lec20-23 - Systems of Linear Equations Introduction to Linear Algebra Math115 Ghazal Geshnizjani \u2014\u2014 Reference for some of the slides

lec20-23 - Systems of Linear Equations Introduction to...

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Unformatted text preview: Systems of Linear Equations Introduction to Linear Algebra, Math115 Ghazal Geshnizjani —— Reference for some of the slides: Introduction to Linear Algebra for Science and Engineering by Norman and Wolczuk Chapter 2 Linear Equations and Their Solutions Definition (Linear Equation) A linear equation in n variables x1 , . . . , xn is an equation that can be written in the form a1 x1 + a2 x2 + a3 x3 + · · · + an xn = b, where a1 , a2 , . . . , an ∈ R. Definition (Solution) s1 .. A vector . in Rn is a solution of equation (1) if the equation is sn satisfied when we make the substitution x1 = s1 , x2 = s2 , . . . , xn = sn . (1) A System of Linear Equations A system of linear equations is a collection of finitely many linear equations. Example: 3x1 + 2x2 − x3 = 3 2x1 + x3 = −1 3x2 − 4x3 = 4 is a system of three linear equations in three variables. How-many solutions can a system of linear equations have? A System of Linear Equations Example: ( a x1 + b x2 = c d x1 + e x2 = f a, b, c, d, e, f ∈ R is a system of 2 linear equations in 2 variables. Each equation is a line in R2 =⇒ Solution should belong to intersection of the two line. x2 x2 x2 P nonparallel lines intersect at a point ONE solution x1 x1 x1 parallel lines don’t intersect NO solution same line intersect at every point on the line INFINITELY many What about a system of 3 linear equations in 3 variables such as 3x1 + 2x2 − x3 = 3 2x1 + x3 = −1 3x2 − 4x3 = 4 ? Each equations describes one plane. P1 P2 P3 point of intersection line of intersection P1 P3 P2 P1 P2 P1 P2 P3 P3 solution is the intersection of all the three planes:a point (ONE solution), one line ( INFINITELY many solutions), parallel lines (NO solutions) Consistent Systems and Unique Solutions Theorem Given any system of linear equations, there can be just ONE solution, INFINITELY many solutions or NO solutions. Terminology A system is consistent if it has at least one solution. A system is inconsistent if it does not have any solutions. The Matrix Representation of a System of Linear Equations To the linear system a11 x1 a21 x1 + + a12 x2 a22 x2 + ··· + + ··· + a1n xn a2n xn = = .. . b1 b2 am1 x1 + am2 x2 + · · · + amn xn = bm we associate three matrices: coefficient matrix constant matrix a11 a12 · · · a1n b1 a21 a22 · · · a2n b2 A= . .. .. ~b = .. .. . . . am1 am2 · · · amn bm augmented matrix, [A|~b] a11 a21 .. . a12 a22 .. . ··· ··· a1n a2n .. . am1 am2 · · · amn b1 b2 bm Example x1 + 3x2 = −1 x1 + x2 = 3  Subtract 1st Eq. from 2nd Eq. x1 + 1 3 −1 1 1 3  R2 − R1 3x2 = −1 −2x2 = 4  multiply 2nd Eq. by − 12 1 3 −1 0 −2 4  − 12 R2 x1 + 3x2 = −1 x2 = −2  Subtract 3 times 2nd Eq. from 1st Eq. 1 3 −1 0 1 −2  R1 − 3R2  1 0 5 x1 = 5 0 1 −2 x2 = −2     x 5 So we obtain the solution in vector form 1 = . In point form or x2 −2 n-tulip form it’s (5,-2).  