DOCSLIB.ORG

Algorithms to Compute Bases and the Rank of a Matrix Subspaces

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Algorithms to Compute Bases and the Rank of a Matrix Subspaces Algorithms to Compute Bases and the Rank of a Matrix Subspaces associated to a matrix Suppose that A is an m × n matrix. The row space of A is the subspace of Rn spanned m by the rows of A. The column space of A is the subspace of R spanned by the columns of A. The null space of A is the subspace n ~ N(A) = f~x 2 R j A~x = 0mg n n of R . N(A) is a subspace of R by the Subspace Theorem (Theorem 1 of Lecture Note 4). The row rank of A is the dimension of the row space of A. The column rank of A is the dimension of the column space of A. The nullity of A is the dimension of the null space N(A). More on row equivalence We give some theorems on row equivalence, showing that some subspaces constructed from matrices do not change under elementary row operations. This allows us to find important information about these subspaces from the RRE form of a matrix. These interpretations will be given later in this note. Theorem 0.1. Suppose that the m × n matrices 0 1 0 1 A1 B1 B . C B . C A = @ . A and B = @ . A Am Bm are row equivalent (with A1;:::;Am;B1;:::;Bm 2 Rn). Then the subspaces Span(A1;:::;Am) and Span(B1;:::;Bm) of Rn are equal. To prove Theorem 0.1, we need only verify it for a single elementary row operation. Theorem 0.2. Suppose that A and B are row equivalent matrices. Then N(A) = N(B). Theorem 0.2 is just a restatement of Theorem 6 of Lecture Note 1. Theorem 0.3. Suppose that m × n matrices A = (A1;:::;An) and B = (B1;:::;Bn) 1 n 1 n m with A ;:::;A ;B ;:::;B 2 R are row equivalent. Then the subspaces T n 1 n ~ f(x1; : : : ; xn) 2 R j x1A + ··· + xnA = 0g and T n 1 n ~ f(x1; : : : ; xn) 2 R j x1B + ··· + xnB = 0g n 1 n 1 n of R are equal. That is, A ;:::;A and B ;:::;B have the same dependence relations. Theorem 0.3 follows from Theorem 0.2 and the identities 0 1 x1 B . C 1 n A @ . A = x1A + ··· + xnA xn and 0 1 x1 B . C 1 n B @ . A = x1B + ··· + xnB xn 1 of formula (1) of Lecture Note 2. Algorithm to compute a basis of the row space of a matrix. Suppose that 0 1 A1 B A2 C B C A = B . C @ . A Am is an m × n matrix. Transform A into its reduced row echelon form B. Then the nonzero rows of B form a basis of the row space Span(A1;:::;Am) of A. To establish this algorithm, apply Theorem 0.1 to conclude that the span of the rows of B is equal to the row space of A. Using the fact that every nonzero row of the RRE form B contains a leading one, it follows from the definition of linear independence that the nonzero rows of B are linearly independent. Example 0.4. Let u1 = (1; 2; 1; 5); u2 = (2; 4; 1; 8); u3 = (3; 6; 2; 13) Find a basis of the subspace X = Span(u1; u2; u3) of R4. Solution: Let 0 1 0 1 u1 1 2 1 5 A = @ u2 A = @ 2 4 1 8 A : u3 3 6 2 13 We compute the RRE form of A which is 0 1 2 0 3 1 @ 0 0 1 2 A : 0 0 0 0 The first two rows are nonzero. Thus f(1; 2; 0; 3); (0; 0; 1; 2)g is a basis of X Algorithm to refine the columns of a matrix to a basis of its column space. Sup- pose that A = (A1;A2; ··· ;An) is an m×n matrix. Transform A into its reduced row ech- elon form B = (B1;B2; ··· ;Bn). Let Bi1 ;Bi2 ;:::;Bir be the columns of B which contain a leading 1. Then fAi1 ;Ai2 ;:::;Air g is a basis of the column space Span(A1;A2;:::;An) of A. To establish this algorithm, observe using Theorem 0.3 that the RRE form B of A gives the standard form solutions to the equation 1 n x1A + ··· + xnA = ~0: i i ir Substituting the standard form solution for the xi, we obtain first that A 1 ;A 2 ;:::;A are linearly independent, and then that