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Solving Linear Systems, Continued and the Inverse of a Matrix

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Solving Linear Systems Math 240 Solving Linear Systems Gauss-Jordan elimination Rank Solving Linear Systems, Continued Inverse matrices and Definition Computing inverses The Inverse of a Matrix Properties of inverses Using inverse matrices Conclusion Math 240 | Calculus III Summer 2013, Session II Monday, July 15, 2013 Solving Linear Systems Agenda Math 240 Solving Linear Systems Gauss-Jordan elimination Rank 1. Solving Linear Systems Inverse matrices Gauss-Jordan elimination Definition Computing The rank of a matrix inverses Properties of inverses Using inverse matrices Conclusion 2. The inverse of a square matrix Definition Computing inverses Properties of inverses Using inverse matrices Conclusion Solving Linear Systems Gaussian elimination Math 240 Solving Linear Gaussian elimination solves a linear system by reducing to Systems Gauss-Jordan REF via elementary row ops and then using back substitution. elimination Rank Example Inverse matrices 2 3 Definition 3x1 − 2x2 + 2x3 = 9 3 −2 2 9 Computing inverses x − 2x + x = 5 1 −2 1 5 Properties of 1 2 3 4 5 inverses Using inverse 2x1 − x2 − 2x3 = −1 2 −1 −2 −1 matrices Conclusion 2 3 1 −2 1 5 x1 − 2x2 + x3 = 5 ! 40 1 3 55 x2 + 3x3 = 5 0 0 1 2 x3 = 2 Steps 1. P12 3. A13(−2) 5. A23(−3) −1 2. A12(−3) 4. A32(−1) 6. M3 13 Back substitution gives the solution (1; −1; 2). Solving Linear Systems Gauss-Jordan elimination Math 240 Solving Linear Reducing the augmented matrix to RREF makes the system Systems Gauss-Jordan even easier to solve. elimination Rank Example Inverse matrices 2 3 2 3 Definition 1 −2 1 5 1 0 0 1 x1 = 1 Computing inverses 0 1 3 5 ! 0 1 0 −1 x = −1 4 5 4 5 2 Properties of inverses 0 0 1 2 0 0 1 2 x3 = 2 Using inverse matrices Conclusion Steps 1. A32(−3) 2. A31(−1) 3. A21(2) Now, without any back substitution, we can see that the solution is (1; −1; 2). The method of solving a linear system by reducing its augmented matrix to RREF is called Gauss-Jordan elimination. Solving Linear Systems The rank of a matrix Math 240 Solving Linear Definition Systems Gauss-Jordan The rank of a matrix, A, is the number of nonzero rows it has elimination Rank after reduction to REF. It is denoted by rank(A). Inverse matrices If A the coefficient matrix of an m × n linear system and Definition # Computing rank(A ) = rank(A) = n then the REF looks like inverses Properties of inverses 2 3 Using inverse 1 ∗∗···∗ matrices x1 = ∗ Conclusion 6 1 ∗ ::: ∗7 6 . .7 x2 = ∗ 6 .. .7 . 6 7 . 6 1 ∗7 40 5 x = ∗ 0 ::::::::::: 0 n Lemma Suppose Ax = b is an m × n linear system with augmented matrix A#. If rank(A#) = rank(A) = n then the system has a unique solution. Solving Linear Systems The rank of a matrix Math 240 Solving Linear Example Systems Gauss-Jordan elimination Determine the solution set of the linear system Rank Inverse x1 + x2 − x3 + x4 = 1; matrices Definition 2x1 + 3x2 + x3 = 4; Computing inverses 3x1 + 5x2 + 3x3 − x4 = 5: Properties of inverses Using inverse matrices Reduce the augmented matrix. Conclusion 21 1 −1 1 13 A12(−2) 21 1 −1 1 13 A13(−3) 42 3 1 0 45 −−−−−! 40 1 3 − 2 25 3 5 3 −1 5 A23(−2) 0 0 0 0 − 2 The last row says 0 = −2; the system is inconsistent. Lemma Suppose Ax = b is a linear system with augmented matrix A#. If rank(A#) > rank(A) then the system is inconsistent. Solving Linear Systems The rank of a matrix Math 240 Solving Linear Systems Example Gauss-Jordan elimination Determine the solution set of the linear system Rank Inverse 5x1 − 6x2 + x3 = 4; matrices Definition 2x1 − 3x2 + x3 = 1; Computing inverses 4x1 − 3x2 − x3 = 5: Properties of inverses Using inverse Reduce the augmented matrix. matrices Conclusion 25 −6 1 43 21 0 − 1 23 x − x = 2 2 −3 1 1 ! 