Example So we can solve the system of equations using the augmented matrix. x1 + 3x2 = −1 x1 + x2 = 3 We write the augmented matrix      1 3 −1 1 3 −1 1 3 −1 ∼ ∼ 1 1 3 0 −2 4 0 1 −2 R2 −R1 − 12 R2   1 0 5 R1 −3R2 ∼ 0 1 −2  So we obtain the solution in vector form     x1 5 = . In point form or x2 −2 n-tulip form it’s (5,-2). Note: this system is consistent since it has a solution. Row Operations Types of Elementary Row Operations (EROs) (1) (2) (3) Multiply one row by a non-zero constant. tRi Interchange two rows. Ri ↔ Rj Add a multiple of one row to another row. Ri + tRj Terminology Row reduction: the process of performing elementary row operations on a matrix to bring it into some simpler form. Row equivalent matrices: A matrix M is row equivalent to another matrix N if M can be reduced to N by a sequence of elementary row operations. Example 2x1 + x2 + 9x3 = 31 x2 + 2x3 = 8 x1 + 3x3 = 10 We write the augmented matrix 2 1 9 31 0 1 2 8 1 0 3 10 1 0 0 1 ∼ R3 −R2 0 0 1 0 3 10 ∼ 0 1 2 8 R3 ↔R1 2 1 9 31 3 10 R1 − 3R3 1 2 8 R2 − 2R3 0 1 3 ∼ 0 1 0 3 10 ∼ 0 1 2 8 R3 −2R1 0 1 3 11 0 0 1 x1 = 1 1 0 2 =⇒ x2 = 2 0 1 3 x3 = 3 x1 1 So we obtain the solution in vector form x2 = 2 . In point form or x3 3 n-tulip form it’s (1,2,3). Note: this system is consistent since it has a Remark x1 1 x2 = 2, x3 3 P1 P2 P3 point of intersection we could have stopped at this step 1 0 3 10 0 1 2 8 0 0 1 3 turn the augmented matrix back to the system : x1 x2 + 3x3 = 10 + 2x3 = 8 x3 = 3 substitute x3 = 3 to 1st and 2nd equation to get x2 and x1 ; This technique is called Back Substitution. Remarks about doing multiple EROs in one step 1 2 No row should be modified more than once in one step. No row should be both modified and be used to modify other rows in one step. ∼ · · · R2 − 2R1 This is ok. R3 + 3R1 ··· ··· R1 + R2 ∼ R3 + 3R1 This is not ok. 2R1 + 4R2 ∼ This is not ok. You have to do 2R1 first then in the next step do R1 + 4R2 . Row Echelon Form We write −1 −2 −1 ∼ R3 −R2 the 5 6 1 −1 0 0 −x1 + 5x2 = −2x1 + 6x2 + x3 = −x1 x2 + x3 = augmented matrix 0 5 ∼ −1 1 3 R2 − 2R1 0 1 −2 R3 − R1 0 −R1 1 5 0 5 −4 1 −7 − 14 R2 0 0 0 0 ∼ 0 5 3 −2 5 0 5 −4 1 −7 −4 1 −7 −5 0 1 − 14 0 0 REF 5 4 1 0 R1 + 5R2 0 1 − 14 ∼ 0 0 0 RREF − 15 4 7 4 0 REF −5 0 7 4 Row Echelon Form −1 5 0 5 0 −4 1 −7 , 0 0 0 0 0 5 0 5 0 0 −3 2 , 0 0 0 1 REF REF  0 2 0 5 −1 0 0 −5 2 0 REF Definition (Row Echelon Form) A matrix is in row echelon form (REF) if (1) when all entries in a row are zeros, this row appears below all rows that contain a non-zero entry, and (2) when two non-zero rows are compared, the first non-zero entry, called the leading entry(or pivot), in the upper row is to the left of the leading entry in the lower row. Note: A matrix can have many REFs!  