they span the column space of A. 2 Example 0.5. Find a subset of the vectors 0 1 1 0 2 1 0 1 1 0 5 1 v1 = @ 2 A ; v2 = @ 4 A ; v3 = @ 1 A ; v4 = @ 8 A 3 6 2 13 3 which is a basis of the subspace W = Span(v1; v2; v3; v4) of R . Solution: Let 0 1 2 1 5 1 A = (v1; v2; v2; v4) = @ 2 4 1 8 A : 3 6 2 13 We compute the RRE form of A which is 0 1 2 0 3 1 @ 0 0 1 2 A : 0 0 0 0 There are leading ones in the first and third columns. Thus 8 0 1 0 19 < 1 1 = v1 = @ 2 A ; v3 = @ 1 A : 3 2 ; is a basis of W Algorithm to compute a basis of the null space of a matrix. Let A = (aij) be an m × n matrix, and 0 1 x1 B x2 C B C ~x = B . C @ . A xn be a n × 1 vector of indeterminates. Let N(A) be the null space of the matrix A. By Theorem 0.2, a basis for N(A) can be found by solving the system A~x = ~0m using Gaussian elimination to find the standard form solution, putting the standard form solution into a column vector and expanding with indeterminate coefficients. The vectors in this expansion are a basis of N(A). Example 0.6. Find a basis of N(A) if 0 1 2 3 4 1 A = @ 1 3 5 6 A : 2 5 8 10 Solution: We compute that the RRE form of A is 0 1 0 −1 0 1 @ 0 1 2 2 A : 0 0 0 0 3 We write the standard form solution as 0 1 0 1 0 1 0 1 x1 s 1 0 B x2 C B −2s − 2t C B −2 C B −2 C B C = B C = s B C + t B C @ x3 A @ s A @ 1 A @ 0 A x4 t 0 1 with s; t 2 R. Thus 80 1 1 0 0 19 > > <B −2 C B −2 C= B C ; B C >@ 1 A @ 0 A> :> 0 1 ;> is a basis of N(A). Algorithm to extend a set of linearly independent row vectors to a basis of Rn. Suppose that w1; : : : ; wm 2 Rn are linearly independent. Let fe1; : : : ; eng be the standard basis of Rn. Transform the m × n matrix 0 1 w1 B w2 C B C B . C @ . A wm into a reduced row echelon form B. Let Bi1 ;Bi2 ;:::;Bin−m be the indices of the columns of B which do not contain a leading 1. Then fw1; : : : ; wm; ei1 ; : : : ; ein−m g is a basis of Rn. To establish this algorithm, recall that the span of the rows of the RRE form of B is equal to Span(w1; : : : ; wm) by Theorem 0.1. Since w1; : : : ; wm are linearly independent, B contains m leading ones. One then checks that the rows of B and ei1 ; : : : ; ein−m are linearly independent. We then have that w1; : : : ; wm; ei1 ; : : : ; ein−m span an n-dimensional subspace of Rn, so they are a basis of Rn by 2) of Theorem 4.10 of Lecture Note 4. The rank of a matrix Theorem 0.7. rowrank(A) = columnrank(A). Proof. Let r be the number of leading ones in the RRE form of A. By the algorithm to compute a basis of a row space of a matrix, a basis of the row space of A has r vectors. By the algorithm to refine the columns of a matrix to a basis of its column space, a basis of the column space of A also has r vectors. Thus the row space and column space of A have the same dimension. We define the rank of A to be the row rank of A (which is also the column rank of A). Theorem 0.8. (Rank Nullity Theorem) Suppose that A is an m × n matrix. Then rank(A) + nullity(A) = n: Proof. Let r be the number of leading ones in the RRE form of A. Then rank(A) = r (by the algorithm to compute a basis of the row space of a matrix or the algorithm to refine the columns of a matrix to a basis of its column space). The nullity of A is n − r, by the algorithm to compute a basis of the null space of a matrix, since there are n − r free ~ variables in the general solution to A~x = 0m. 4 As explained in Theorem 2 of Lecture Note 5, we have the following theorem. Theorem 0.9. Suppose that A is an n × n matrix. Then the columns of A are linearly independent if and only if Det(A) 6= 0. Suppose that A is an m × n matrix. An r × r submatrix of A is a matrix obtained from A by removing m − r rows and n − r columns.