0 1 − 1 1 1 3 4 5 4 5 x − x = 1 4 −3 −1 5 0 0 0 0 2 3 The unknown x3 can assume any value. Let x3 = t. Then by back substitution we get x2 = t + 1 and x1 = t + 2. Thus, the solution set is the line f(t + 2; t + 1; t): t 2 Rg : Solving Linear Systems The rank of a matrix Math 240 Solving Linear Systems Gauss-Jordan elimination Rank Definition Inverse When an unknown variable in a linear system is free to assume matrices Definition any value, we call it a free variable. Variables that are not free Computing inverses are called bound variables. Properties of inverses Using inverse The value of a bound variable is uniquely determined by a matrices Conclusion choice of values for all of the free variables in the system. Lemma Suppose Ax = b is an m × n linear system with augmented matrix A#. If rank(A#) = rank(A) < n then the system has an infinite number of solutions. Such a system will have n − rank(A#) free variables. Solving Linear Systems Solving linear systems with free variables Math 240 Solving Linear Example Systems Gauss-Jordan Use Gaussian elimination to solve elimination Rank x1 + 2x2 − 2x3 − x4 = 3; Inverse matrices 3x1 + 6x2 + x3 + 11x4 = 16; Definition Computing 2x1 + 4x2 − x3 + 4x4 = 9: inverses Properties of inverses Reducing to row-echelon form yields Using inverse matrices Conclusion x1 + 2x2 − 2x3 − x4 = 3; x3 + 2x4 = 1: Choose as free variables those variables that do not have a pivot in their column. In this case, our free variables will be x2 and x4. The solution set is the plane f(5 − 2s − 3t; s; 1 − 2t; t): s; t 2 Rg : Solving Linear Systems The inverse of a square matrix Math 240 Can we divide by a matrix? What properties should the inverse Solving Linear Systems matrix have? Gauss-Jordan elimination Rank Definition Inverse Suppose A is a square, n × n matrix. An inverse matrix for A matrices Definition is an n × n matrix, B, such that Computing inverses Properties of inverses AB = In and BA = In: Using inverse matrices Conclusion If A has such an inverse then we say that it is invertible or nonsingular. Otherwise, we say that A is singular. Remark Not every matrix is invertible. If you have a linear system Ax = b and B is an inverse matrix for A then the linear system has the unique solution x = Bb: Solving Linear Systems The inverse of a square matrix Math 240 Solving Linear Systems Example Gauss-Jordan elimination If Rank 2 3 2 3 Inverse 1 −1 2 0 −1 3 matrices −1 Definition A = 42 −3 35 and B = 41 −1 15 = A Computing inverses 1 −1 1 1 0 −1 Properties of inverses Using inverse then B is the inverse of A. matrices Conclusion Theorem (Matrix inverses are well-defined) Suppose A is an n × n matrix. If B and C are two inverses of A then B = C. Thus, we can write A−1 for the inverse of A with no ambiguity. Useful Example a b d −b If A = and ad − bc 6= 0 then A−1 = 1 . c d ad−bc −c a Solving Linear Systems Finding the inverse of a matrix Math 240 Solving Linear Inverse matrices sound great! How do I find one? Systems Suppose A is a 3 × 3 invertible matrix. If A−1 = x x x Gauss-Jordan 1 2 3 elimination Rank then Inverse 2 3 2 3 2 3 matrices 1 0 0 Definition Computing Ax1 = 405 ;Ax2 = 415 ; and Ax3 = 405 : inverses Properties of 0 0 1 inverses Using inverse matrices −1 Conclusion We can find A by solving 3 linear systems at once! In general, form the augmented matrix and reduce to RREF. You end up with A−1 on the right. −1 A In In A Solving Linear Systems Finding the inverse of a matrix Math 240 Solving Linear Example Systems 2 3 Gauss-Jordan 1 −1 2 elimination Rank Let's find the inverse of A = 42 −3 35. Inverse 1 −1 1 matrices Definition Computing Take the augmented matrix and row reduce. inverses Properties of inverses 2 3 2 3 Using inverse 1 −1 2 1 0 0 1 0 0 0 −1 3 matrices Conclusion 42 −3 3 0 1 05 40 1 0 1 −1 15 1 −1 1 0 0 1 0 0 1 1 0 −1 | {z } A−1 Steps 1. A12(−2) 5. A32(−1) 2. A13(−1) 6. A31(−2) 3. M2(−1) 7. A21(1) 4. M3(−1) Solving Linear Systems Finding the inverse of a matrix Math 240 Solving Linear In order to find the inverse of a matrix, A, we row reduced an Systems Gauss-Jordan augmented matrix with A on the left. What if we don't end up elimination Rank with In on the left? Inverse matrices Theorem Definition Computing inverses An n × n matrix, A, is invertible if and only if rank(A) = n. Properties of inverses Using inverse matrices Example Conclusion 1 3 Find the inverse of the matrix A = . 2 6 Try to reduce the matrix to RREF. 1 3 A12(−2) 1 3 −−−−−! 2 6 0 0 Since rank(A) < 2, we conclude that A is not invertible. Notice that (1)(6) − (3)(2) = 0. Solving Linear Systems Finding the inverse of a matrix Math 240 Solving Linear Diagonal matrices have simple inverses. Systems Gauss-Jordan elimination Proposition Rank Inverse The inverse of a diagonal matrix is the diagonal matrix with matrices Definition reciprocal entries.
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