Reduced Row Echelon Form 1 0 54 0 1 −1 4 0 0 0 RREF − 15 4 7 4 0 , 1 0 0 1 0 1 0 2 , 0 0 1 3 RREF  0 1 2 0 3 0 0 0 1 −4 RREF Definition (Reduced Row Echelon Form) A matrix is in reduced row echelon form (RREF) if (1) it is in row echelon form, (2) all leading entries are 1, called a leading 1, and (3) in a column with a leading 1, all the other entries are zeros. Theorem For any given matrix A there is a unique matrix in reduced row echelon form that is row equivalent to A. Note: If a matrix is in RREF, then it is in REF too.  Back to our example, We got the RREF: 1 0 0 x1 0 1 0 x2 5 4 − 14 −x1 + 5x2 = 5 −2x1 + 6x2 + x3 = 3 −x1 + x2 + x3 = −2 − 15 4 7 4 0 x3 0 x1 =⇒ x2 5 + = − 15 4 x3 4 1 7 , + (− 4 )x3 = 4 0 = 0 ‘ let x3 = t, t ∈ R =⇒ x1 = − 15 4 7 x2 = 4 x3 = 0 + (− 54 ) t + ( 14 )t , + (1)t 15 5 −4 x1 −4 =⇒ x2 = 74 + t 14 , t ∈ R x3 0 1 x1 , x2 are leading variables and x3 is a free variable. We assign parameters like t to free variables to solve for leading variables. 3 − 2 = 1 , example continued, −x1 + 5x2 = 5 −2x1 + 6x2 + x3 = 3 −x1 + x2 + x3 = −2 15 5 −4 x1 −4 7 =⇒ x2 = + t 14 , t ∈ R 4 x3 0 1 line of intersection P1 P3 P2 System is consistent and solution corresponds to a line in R3 Existence of a parameter in the solution means we have infinitely many solutions. (one point for each value of parameter.) # of parameters = # of variables − # leading 1s. Another example We get the following RREF:  0 1 2 0 3 0 0 0 1 −4 x1 # of variables= 4, x2 + 2x3 = 3 x4 = −4  x2 x3 x4 # leading 1s= 2 =⇒ # of parameters = 4 − 2 = 2. =⇒ x1 = t, x3 = s, x2 = 3 − 2s, x4 = −4, s, t ∈ R 0 1 0 x1 t 0 −2 x2 3 − 2s 3 =⇒ x3 = s = 0 + t 0 + s 1 , t, s ∈ R −4 0 0 x4 −4 Plane in R4 Rank of a Matrix We saw # of parameters = # of variables − # leading 1s. Definition (Rank) The rank of a matrix is the number of leading 1s in its reduced row echelon form (RREF) and is denoted by rank(A). # of parameters = n − rank(A). −1 5 0 5 1 0 54 − 15 n=3 4 A| ~b = −2 6 1 3 ∼ 0 1 − 14 74 =⇒ rank(A) = 2 −1 1 1 −2 1 parameter 0 0 0 0 h i Remark rank(A) =# of leading 1s in RREF = the number of leading entries in any REF. Example for Inconsistent System 2x1 + 12x2 − 8x3 = −4 2x1 + 13x2 − 6x3 = −5 −2x1 − 14x2 + 4x3 = 7 We write the augmented matrix ∼ 2 12 −8 −4 2 12 −8 −4 2 R2 − R1 0 1 2 −1 13 −6 −5 R3 + R1 0 −2 −4 3 −2 −14 4 7 2 12 −8 −4 2x1 + 12x2 − 8x3 0 1 2 −1 x2 + 2x3 ∼ =⇒ R3 +2R2 0 0 0 1 0 P1 P2 P3 P1 P2 P3 = −4 = −1 = 1 System is inconsistent and has no solutions. Equations describe planes in R3 that don’t have a common intersection Theorem h i Suppose that the augmented matrix A ~b of a linear system is row   ~ S c equivalent to , which is in row echelon    form. If some row of S ~c is of the form 0 0 · · · 0 c with c 6= 0 then the system is inconsistent. Proof: There are no values of x1 , x2 , . . . , xn ∈ R that can