Load more

Recommended publications
  • Implementation of Gaussian- Elimination
    International Journal of Innovative Technology and Exploring Engineering (IJITEE) ISSN: 2278-3075, Volume-5 Issue-11, April 2016 Implementation of Gaussian- Elimination Awatif M.A. Elsiddieg Abstract: Gaussian elimination is an algorithm for solving systems of linear equations, can also use to find the rank of any II. PIVOTING matrix ,we use Gaussian Jordan elimination to find the inverse of a non singular square matrix. This work gives basic concepts The objective of pivoting is to make an element above or in section (1) , show what is pivoting , and implementation of below a leading one into a zero. Gaussian elimination to solve a system of linear equations. The "pivot" or "pivot element" is an element on the left Section (2) we find the rank of any matrix. Section (3) we use hand side of a matrix that you want the elements above and Gaussian elimination to find the inverse of a non singular square matrix. We compare the method by Gauss Jordan method. In below to be zero. Normally, this element is a one. If you can section (4) practical implementation of the method we inherit the find a book that mentions pivoting, they will usually tell you computation features of Gaussian elimination we use programs that you must pivot on a one. If you restrict yourself to the in Matlab software. three elementary row operations, then this is a true Keywords: Gaussian elimination, algorithm Gauss, statement. However, if you are willing to combine the Jordan, method, computation, features, programs in Matlab, second and third elementary row operations, you come up software.
    [Show full text]
  • Linear Algebra and Matrix Theory
    Linear Algebra and Matrix Theory Chapter 1 - Linear Systems, Matrices and Determinants This is a very brief outline of some basic definitions and theorems of linear algebra. We will assume that you know elementary facts such as how to add two matrices, how to multiply a matrix by a number, how to multiply two matrices, what an identity matrix is, and what a solution of a linear system of equations is. Hardly any of the theorems will be proved. More complete treatments may be found in the following references. 1. References (1) S. Friedberg, A. Insel and L. Spence, Linear Algebra, Prentice-Hall. (2) M. Golubitsky and M. Dellnitz, Linear Algebra and Differential Equa- tions Using Matlab, Brooks-Cole. (3) K. Hoffman and R. Kunze, Linear Algebra, Prentice-Hall. (4) P. Lancaster and M. Tismenetsky, The Theory of Matrices, Aca- demic Press. 1 2 2. Linear Systems of Equations and Gaussian Elimination The solutions, if any, of a linear system of equations (2.1) a11x1 + a12x2 + ··· + a1nxn = b1 a21x1 + a22x2 + ··· + a2nxn = b2 . am1x1 + am2x2 + ··· + amnxn = bm may be found by Gaussian elimination. The permitted steps are as follows. (1) Both sides of any equation may be multiplied by the same nonzero constant. (2) Any two equations may be interchanged. (3) Any multiple of one equation may be added to another equation. Instead of working with the symbols for the variables (the xi), it is eas- ier to place the coefficients (the aij) and the forcing terms (the bi) in a rectangular array called the augmented matrix of the system. a11 a12 .
    [Show full text]
  • Math 204, Spring 2020 About the First Test
    Math 204, Spring 2020 About the First Test The first test for this course will be given in class on Friday, February 21. It covers all of the material that we have done in Chapters 1 and 2, up to Chapter 2, Section II. For the test, you should know and understand all the definitions and theorems that we have covered. You should be able to work with matrices and systems of linear equations. The test will include some \short essay" questions that ask you to define something, or discuss something, or explain something, and so on. Other than that, you can expect most of the questions to be similar to problems that have been given on the homework. You can expect to do a few proofs, but they will be fairly straightforward. Here are some terms and ideas that you should be familiar with for the test: systems of linear equations solution set of a linear system of equations Gaussian elimination the three row operations notations for row operations: kρi + ρj, ρi $ ρj, kρj row operations are reversible applying row operations to a system of equations echelon form for a system of linear equations matrix; rows and columns of a matrix; m × n matrix representing a system of linear equations as an augmented matrix row operations and echelon form for matrices applying row operations to a matrix n vectors in R ; column vectors and row vectors n vector addition and scalar multiplication for vectors in R n linear combination of vectors in R expressing the solution set of a linear system in vector form homogeneous system of linear equations associated homogeneous
    [Show full text]
  • The Rank of Recurrence Matrices Christopher Lee University of Portland, [email protected]
    University of Portland Pilot Scholars Mathematics Faculty Publications and Presentations Mathematics 5-2014 The Rank of Recurrence Matrices Christopher Lee University of Portland, [email protected] Valerie Peterson University of Portland, [email protected] Follow this and additional works at: http://pilotscholars.up.edu/mth_facpubs Part of the Applied Mathematics Commons, and the Mathematics Commons Citation: Pilot Scholars Version (Modified MLA Style) Lee, Christopher and Peterson, Valerie, "The Rank of Recurrence Matrices" (2014). Mathematics Faculty Publications and Presentations. 9. http://pilotscholars.up.edu/mth_facpubs/9 This Journal Article is brought to you for free and open access by the Mathematics at Pilot Scholars. It has been accepted for inclusion in Mathematics Faculty Publications and Presentations by an authorized administrator of Pilot Scholars. For more information, please contact [email protected]. The Rank of Recurrence Matrices Christopher Lee and Valerie Peterson Christopher Lee ([email protected]), a Wyoming native, earned his Ph.D. from the University of Illinois in 2009; he is currently a Visiting Assistant Professor at the University of Portland. His primary field of research lies in differential topology and geometry, but he has interests in a variety of disciplines, including linear algebra and the mathematics of physics. When not teaching or learning math, Chris enjoys playing hockey, dabbling in cooking, and resisting the tendency for gravity to anchor heavy things to the ground. Valerie Peterson ([email protected]) received degrees in mathematics from Santa Clara University (B.S.) and the University of Illinois at Urbana-Champaign (Ph.D.) before joining the faculty at the University of Portland in 2009.
