
I There exist invertible matrices P and Q such that R = PA and S = QA. −1 −1 −1 I Hence A = Q S and A = P R, so S = QP R. I That is, S = UR for some invertible matrix U.
Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 24 / 24