satisfy 0 = 1 , Note: This indicates that there is a leading entry/pivot in the last column of augmented matrix. Solving a complex system j z1 − z2 − z3 + (−1 + j) z4 = −1 − (1 + j)z3 − 2j z4 = −1 − 3j 2j z1 − 2z2 − z3 − (1 − 3j) z4 = j Solution: j −1 −1 −1 + j −1 j −1 −1 −1 + j −1 0 0 −1 − j −2j −1 − 3j 0 0 −1 − j −2j −1 − 3j ∼ 2j −2 −1 −1 + 3j j R3 − 2R1 0 0 1 1+j 2+j −jR1 1 j j 1+j j 1 j j 1+j j 0 0 1 1+j 2+j R2 ↔ R3 0 0 1 1+j 2+j ∼ ∼ 0 0 (−1 − j) −2j −1 − 3j R3 + (1 + j)R2 0 0 0 0 0 1 j 0 2 1−j z , z leading R1 − jR2 0 0 1 1 + j 2 + j =⇒ 1 3 z2 , z4 free, ∼ 0 0 0 0 0 z2 =s, z4 =t =⇒ z1 z2 z3 z4 = (1 − j) − js − 2t = s = (2 + j) − (1 + j)t = t z1 1−j −j −2 z2 0 1 0 =⇒ z3 = 2 + j + s 0 + t −1 − j , s, t ∈ C z4 0 0 1 Gaussian Elimination Any matrix can be reduced to row echelon form by using the following steps: (1) (2) (3) (4) (5) (6) Consider the first non-zero column of the matrix (i.e. the first column that does not consist entirely of zero entries). If necessary, interchange rows so that the top entry in the column is non-zero. This entry is called a pivot. Use elementary row operations of type (3) (i.e.Ri + tRj ) to make all entries below the pivot into zeros. Consider the submatrix consisting of all columns to the right of the column you have just worked on and all rows below the row with the most recently obtained leading entry. Repeat the procedure for this submatrix to obtain the next pivot with zeros below it. Keep repeating the procedure until you have “used up” all rows and columns of the original matrix. The resulting matrix is in row echelon Examples         1 −6 4 4 , ~y = , ~u = , ~v = . Express ~u and ~v as 2 −12 8 5 linear combination of ~x and ~y (if possible) using the augmented matrix framework.   123 Let A = . What is rank of A? 246 Let ~x = Consistent Systems and Unique Solutions 2x1 2x1 −2x1 12x2 13x2 14x2 + + − − − + 8x3 6x3 4x3 = = = −4 −5 7 2 ∼ 0 0 =⇒ h A ~b i 12 1 0 −8 2 0 −4 −1 1 Theorem h Suppose that the augmented matrix A   S ~c , which is in row echelon form. ~b i of a linear system is row equivalent to (1) The system is inconsistent if and only if some row of   0 0 · · · 0 c with c 6= 0. (2) If the system is consistent, there are two possibilities: (a) (b) −1 0 0  S ~c  is of the form The number of pivots in S is equal to the number of variables, and the system has a unique solution. The number of pivots in S is less than the number of variables, and the system has infinitely many solutions. 5 −4 0 0 1 0 5 −7 =⇒ 0 n=3 rank(A) = 2 1 parameter Proof: exercise for you. System Rank Theorem 2 1 9 31 1 0 0 1 A| ~b = 0 1 2 8 ∼ 0 1 0 2 1 0 3 10 0 0 1 3 h i rank(A) = 3, rank( A ~b ) = 3, 3 = 3 System is consistent and has a unique solution. 5 −1 5 0 5 1 0 − 15 h i 4 4 7 3 ∼ 0 1 − 14 A| ~b = −2 6 1 4 −1 1 1 −2 0 0 0 0 h i rank(A) = 2, rank( A ~b ) = 2, 2 = 2 System is consistent and has infinite solutions. 