    [Show full text]
  • Math 2331 – Linear Algebra 4.6 Rank
    4.6 Rank Math 2331 { Linear Algebra 4.6 Rank Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/∼jiwenhe/math2331 Jiwen He, University of Houston Math 2331, Linear Algebra 1 / 16 4.6 Rank Row Space Rank Theorem IMT 4.6 Rank Row Space Row Space and Row Equivalence Row Space: Examples Rank: Defintion Rank Theorem Rank Theorem: Examples Visualizing Row A and Nul A The Invertible Matrix Theorem (continued) Jiwen He, University of Houston Math 2331, Linear Algebra 2 / 16 4.6 Rank Row Space Rank Theorem IMT Row Space Row Space The set of all linear combinations of the row vectors of a matrix A is called the row space of A and is denoted by Row A. Example Let 2 3 −1 2 3 6 r1 = (−1; 2; 3; 6) A = 4 2 −5 −6 −12 5 and r2 = (2; −5; −6; −12) : 1 −3 −3 −6 r3 = (1; −3; −3; −6) 4 Row A = Spanfr1; r2; r3g (a subspace of R ) While it is natural to express row vectors horizontally, they can also be written as column vectors if it is more convenient. Therefore Col AT = Row A . Jiwen He, University of Houston Math 2331, Linear Algebra 3 / 16 4.6 Rank Row Space Rank Theorem IMT Row Space and Row Equivalence When we use row operations to reduce matrix A to matrix B, we are taking linear combinations of the rows of A to come up with B. We could reverse this process and use row operations on B to get back to A.
    [Show full text]
  • Section 3.3. Matrix Rank and the Inverse of a Full Rank Matrix
    3.3. Matrix Rank and the Inverse of a Full Rank Matrix 1 Section 3.3. Matrix Rank and the Inverse of a Full Rank Matrix Note. The lengthy section (21 pages in the text) gives a thorough study of the rank of a matrix (and matrix products) and considers inverses of matrices briefly at the end. Note. Recall that the row space of a matrix A is the span of the row vectors of A and the row rank of A is the dimension of this row space. Similarly, the column space of A is the span of the column vectors of A and the column rank is the dimension of this column space. You will recall that the dimension of the column space and the dimension of the row space of a given matrix are the same (see Theorem 2.4 of Fraleigh and Beauregard’s Linear Algebra, 3rd Edition, Addison- Wesley Publishing Company, 1995, in 2.2. The Rank of a Matrix). We now give a proof of this based in part on Gentle’s argument and on Peter Lancaster’s Theory of Matrices, Academic Press (1969), page 42. First, we need a lemma. i k i i i k n Lemma 3.3.1. Let {a }i=1 = {[a1, a2, . , an]}i=1 be a set of vectors in R and let i k π ∈ Sn. Then the set of vectors {a }i=1 is linearly independent if and only if the set i i i k of vectors {[aπ(1), aπ(2), . , aπ(n)]}i=1 is linearly independent.
    [Show full text]
  • Math 290 Linear Modeling for Engineers
    Math 290 Linear Modeling for Engineers Instructor: Minnie Catral Text: Linear Algebra and Its Applications, 3rd ed. Update, by David Lay Have you ever used Matlab as a computing tool? 1. Yes 2. No 2 1.1 Systems of Equations System of 2 equations in 2 unknowns: Three possibilities: 1. The lines intersect. 2. The lines coincide. 3. The lines are parallel. 3 More generally: System of m equations in n unknowns A system is consistent if it has at least one solution. Otherwise, it is said to be inconsistent. 4 FACT: A system of linear equations has either (1) exactly one solution or (2) infinitely many solutions or (3) no solutions. The solution set is the set of all possible solutions of a linear system. Two systems are equivalent if they have the same solution set. STRATEGY FOR SOLVING A SYSTEM: Replace a system with an equivalent system that is easier to solve. 5 • EXAMPLE: 6 Matrix Notation 7 Elementary Row Operations: 1. (Interchange) Interchange two rows. 2. (Scaling) Multiply a row by a nonzero constant. 3. (Replacement) Replace a row by the sum of itself and a multiple of another row. Row equivalent Matrices: Two matrices where one matrix can be transformed into the other matrix by a sequence of elementary row operations. Notation: Fact about Row Equivalence: If the augmented matrices of two linear systems are row equivalent, then the two systems have the same solution set. 8 • EXAMPLE 9 Two Fundamental Questions (Existence and Uniqueness) 1. Is the system consistent? (i.e. does a solution exist?) 2.
    [Show full text]
  • Notes on Row Reduction
    Notes on Row Reduction Francis J. Narcowich Department of Mathematics Texas A&M University May 2012 1 Notation and Terms The row echelon form of a matrix contains a great deal of information, both about the matrix itself and about systems of equations that may be associated with it. To talk about row reduction, we need to define several terms and introduce some notation. 1. Notation for elementary row operations. Row reduction is made easier by having notation for various row operations. This is the notation that we will use here. (The equals sign “‘=” used in the row operations below means “replacement,” which is common usage in computer languages.) I. R ↔ R′ means interchange rows R and R′. II. R = cR means multiply the current row R by c = 0 and make the result the new row R. III. R = R + cR′ means multiply R′ by c, then add cR′ to R, and make the result the new row R. 2. Row equivalence. A matrix A is row equivalent to a matrix B if A can be transformed into B using elementary row operations. When this happens, we write A ⇔ B. 3. Leading entry. The leading entry1 in a row is the first non-zero entry in a row. The leading entries in each row of M are in boldface type. 0 1 3 2 M = 2 4 0 −1 0 0 6 5 4. Row echelon form of a matrix. A matrix is in row echelon form if these hold: 1Leon does not have a name for this entry.
    [Show full text]
  • Gaussian Elimination Notation for Linear Systems
    Gaussian Elimination Notation for Linear Systems Last time we studied the linear system x + y = 27 (1) 2x − y = 0 (2) and found that x = 9 (3) y = 18 (4) We learned to write the linear system using a matrix and two vectors like so: ! ! ! 1 2 x 27 = 2 −1 y 0 Likewise, we can write the solution as: ! ! ! 1 0 x 9 = 0 1 y 18 ! 1 0 The matrix I = is called the Identity Matrix. You can check 0 1 that for any vector v, then Iv = v. A useful shorthand for a linear system is an Augmented Matrix, which looks like this for the linear system we've been dealing with: ! 1 1 27 2 −1 ! x We don't bother writing the vector , since it will show up in any y linear system we deal with, and we just leave blank spaces for 0's. The solution to the linear system looks like this: 1 ! 1 9 1 18 Here's another example of an augmented matrix, for a linear system with three equations and four unknowns: 0 1 3 2 91 B C @ 6 2 −2 A −1 1 1 3 And finally, here's the general case. The number of equations in the linear system is the number of rows r in the augmented matrix, and the number of columns k in the matrix left of the vertical line is the number of unknowns. 0 1 1 1 11 a1 a1 : : : a1 b B 2 2 2 2C Ba1 a2 : : : ak b C B .