2 12 −8 −4 2 12 −8 −4 h i 13 −6 −5 ∼ 0 1 2 −1 A ~b ∼ 2 −2 −14 4 7 0 0 0 1 h i rank(A) = 2, rank( A ~b ) = 3, 2 < 3 System is inconsistent and has no solutions. h P1 P2 P3 point of intersection line of intersection P1 P1 P2 P3 P3 P2 P1 P2 P3 i System Rank Theorem (part 1) Theorem Let (1) (2) h be a system of m linear equations in n variables. h The system is consistent if and only if rank(A) = rank A A ~b i ~b i . If the system is consistent, then the number of parameters ( or number of free variables) in the general solution is the number of variables minus the rank of the matrix: # of parameters = n − rank(A). h i Proof (1):First n columns in RREF of A ~b form the RREF for A. We use proof by contraposition.   System is inconsistent ⇐⇒ RREF has some row 0 0 · · · 0 c with c 6= 0. h i ⇐⇒ rank(A) < rank . A ~b Proof (2): for you More Examples (for you) h A ~b i x + 3x2 = −1 = 1 x1 + x2 = 3  =⇒ 1 3 −1 1 1 3   ∼ 1 0 5 0 1 −2 h i rank(A) = 2 = rank( A ~b ) consistent X total # of variables = 2, # of leading 1’s in RREF = 2 =⇒ # of free variables = 2 − 2 = 0 =⇒ unique solution. X     x 5 solution in vector form 1 = . x2 −2   2 1 2 0 3 0 0 0 1 −4  rank(A) = 2, m = 2, n = 4, rank(A) ≤ min{2, 4} Theorem h i Let A ~b be a system of m linear equations in n variables. Then rank(A) ≤ min{ m, n}. Proof: There can be at most one leading entry in each row of any REF of A =⇒ rank(A) ≤ m There can be at most one leading entry in each column of any REF of A =⇒ rank(A) ≤ n. =⇒ rank(A) ≤ min{ m, n} 2 A| ~b = 0 1 h rank(A) = 3, rank( A 9 31 1 0 0 1 2 8 ∼ 0 1 0 2 3 10 0 0 1 3 i ~b ) = 3, 3 = 3 System is consistent. If we change ~b, it will still be consistent! rank(A) = 3, m = 3 (m for number of rows/equations) h P1 P2 P3 point of intersection 1 1 0 1 −1 5 0 5 3 ∼ 0 A| ~b = −2 6 1 −1 1 1 −2 0 h i rank(A) = 2, rank( A ~b ) = 2, 2 = 2 System is consistent but if we change ~b it may 0 1 0 line of intersection P1 i h P3 P2 rank(A) = 2, m = 3 , i 5 4 − 15 4 0 0 − 14 7 4 become inconsistent! 3 6= 2 System-Rank Theorem (part 2) (3) Let be a system of m linear equations in n variables. The system is consistent for all ~b ∈ Rm if and only if rank(A) = m. h A ~b i Remark: rank(A) = m doesn’t mean the solution is always unique. 2x1 + x2 + − 2x3 x3 + x4 = = 3 −4  2 0 1 0 2 −1 0 1 3 −4  , m = 2, n = 4 Definition Let h A ~b i be a system of m linear equations in n variables. The system is underdetermined if n > m, that is it has more variables than equations. The system is overderdetermined if n < m, that is it has more equations than variables. Remark: rank(A) ≤ m =⇒ n − rank(A) ≥ n − m =⇒ # of free variables ≥ n − m So if we have a consistent underdetermined system (n > m) then it has infinitely many solutions. Remark: An overderdetermined system is often inconsistent. Definition (Homogeneous Linear Equations) A linear equation is homogeneous if the right-hand side (constant term) is zero. A system of equations is homogeneous if all of its equations are homogeneous. Augmented Matrix is h A ~0 i Remark The zero vector ~0 is always a solution of any homogeneous system. This is called the trivial solution. Example:  1 1 1 0 0 3 −1 0 x1 +  ∼ 1 3 R2 x2 + x3 = 0 3x2 − x3 = 0    1 1 1 0 1 0 43 R1 − R2 ∼ 0 1 − 13 0 0 1 − 13 0 0  We often omit zero column and since only need to row reduce, the coefficient matrix (A).   4 1 1 0 43 0 =⇒ x3 = t, x1 = − t, x2 = t, t ∈ R 0 1 − 13 0 3 3 4 4 −3t −3 x1 1 =⇒ x2 = 3 t = t 13 , t ∈ R, line through origin x3 t 1 Remarks: you can multiply the direction vector by a number (3 here )to simplify it. This system was underdetermined, so it is not a surprise that has infinitely many solution. Solution to the homogeneous system corresponded to a subspace of R3 Theorem Let S be the set of solutions to a homogeneous system of m linear equation and n variables. Then S is a subspace of Rn . We call S solution space of the homogeneous system. Proof: Since system has n variables, S ∈ Rn ~ Since system is homogeneous 0 ∈ S =⇒ non-empty z1 y1 .. .. Let ~z = . and ~y = . be vectors in S. Then for any arbitrary zn yn equation in the system: a1 z1 + · · · + an zn = 0 and a1 y1 + · · · + an yn = 0 =⇒ a1 (z1 + y1 ) + · · · + an (zn + yn ) = 0 =⇒ ~z + ~y ∈ S =⇒ closed underaddition. also ∀k ∈ R we have k(a1 y1 + · · · + an yn ) = k0 =⇒ a1 (ky1 ) + · · · + an (kyn ) = 0 =⇒ k~y ∈ S so it’s closed under scalar multiplication =⇒ S is a sub space of Rn . now lets solve an non-homogeneous system with same coefficient matrix: x1 +  x + x2 + x3 = 1 x2 + x3 = 0 −→ 1 3x2 − x3 = 3 3x2 − x3 = 0 1 1 1 1 0 3 −1 3  ∼ 1 3 R2  1 1 1 0 1 − 13 1 1  R1 − R2 ∼  1 0 43 0 1 − 13 0 1 4 1 =⇒ x3 = t, x1 = − t, x2 = 1 + t, t ∈ R 3 3 4 4 −3t −3 x1 0 1 =⇒ x2 = 1 + 3 t = 1 + t 13 , t ∈ R x3 0 t 1 4 −3 x1 solution to associated homogeneous was: x2 = t 13 , t ∈ R x3 1 0 1 is a particular solution to the non-homogeneous system. 0  Relation between non-homogeneous and its associated homogeneous system The complete solution to a consistent non-homogeneous system is a particular solution to the system plus the complete solution to corresponding homogeneous system. 4 −3 x1 0 x2 = 1 + t 1 , t ∈ R 3 x3 0 1 Example: Solve the following system and find a basis for solution space. 4x1 − 2x2 + 3x3 + 5x4 = 0 8x1 − 4x2 + 6x3 + 11x4 = 0 −4x1 + 2x2 − 3x3 − 7x4 = 0 Solution: 4 −2 3 5 ∼ 4 −2 3 5 8 −4 6 11 R2 − 2R1 0 0 0 1 ∼ −4 2 −3 −7 R3 + R1 0 0 0 −2 R3 + 2R2 4 −2 3 5 4 −2 3 0 0 0 0 1 R1 − 5R2 0 0 0 1 ∼ 0 0 0 0 0 0 0 0 continued... 