    [Show full text]
  • Row Equivalence of Matrices
    Row Equivalence of matrices March 1, 2007 0.1 Row equivalence Let F be a field and let m and n be positive integers. Two m by n matrices are said to be row equivalent if there is an invertible matrix S such that B = SA. (Check that this is indeed an equivalence relation.) The textbook shows that any two row equivalent matrices have the same null space. In fact the converse is also true, so that we have the following theorem: Theorem: If A and B are two m by n matrices, then the following conditions are equivalent: 1. There exists an invertible matrix S such that B = SA 2. The matrices A and B have the same nullspace. This theorem can be translated into a statement about linear transfor- mations by considering the linear transformations LA, LB, and LS associated with the matrices above. It seems more revealing to treat this more abstract looking version. Theorem Let V and W be finite dimensional vector spaces, and let T and T 0 be linear transformations from V to W . Then the following are equivalent: 1. There exists an isomorphism S: T → T 0 such that T 0 = S ◦ T . 2. The transformations T and T 0 have the same kernel. Proof: The proof that (1) implies (2) is relatively easy (and is in the book) and we omit it. Let us prove that (2) implies (1). Let K := Ker(T ) = Ker(T 0), a linear subspace of V . By the rank-nulity theorem, the dimension k of K is n−r, where r is the rank of T and hence also the rank of T 0.
    [Show full text]
  • LINEAR ALGEBRA Rudi Weikard
    LINEAR ALGEBRA Lecture notes for MA 434/534 Rudi Weikard 250 200 150 100 50 0 Mar 30 Apr 06 Apr 13 Apr 20 Apr 27 Version of December 1, 2020 Contents Preface iii Chapter 1. Systems of linear equations1 1.1. Introduction1 1.2. Solving systems of linear equations2 1.3. Matrices and vectors3 1.4. Back to systems of linear equations5 Chapter 2. Vector spaces7 2.1. Spaces and subspaces7 2.2. Linear independence and spans8 2.3. Direct sums 10 Chapter 3. Linear transformations 13 3.1. Basics 13 3.2. The fundamental theorem of linear algebra 14 3.3. The algebra of linear transformation 15 3.4. Linear transformations and matrices 15 3.5. Matrix algebra 16 Chapter 4. Inner product spaces 19 4.1. Inner products 19 4.2. Orthogonality 20 4.3. Linear functionals and adjoints 21 4.4. Normal and self-adjoint transformations 22 4.5. Least squares approximation 23 Chapter 5. Spectral theory 25 5.1. Eigenvalues and Eigenvectors 25 5.2. Spectral theory for general linear transformations 26 5.3. Spectral theory for normal transformations 27 5.4. The functional calculus for general linear transformations 28 Appendix A. Appendix 33 A.1. Set Theory 33 A.2. Algebra 34 List of special symbols 37 Index 39 i ii CONTENTS Bibliography 41 Preface We all know how to solve two linear equations in two unknowns like 2x − 3y = 5 and x + 3y = −2: Linear Algebra grew out of the need to solve simultaneously many such equations for perhaps many unknowns. For instance, the frontispiece of these notes shows a number of data points and an attempt to find the \best" straight line as an approximation.
    [Show full text]
  • Bc = 0, Then the Matrix a = (Abcd ) Is Inve
    Math 60. Rumbos Spring 2009 1 Solutions to Assignment #17 a b 1. Prove that if ad − bc 6= 0, then the matrix A = is invertible and c d compute A−1. a b Solution: Let A = and assume that ad − bc 6= 0. c d First consider the case a 6= 0. Perform the elementary row operations: (1=a)R1 ! R1 and −cR1 + R2 ! R2, successively, on the augmented matrix a b j 1 0 ; (1) c d j 0 1 we obtain the augmented matrix 1 b=a j 1=a 0 ; 0 −(cb=a) + d j −c=a 1 or 1 b=a j 1=a 0 ; (2) 0 ∆=a j −c=a 1 where we have set ∆ = ad − bc; thus ∆ 6= 0. a Next, perform the elementary row operation R ! R on the aug- ∆ 2 2 mented matrix in (2) to get 1 b=a j 1=a 0 : (3) 0 1 j −c=∆ a=∆ b Finally, perform the elementary row operation − R + R ! R on a 2 1 1 the augmented matrix in (3) to get 1 0 j d=∆ −b=∆ : (4) 0 1 j −c=∆ a=∆ From (4) we then see that, if ∆ = ad − bc 6= 0, then A is invertible and d=∆ −b=∆ A−1 = ; −c=∆ a=∆ or 1 d −b A−1 = : (5) ∆ −c a Math 60. Rumbos Spring 2009 2 Observe that the formula for A−1 in (5) also works for the case a = 0. In this case, ∆ = −bc; so that b 6= 0 and c 6= 0, and 1 d −b A−1 = ; −bc −c 0 or −d=bc 1=c A−1 = ; 1=b 0 and we can check that −d=bc 1=c 0 b 1 0 = ; 1=b 0 c d 0 1 and therefore A is invertible in this case as well.
    [Show full text]