1 4 R1 ∼ 1 − 12 0 0 0 0 3 4 0 0 1 0 0 =⇒ x1 , x4 leading, x2 , x3 are free. we take x2 = s, x3 = t =⇒ x1 = 12 s − 34 t, we also have x4 = 0 1 1 3 3 x1 −4 2s − 4t 2 x2 s 1 0 = s + t s, t ∈ R =⇒ x3 = t 0 1 x4 0 0 0 Solution is plane through origin, a subspace in R4 and space 1 −3 2 4 1 0 Span , 0 1 0 0 1 3 −4 2 1 0 Also B = , is linearly independent, so together it implies B 0 1 0 0 is a basis for solution space. Example Consider: x1 + x2 + x3 x4 + 4x5 + 5x6 = 0 + 2x5 + 3x6 = 0 homogenous system of equations. the augmented matrix:   1 1 1 0 4 5 0 0 0 1 2 3 Already RREF so we have x2 = t1 , x3 = t2 , x5 = t3 , x6 = t4 − 3t4 =⇒ x1 =−t1 −t2 −4t3 −5t4 , x4 = −2t 3 x1 −1 −1 −4 −5 x2 1 0 0 0 x3 0 1 0 0 =⇒ = t1 + t2 + t3 + t4 −3where x4 0 0 −2 x5 0 0 1 0 x6 0 0 0 1 t1 , t2 , t3 , t4 ∈ R −1 −1 −4 −5 1 0 0 0 0 1 0 0 , , , =⇒ B = spanning set for solution 0 0 −2 −3 0 0 1 0 0 0 0 1 space. Every vector in B has a 1 for one of its entries that others have zeros there. =⇒ B is linearly independent =⇒ B is a Basis for Solution space. Theorem Let [A|~b] be the augmented matrix for a consistent system of m equations in n variables. If rank(A) = k < n, then the solution of the system has the form ~x = ~b + t1~v1 + · · · + tn−k ~vn−k where ~b ∈ Rn , t1 , . . . , tn−k ∈ R and {~v1 , . . . , ~vn−k } is linearly independent in Rn . In particular the solution set is an (n − k)−flat in Rn . Applications of Systems of Linear Equations: Chemical Reactions Example: Hydrogen molecules (H2 ) and Oxygen molecules (O2 ) can combine to produce water (H2 O) ( )H2 + ( )O2 −→ ( )H2 O balance chemical reaction: the same number of atoms of each type before and after the chemical reaction. 2H2 + 1O2 −→ 2H2 O inspection becomes difficult for more complex molecules! Example, the chemical reaction photosynthesis Plants combine carbon dioxide (CO2 ) and water (H2 O) to produce glucose (C6 H12 O6 ) and oxygen (O2 ): ( )CO2 + ( )H2 O −→ ( )C6 H12 O6 + ( )O2 New method: x1 CO2 + x2 H2 O −→ x3 C6 H12 O6 + x4 O2 number of atoms of each type before and after the reaction: C: x1 = 6x3 O : 2x1 + x2 = 6x3 + 2x4 H: 2x2 = 12x3 =⇒ x1 2x1 + x2 2x2 − 6x3 = 0 − 6x3 − 2x4 = 0 − 12x3 = 0 1 =⇒ 2 0 1 0 0 1 0 0 x1 2x1 + 0 1 2 −6 −6 −12 0 −2 0 0 1 0 −6 6 −12 0 −2 2 0 1 0 0 0 6 −1 −1 −1 x2 2x2 0 0 0 − − − 6x3 6x3 12x3 − 2x4 = = = 0 0 0 0 ∼ 1 0 −6 0 0 ∼ 0 R2 −2R1 0 1 6 −2 0 1R 0 0 1 −6 0 0 R3 −R2 2 3 0 ∼ 1 0 −6 0 0 R1 +R3 0 1 0 6 −2 0 R2 −R3 0 − 12 R3 0 0 6 −1 0 ∼ ∼ 1 0 0 −1 0 0 1 0 −1 0 1R 0 0 1 −1/6 0 6 3 =⇒ x4 = t =⇒ x1 = t, x2 = t, x3 = t/6, t ∈ R infinitely many solutions! Cannot have a fractional or negative number of molecules! =⇒ x1 , x2 , x3 and x4 nonnegative integers. =⇒ t is positve multiple of 6. Take the simplest (or smallest) solution, t = 6. =⇒ x1 = x2 = x4 = 6 and x3 = 1 . Thus, 6CO2 + 6H2 O −→ C6 H12 O6 + 6O2 Example, the chemical reaction The fermentation of sugar Balance this chemical reaction: C6 H12 O6 −→ CO2 + C2 H5 OH where C6 H12 O6 is glucose, CO2 is carbon dioxide and C2 H5 OH is ethanol. Solution: x1 C6 H12 O6 −→ x2 CO2 + x3 C2 H5 OH =⇒ x2 + 2x3 2x2 + x3 =⇒ 6x3 6 −1 −2 0 ∼ 6 −2 −1 0 R2 −R1 12 0 −6 0 R3 −2R1 1R 1 6 0 −3 0 6 1 0 −1 1 0 −R2 0 ∼ 0 0 0 0 0 C: O:...
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